Eigenvalues And Eigenvectors

John Constable (1776-1837) was an English painter.

We have sections

\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}

Eigenspaces.

Definition. Let $V$ be a vector space over the scalar field ${\cal F}$, and let $T:V\rightarrow V$ be a linear operator on $V$. An element $v\in V$ is called an eigenvector of $T$ when there exists $\lambda\in {\cal F}$ satisfying $T(v)=\lambda v$. $\sharp$

Suppose that $v\neq\theta$ and $v$ is an eigenvector of a linear operator $T$. Then, there exists $\lambda\in {\cal F}$ such that $T(v)=\lambda v$. If there exists another $\lambda^{*}\in {\cal F}$ satisfying $T(v)=\lambda^{*}v$, then we have $\lambda v-\lambda^{*}v=\theta$. This says that $\lambda =\lambda^{*}$, since $v\neq\theta$. In this case, the scalar $\lambda$ is unique and is called the eigenvalue of $T$ with respect to the eigenvector $v$. We can also says that $v$ is an eigenvector with eigenvalue $\lambda$.

If $A$ is an $n\times n$ square matrix with coefficients in ${\cal F}$, then an eigenvector of $A$ is defined as an eigenvector of the linear operator $T:{\cal F}^{n}\rightarrow {\cal F}^{n}$ represented by this matrix relative to the standard basis. Therefore, an eigenvector $X$ of $A$ is a column vector in ${\cal F}^{n}$ for which there exists $\lambda\in {\cal F}$ satisfying $AX=\lambda X$.

Example. Let $V$ be a vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. If $v$ is an eigenvector with eigenvalue $\lambda$, then, for any $\alpha\in {\cal F}$, we have
\begin{align*} T(\alpha v) & =\alpha T(v)\\ & =\alpha\lambda v\\ & =\lambda (\alpha v),\end{align*}
which says that $\alpha v$ is also an eigenvector with the same eigenvalue $\lambda$. $\sharp$

\begin{equation}{\label{lap30}}\mbox{}\tag{1}\end{equation}

Proposition \ref{lap30}. Let $V$ be a vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$ with eigenvalue $\lambda$. We define the set
\[V_{\lambda}=\left\{v\in V:T(v)=\lambda v\right\},\]
which consists of all eigenvectors of $T$ having the same eigenvalue $\lambda$. Then $V_{\lambda}$ is a subspace of $V$.

Proof. Let $v_{1},v_{2}\in V_{\lambda}$ and $\alpha\in {\cal F}$. Then, we have
\begin{align*} T(v_{1}+v_{2}) & =T(v_{1})+T(v_{2})\\ & =\lambda v_{1}+\lambda v_{2}\\ & =\lambda (v_{1}+v_{2})\end{align*}
and
\begin{align*} T(\alpha v_{1}) & =\alpha T(v_{1})\\ & =\alpha\lambda v_{1}\\ & =\lambda (\alpha v_{1}).\end{align*}
This completes the proof. $\blacksquare$

The subspace $V_{\lambda}$ in Proposition \ref{lap30} is also called the eigenspace of $T$ with eigenvalue $\lambda$.

\begin{equation}{\label{lap217}}\tag{2}\mbox{}\end{equation}

Proposition \ref{lap217}. Let $V$ be a vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Let $v_{1},\cdots ,v_{m}$ be eigenvectors of $T$ with the corresponding eigenvalues $\lambda_{1},\cdots ,\lambda_{m}$. If $\lambda_{i}\neq\lambda_{j}$ for $i\neq j$, then $v_{1},\cdots ,v_{m}$ are linearly independent.

Proof. We are going to prove it by induction on $m$. For $m=1$, an element $v_{1}\in V$ with $v_{1}\neq\theta$ is linearly independent. Suppose that $\{v_{2},\cdots ,v_{m}\}$ are linearly independent. If
\begin{equation}{\label{laeq31}}\tag{3}
\alpha_{1}v_{1}+\cdots +\alpha_{m}v_{m}=\theta ,
\end{equation}
then we want to prove $\alpha_{i}=0$ for all $i=1,\cdots ,m$. We multiply the equality (\ref{laeq31}) by $\lambda_{1}$ on both sides to obtain
\begin{equation}{\label{laeq32}}\tag{4}
\lambda_{1}\alpha_{1}v_{1}+\cdots +\lambda_{1}\alpha_{m}v_{m}=\theta .
\end{equation}
We also apply $T$ to the equality (\ref{laeq31}) to obtain
\begin{equation}{\label{laeq33}}\tag{5}
\lambda_{1}\alpha_{1}v_{1}+\cdots +\lambda_{m}\alpha_{m}v_{m}=\theta .
\end{equation}
We subtract the expressions (\ref{laeq32}) and (\ref{laeq33}) to obtain
\[\alpha_{2}(\lambda_{2}-\lambda_{1})v_{2}+\cdots +\alpha_{m}(\lambda_{m}-\lambda_{1})v_{m}=\theta .\]
Since $\lambda_{i}-\lambda_{1}\neq 0$ for $i=2,\cdots ,m$ and $\{v_{2},\cdots ,v_{m}\}$ are linearly independent, we must have $\alpha_{i}=0$ for $i=2,\cdots ,m$. Therefore, from (\ref{laeq31}), we obtain $\alpha_{1}v_{1}=\theta$, which also implies $\alpha_{1}=0$. This completes the proof. $\blacksquare$

\begin{equation}{\label{lac34}}\tag{6}\mbox{}\end{equation}

Lemma\ref{lac34}. Let $V$ and $W$ be vector spaces over the same scalar field ${\cal F}$, and let $T:V\rightarrow W$ be a linear mapping such that it is surjective and $\mbox{Ker}(T)=\{\theta_{U}\}$. Then $T$ has an inverse linear mapping.

Proof. By Proposition \ref{lap13}, we see that $T$ is injective and surjective. This says that $T$ has an inverse. By Proposition \ref{lap14}, this inverse mapping is linear, and the proof is complete. $\blacksquare$

The above Lemma \ref{lac34} can refer to the page Linear Mappings.

\begin{equation}{\label{lap35}}\tag{7}\mbox{}\end{equation}

Proposition \ref{lap35}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Then $\lambda$ is an eigenvalue of $T$ if and only if $T-\lambda I$ is not invertible.

Proof. Let $\lambda$ be an eigenvalue of $T$. Then there exists an element $v\in V$ with $v\neq\theta$ such that $T(v)=\lambda v$, i.e., $T(v)-\lambda v=\theta$. This shows that the kernel of $T-\lambda I$ cannot be invertible. Conversely, we assume that $T-\lambda I$ is not invertible. Using Lemma \ref{lac34}, we see that $T-\lambda I$ must have a non-zero kernel, which means that there exists $v\neq\theta$ satisfying $(T-\lambda I)(v)=\theta$. Therefore, we obtain $T(v)=\lambda v$. This completes the proof. $\blacksquare$

\begin{equation}{\label{lap257}}\tag{8}\mbox{}\end{equation}

Proposition \ref{lap257}. Let $V$ be a finite-dimensional vector space over the complex numbers with $\dim (V)\geq 1$, and let $T$ be a linear operator on $V$. Then, there exists a non-zero eigenvector of $T$. $\sharp$

Proposition \ref{lap257} is false if the vector space $V$ is assumed to be over $\mathbb{R}$ instead of being over $\mathbb{C}$.

\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}

Diagonalization.

Definition. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. We say that $T$ is diagonalizable when there is a basis $\mathfrak{B}$ for $V$ such that $[T]_{\mathfrak{B}}$ is a diagonal matrix. $\sharp$

If $A$ is an $n\times n$ square matrix, then we say that $A$ is diagonalizable if and only if $T_{A}$ is diagonalizable.

\begin{equation}{\label{lap106}}\tag{9}\mbox{}\end{equation}

Proposition \ref{lap106}. Let $V$ be an $n$-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Suppose that the set of eigenvectors $\{v_{1},\cdots ,v_{n}\}$ with the corresponding eigenvalues $\lambda_{1},\cdots ,\lambda_{n}$ forms an ordered basis $\mathfrak{B}$ for $V$. Then, the matrix associated with $T$ with respect to $\mathfrak{B}$ is the diagonal matrix given by
\[[T]_{\mathfrak{B}}=\left [\begin{array}{rrrr}
\lambda_{1} & 0 & \cdots & 0\\
0 & \lambda_{2} & \cdots & 0\\
\vdots & \vdots & \cdots & \vdots\\
0 & 0 & \cdots & \lambda_{n}
\end{array}\right ].\]
\end{Pro}

Proof. It is obvious by the fact of $T(v_{i})=\lambda_{i}v_{i}$ for all $i=1,\cdots ,n$. $\blacksquare$

Corollary. Let $V$ be an $n$-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Suppose that $T$ has $n$ distinct eigenvalues. Then $V$ has a basis consisting of eigenvectors for $A$, and $T$ is diagonalizable.

\begin{equation}{\label{lac105}}\tag{10}\mbox{}\end{equation}

Corollary \ref{lac105}. Let $A$ be an $n\times n$ matrix over the scalar field ${\cal F}$. Suppose that $A$ has $n$ distinct eigenvalues in ${\cal F}$. Then, there exists a non-sigular matrix $B$ in ${\cal F}$ such that $B^{-1}AB$ is a diagonal matrix in which the diagonal entries consist of the corresponding eigenvalues.

\begin{equation}{\label{lap214}}\tag{11}\mbox{}\end{equation}

Proposition \ref{lap214}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Then $T$ is diagonalizable if and only if there is a basis $\mathfrak{B}$ for $V$ such that each vector of which is an eigenvector of $T$. Moreover, the diagonal matrix $[T]_{\mathfrak{B}}$ consists of eigenvalues that are corresponding to the eigenvectors in $\mathfrak{B}$.

Proof. If $T$ is diagonalizable, by definition, there exists a basis $\mathfrak{B}=\{v_{1},\cdots ,v_{n}\}$ for $V$ such that $[T]_{\mathfrak{B}}\equiv D$ is a diagonal matrix. Therefore, we have
\begin{align*} T(v_{j}) & =\sum_{i=1}^{n}d_{ij}v_{i}\\ & =d_{jj}v_{j}\equiv\lambda_{j}v_{j},\end{align*}
which says that each vector of $\mathfrak{B}$ is an eigenvector of $T$. The converse follows from Proposition \ref{lap106} immediately, and the proof is complete. $\blacksquare$

Example. We consider the following matrix
\[A=\left [\begin{array}{cc}
1 & 3\\ 4 & 2
\end{array}\right ]\]
and the vectors
\[v_{1}=\left [\begin{array}{r}
1 \\ -1
\end{array}\right ]\mbox{ and }
v_{2}=\left [\begin{array}{r}
3 \\ 4
\end{array}\right ].\]
Then, we have
\[T_{A}(v_{1})=Av_{1}=\left [\begin{array}{rr}
1 & 3\\ 4 & 2
\end{array}\right ]\left [\begin{array}{r}
1 \\ -1
\end{array}\right ]=\left [\begin{array}{r}
2 \\ -2
\end{array}\right ]=2\left [\begin{array}{r}
1 \\ -1
\end{array}\right ]=-2v_{1},\]
which says that $v_{1}$ is an eigenvector of $A$ with eigenvalue $-2$. We also have
\begin{align*} T_{A}(v_{2}) & =Av_{2}\\ & =\left [\begin{array}{rr}
1 & 3\\ 4 & 2
\end{array}\right ]\left [\begin{array}{r}
3 \\ 4
\end{array}\right ]\\ & =\left [\begin{array}{r}
15 \\ 20
\end{array}\right ]\\ & =5\left [\begin{array}{r}
3 \\ 4
\end{array}\right ]\\ & =5v_{2},\end{align*}
which says that $v_{2}$ is an eigenvector of $A$ with eigenvalue $5$. Since $\mathfrak{B}=\{v_{1},v_{2}\}$ is an ordered basis for $\mathbb{R}^{2}$, Proposition~ ref{lap214} says that $A$ is diagonalizable satisfying
\[[T_{A}]_{\mathfrak{B}}=\left [\begin{array}{rr}
-2 & 0\\ 0 & 5
\end{array}\right ].\]
Suppose that we take
\[B=\left [\begin{array}{rrr}
1 & -1\\ 3 & 4
\end{array}\right ],\]
where the columns are the eigenvectors for the different eigenvalues $-2$ and $5$. Then, we have
\[B^{-1}AB=\left [\begin{array}{rrr}
-2 & 0\\ 0 & 5
\end{array}\right ],\]
where the diagonal consists of all eigenvalues of $A$. $\sharp$

\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}

Characteristic Polynomials.

Let $A$ be an $n\times n$ matrix in a scalar field ${\cal F}$. The characteristic polynomial $P_{A}$ of $A$ is defined by
\[P_{A}(\lambda )=\det (\lambda I-A).\]

\begin{equation}{\label{lat36}}\tag{12}\mbox{}\end{equation}

Lemma \ref{lat36}. Let $A$ be an $n\times n$ matrix such that $\det (A)\neq 0$. Then $A$ is invertible. The $ij$th entry $a_{ij}^{*}$ of $A^{-1}$ is given by
\[a_{ij}^{*}=\frac{(-1)^{i+j}\cdot\det (A^{(ji)})}{\det (A)}.\]

The above Lemma \ref{lat36} can refer to the page Arithmetic Operations of Matrices.

Theorem. Let $A$ be an $n\times n$ matrix in a scalar field ${\cal F}$. A scalar $\lambda\in {\cal F}$ is an eigenvalue of $A$ if and only if $\lambda$ is a root of the characteristic polynomial $P_{A}$ of $A$. In other words, a scalar $\lambda\in {\cal F}$ is an eigenvalue of $A$ if and only if $\det (\lambda I_{n}-A)=0$.

Proof. Assume that $\lambda$ is an eigenvalue of $A$. Then, using Proposition \ref{lap35}, $\lambda I-A$ is not invertible. According to Lemma \ref{lat36}, we have $\det (\lambda I-A)=0$, which says that $\lambda$ is a root of the characteristic polynomial $P_{A}$ of $A$. Conversely, if $\lambda$ is a root of $P_{A}$, i.e., $\det (\lambda I-A)=0$, then $\lambda I-A$ is not invertible. Using Proposition \ref{lap35} again, we see that $\lambda$ is an eigenvalue of $A$. This completes the proof. $\blacksquare$

\begin{equation}{\label{lap37}}\tag{13}\mbox{}\end{equation}

Proposition \ref{lap37}. Let $A$ and $B$ be two $n\times n$ matrices. Suppose that $B$ is invertible. Then the characteristic polynomial of $A$ is equal to the characteristic polynomial of $B^{-1}AB$.

Proof. Using the definition and properties of determinant, we have
\begin{align*} \det (\lambda I-A) & =\det (B^{-1}(\lambda I-A)B)\\ & =\det (\lambda B^{-1}B-B^{-1}AB)\\ & =\det (\lambda I-B^{-1}AB).\end{align*}
This completes the proof. $\blacksquare$

Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Now, we shall define the characteristic polynomial of a linear operator $T$. Let $\mathfrak{B}$ be an ordered basis for $V$, and let $A$ be the matrix representing $T$ with respect to $\mathfrak{B}$. If $\mathfrak{B}^{\prime}$ is another ordered basis of $V$, then there exists an invertible matrix $B$ in ${\cal F}$ such that the matrix $A’$ representing $T$ with respect to $\mathfrak{B}’$ is given by $A’=B^{-1}AB$. Proposition \ref{lap37} says that the characteristic polynomial of $A$ is the same as that of $A’$. Therefore, the formal definition of characteristic polynomial for $T$ is given below.

Definition. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Let $A$ be a matrix representing $T$ with respect to some ordered basis for $V$. The characteristic polynomial, denoted by $P_{T}$, of the linear operator $T$ is defined to be
the characteristic polynomial $P_{A}$ of $A$. $\sharp$

Example. Let $T$ be a linear operator on $\mathbb{C}^{3}$ which is represented in the standard ordered basis $\mathfrak{B}$ by the following $3\times 3$ matrix
\[[T]_{\mathfrak{B}}=A=\left [\begin{array}{rrr}
1 & 1 & -1\\ 0 & 1 & 0\\ 1 & 0 & 1
\end{array}\right ].\]
The characteristic polynomial $P_{T}$ for $T$ (or $P_{A}$ for $A$) is given by
\begin{align*} \det (\lambda I-A) & =\left |\begin{array}{ccc}
\lambda -1 & -1 & 1\\ 0 & \lambda -1 & 0\\ -1 & 0 & \lambda -1
\end{array}\right |\\ & =(\lambda -1)(\lambda^{2}-2\lambda +2).\end{align*}
Therefore, the eigenvalues of $T$ (or $A$) are $1$, $1+i$ and $1-i$. $\sharp$

Example. Let $T$ be a linear operator on $\mathbb{R}^{2}$ which is represented in the standard ordered basis $\mathfrak{B}$ by the following $2\times 2$ matrix
\[[T]_{\mathfrak{B}}=A=\left [\begin{array}{rrr}
0 & -1\\ 1 & 0
\end{array}\right ].\]
The characteristic polynomial $P_{T}$ for $T$ (or $P_{A}$ for $A$) is given by
\begin{align*} \det (\lambda I-A) & =\left |\begin{array}{ccc}
\lambda & 1\\ -1 & \lambda
\end{array}\right |\\ & =\lambda^{2}+1.\end{align*}
Since this polynomial has no roots in $\mathbb{R}$, it says that $T$ has no eigenvalues. However, if we consider $T$ as a linear operator on $\mathbb{C}^{2}$ instead of $\mathbb{R}^{2}$, then the above polynomial has two roots $i$ and $-i$. In this case, the eigenvalues of $T$ (or $A$) are $i$ and $-i$. $\sharp$

Example. We consider the following matrix
\[A=\left [\begin{array}{rrr}
1 & 1\\ 4 & 1
\end{array}\right ].\]
The characteristic polynomial $P_{A}$ for $A$ is given by
\begin{align*} \det (\lambda I-A) & =\left |\begin{array}{ccc}
\lambda -1 & 1\\ 4 & \lambda -1
\end{array}\right |\\ & =(\lambda +1)(\lambda -3).\end{align*}
Therefore, the eigenvalues of $T$ are $-1$ and $3$. Next, we want to find all the eigenvectors of $A$. Given any eigenvalue $\lambda$, the eigenvector $X$ satisfies $AX=\lambda X$, i.e., $(\lambda I_{n}-A)X={\bf 0}$. We first find all the eigenvectors corresponding to eigenvalue $3$. This says that we need to find $X\in\mathbb{R}^{2}$ satisfying
\begin{align*}
\left [\begin{array}{c}
0\\ 0
\end{array}\right ] & =(3I_{2}-A)X\\ & =\left (\left [\begin{array}{rrr}
3 & 0\\ 0 & 3
\end{array}\right ]-\left [\begin{array}{rrr}
1 & 1\\ 4 & 1
\end{array}\right ]\right )\left [\begin{array}{c}
x\\ y
\end{array}\right ]\\
& =\left [\begin{array}{rrr}
2 & -1\\ -4 & 2
\end{array}\right ]\left [\begin{array}{c}
x\\ y
\end{array}\right ]\\ & =\left [\begin{array}{c}
-2x+y\\ 4x-2y
\end{array}\right ].
\end{align*}
The set of all solutions is given by
\[\left\{t\left [\begin{array}{c}
1\\ 2
\end{array}\right ]:t\in\mathbb{R}\right\},\]
which is also the set of all eigenvectors corresponding to eigenvalue $3$. Using the same arguments, we can find all eigenvectors corresponding to eigenvalue
$-1$, which is given by
\[\left\{t\left [\begin{array}{r}
1\\ -2
\end{array}\right ]:t\in\mathbb{R}\right\}.\]
We also observe that
\[\mathfrak{B}=\left\{\left [\begin{array}{c}
1\\ 2
\end{array}\right ],\left [\begin{array}{r}
1\\ -2
\end{array}\right ]\right\}\]
is a basis for $\mathbb{R}^{2}$. Therefore, the matrix $A$ is diagonalizable. Suppose that we take
\[B=\left [\begin{array}{rrr}
1 & 1\\ 2 & -2
\end{array}\right ],\]
where the columns are the eigenvectors for the corresponding eigenvalues $3$ and $-1$. Then, we have
\[B^{-1}AB=\left [\begin{array}{rrr}
3 & 0\\ 0 & -1
\end{array}\right ],\]
where the diagonal consists of all eigenvalues of $A$. $\sharp$

\begin{equation}{\label{laex216}}\tag{14}\mbox{}\end{equation}

Example \ref{laex216}. Let $T$ be a linear operator on $\mathbb{R}^{3}$ defined by
\[T(x,y,z)=\left (4x+z,2x+3y+2z,x+4z\right ).\]
Let $\mathfrak{B}$ be the standard basis for $\mathbb{R}^{3}$. Then, we have
\[A=[T]_{\mathfrak{B}}=\left [\begin{array}{rrr}
4 & 0 & 1\\ 2 & 3 & 2\\ 1 & 0 & 4
\end{array}\right ].\]
The characteristic polynomial of $T$ is given by
\begin{align*} \det (\lambda I-A) & =\left |\begin{array}{ccc}
\lambda -4 & 0 & -1\\ -2 & \lambda -3 & -2\\ -1 & 0 & \lambda -4
\end{array}\right |\\ & =(\lambda -5)(\lambda -3)^{2}.\end{align*}
Therefore, the eigenvalues of $T$ are $5$ and $3$.

For eigenvalue $5$, we consider
\begin{align*}
\left [\begin{array}{c}
0\\ 0 \\ 0
\end{array}\right ] & =(5I_{3}-A)X\\ & =\left (\left [\begin{array}{rrr}
5 & 0 & 0\\ 0 & 5 & 0\\ 0 & 0 & 5
\end{array}\right ]-\left [\begin{array}{rrr}
4 & 0 & 1\\ 2 & 3 & 2\\ 1 & 0 & 4
\end{array}\right ]\right )\left [\begin{array}{c}
x\\ y\\ z
\end{array}\right ]\\
& =\left [\begin{array}{rrr}
1 & 0 & -1\\ -2 & 2 & -2\\ -1 & 0 & 1
\end{array}\right ]\left [\begin{array}{c}
x\\ y\\ z
\end{array}\right ]\\ & =\left [\begin{array}{c}
x-z\\ -2x+2y-2z\\ -x+z
\end{array}\right ].
\end{align*}
The set of all solutions is given by
\[V_{5}=\left\{t\left [\begin{array}{c}
1\\ 2\\ 1
\end{array}\right ]:t\in\mathbb{R}\right\},\]
which is the eigenspace with eigenvalue $5$ and basis
\[\left\{\left [\begin{array}{c}
1\\ 2\\ 1
\end{array}\right ]\right\}.\]
For eigenvalue $-2$, we consider
\begin{align*}
\left [\begin{array}{c}
0\\ 0 \\ 0
\end{array}\right ] & =(3I_{3}-A)X\\ & =\left (\left [\begin{array}{rrr}
3 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 3
\end{array}\right ]-\left [\begin{array}{rrr}
4 & 0 & 1\\ 2 & 3 & 2\\ 1 & 0 & 4
\end{array}\right ]\right )\left [\begin{array}{c}
x\\ y\\ z
\end{array}\right ]\\
& =\left [\begin{array}{rrr}
-1 & 0 & -1\\ -2 & 0 & -2\\ -1 & 0 & -1
\end{array}\right ]\left [\begin{array}{c}
x\\ y\\ z
\end{array}\right ]\\ & =\left [\begin{array}{c}
-x-z\\ -2x-2z\\ -x-z
\end{array}\right ].
\end{align*} Let $y=s$ be a parameter, and solve the system for $x$ and $z$ by introducing another parameter $t$. The set of all solutions is given by\[V_{3}=\left\{s\left [\begin{array}{c}
0\\ 1\\ 0
\end{array}\right ]+t\left [\begin{array}{c}
-1\\ 0\\ 1
\end{array}\right ]:s,t\in\mathbb{R}\right\},\]
which is the eigenspace with eigenvalue $3$ and basis
\[\left\{\left [\begin{array}{c}
0\\ 1\\ 0
\end{array}\right ],\left [\begin{array}{c}
-1\\ 0\\ 1
\end{array}\right ]\right\}.\]

The union of the bases for $V_{5}$ and $V_{3}$ given by
\[\mathfrak{B}=\left\{\left [\begin{array}{c}
1\\ 2\\ 1
\end{array}\right ],\left [\begin{array}{c}
0\\ 1\\ 0
\end{array}\right ],\left [\begin{array}{c}
-1\\ 0\\ 1
\end{array}\right ]\right\}\]
forms a basis of $\mathbb{R}^{3}$. Since basis $\mathfrak{B}$ consists of eigenvectors of $T$, it says that $T$ is diagonalizable. Suppose that we take
\[B=\left [\begin{array}{rrr}
1 & 0 & -1\\ 2 & 1 & 0\\ 1 & 0 & 1
\end{array}\right ],\]
where the columns are the eigenvectors for the corresponding eigenvalues $5$ and $3$. Then, we have
\[B^{-1}AB=\left [\begin{array}{rrr}
5 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 3
\end{array}\right ],\]
where the diagonal consists of all eigenvalues of $T$. $\sharp$

In the sequel, we shall present the general results as given in Example \ref{laex216}. The multiplicity of the eigenvalue $\lambda_{i}$ for $T$ is defined to be the multiplicity of root $\lambda_{i}$ for the characteristic polynomial of $T$. For example, the multiplicity of eigenvalue $3$ in Example \ref{laex216} has multiplicity $2$. Let ${\cal F}$ be a scalar field, we say that a polynomial $f$ with degree $n$ can be factored over ${\cal F}$ if and only if there exists $c,\lambda_{1},\cdots ,\lambda_{n}\in {\cal F}$ satisfying
\[f(t)=c(t-\lambda_{1})(t-\lambda_{2})\cdots (t-\lambda_{n}).\]
It is obvious that $(t^{2}+1)(t-3)$ cannot be factored over $\mathbb{R}$. However, it can be factored over $\mathbb{C}$.

\begin{equation}{\label{lal218}}\tag{15}\mbox{}\end{equation}

Proposition \ref{lal218}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Let $\lambda$ be an eigenvalue of $T$ having multiplicity $m$, and let $V_{\lambda}$ be the eigenspace for $\lambda$. Then $1\leq\dim (V_{\lambda})\leq m$.

Proof. Suppose that $\dim (V)=n$. Let $\{v_{1},\cdots ,v_{p}\}$ be an ordered basis for $V_{\lambda}$. We can extend it to an ordered basis
\[\mathfrak{B}=\left\{v_{1},\cdots ,v_{p},v_{p+1},\cdots ,v_{n}\right\}\]
for $V$. Since each $v_{i}$ for $i=1,\cdots ,p$ is an eigenvector of $T$ corresponding to $\lambda$, we have
\[A=[T]_{\mathfrak{B}}=\left [\begin{array}{cc}
\lambda I_{p} & B\\
{\bf 0} & C
\end{array}\right ].\]
Therefore, the characteristic polynomial of $T$ corresponding to $\lambda$ is given by
\begin{align*}
P_{T}(t) & =\det\left (tI_{n}-A\right )\\ & =\left [\begin{array}{cc}(t-\lambda )I_{p} & -B\\
{\bf 0} & \lambda I_{n-p}-C
\end{array}\right ]\\
& =\det\left ((t-\lambda )I_{p}\right )\cdot\det\left (\lambda I_{n-p}-C\right )
\\ & =(t-\lambda )^{p}\cdot f(t),
\end{align*}
where $f$ is a polynomial. We can see that the multiplicity of $\lambda$ is at least $p=\dim (V_{\lambda})$. This says that $1\leq p\leq m$, and the proof is complete. $\blacksquare$

\begin{equation}{\label{lal219}}\tag{16}\mbox{}\end{equation}

Proposition \ref{lal219}. Let $V$ be a vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Let $\lambda_{1},\cdots ,\lambda_{k}$ be the distinct eigenvalues of $T$ with the corresponding eigenspaces $V_{\lambda_{1}},V_{\lambda_{2}},\cdots ,V_{\lambda_{k}}$. Then, we have the following properties.

(i) For each $v_{i}\in V_{\lambda_{i}}$, $i=1,\cdots ,k$, if
\[v_{1}+v_{2}+\cdots +v_{k}=\theta ,\]
then $v_{i}=\theta$ for all $i=1,\cdots ,k$.

(ii) Let $S_{i}$ be a finite and linearly independent subset of $V_{\lambda_{i}}$. Then, the union
\[S=S_{1}\cup S_{2}\cup\cdots\cup S_{k}\]
is a linearly independent subset of $V$

Proof. Part (i) follows mmediately from Proposition \ref{lap217} . To prove part (ii), for each $i=1,\cdots ,k$, we define
\[S_{i}=\left\{v_{i1},v_{i2},\cdots ,v_{in_{i}}\right\}.\]
Given any scalars $\{\alpha_{ij}\}$ satisfying
\[\sum_{i=1}^{k}\sum_{j=1}^{n_{i}}\alpha_{ij}v_{ij}=\theta ,\]
for each $i=1,\cdots , k$, let
\[w_{i}=\sum_{j=1}^{n_{i}}\alpha_{ij}v_{ij}.\]
Then $w_{i}\in V_{\lambda_{i}}$ for each $i=1,\cdots ,k$ and
\[w_{1}+w_{2}+\cdots +w_{k}=\theta .\]
Using part (i), we must have $w_{i}=\theta$ for all $i=1,\cdots ,k$. Since $S_{i}$ is linearly independent, we must have $\alpha_{ij}=0$ for all $j=1,\cdots ,n_{i}$. This conclude that $S$ is linearly independent, and the proof is complete. $\blacksquare$

\begin{equation}{\label{lat230}}\tag{17}\mbox{}\end{equation}

Theorem \ref{lat230}, Let $V$ be a finite-dimensional vector space over ${\cal F}$, and let $T$ be a linear operator on $V$ such that its characteristic polynomial can be factored over ${\cal F}$, i.e., the characteristic polynomial $P_{T}$ of $T$ is given by
\[P_{T}(\lambda )=(\lambda -\lambda_{1})^{d_{1}}\cdots (\lambda -\lambda_{r})^{d_{r}}.\]
Let $\lambda_{1},\cdots ,\lambda_{r}$ be the distinct eigenvalues of $T$ with the corresponding eigenspaces $V_{\lambda_{1}},V_{\lambda_{2}},\cdots ,V_{\lambda_{r}}$. Then, we have the following properties.

(i) The linear operator is diagonalizable if and only if the multiplicity of $\lambda_{i}$ is equal to the dimension of eigenspace $V_{\lambda_{i}}$ for $i=1,\cdots ,r$.

(ii) If $T$ is diagonalizable and $\mathfrak{B}_{i}$ is an ordered basis for $V_{\lambda_{i}}$ for each $i=1,\cdots ,r$, then the union
\[\mathfrak{B}=\mathfrak{B}_{1}\cup\mathfrak{B}_{2}\cup\cdots\cup\mathfrak{B}_{r}\]
is an ordered basis for $V$ consisting of eigenvectors of $T$.

Proof. Suppose that $\dim (V)=n$. Let $m_{i}$ denote the multiplicity of $\lambda_{i}$ and $d_{i}$ denote the dimension of $V_{\lambda_{i}}$ for $i=1,\cdots ,r$. To prove part (i), we assume that $T$ is diagonalizable. Let $\mathfrak{B}$ be a basis for $V$ consisting of eigenvectors of $T$. Let $\mathfrak{B}_{i}=\mathfrak{B}\cap V_{\lambda}$, and let $n_{i}$ denote the number of eigenvectors in $\mathfrak{B}_{i}$ for $i=1,\cdots ,r$. Using Proposition \ref{lal218}, we have $d_{i}\leq m_{i}$ for $i=1,\cdots ,r$. Since $\mathfrak{B}_{i}$ is a linearly independent subset of $V_{\lambda_{i}}$ and $d_{i}=\dim (V_{\lambda_{i}})$, we must have $n_{i}\leq d_{i}$ for $i=1,\cdots ,r$. Since $\mathfrak{B}$ consists of $n$ vectors, we have $n=\sum_{i=1}^{r}n_{i}$. On the other hand, since the degree of the characteristic polynomial of $T$ is equal to the sum of the multiplicities of the eigenvalues, we also have $n=\sum_{i=1}^{r}m_{i}=n$. Therefore, we obtain
\[n=\sum_{i=1}^{r}n_{i}\leq\sum_{i=1}^{r}d_{i}\leq\sum_{i=1}^{r}m_{i}=n,\]
which implies
\[\sum_{i=1}^{r}\left (m_{i}-d_{i}\right )=0.\]
Since $m_{i}-d_{i}\geq 0$ for $i=1,\cdots ,r$, we must have $m_{i}=d_{i}$ for $i=1,\cdots ,r$.

For the converse, suppose that $m_{i}=d_{i}$ for $i=1,\cdots ,r$. Let $\mathfrak{B}_{i}$ be an ordered basis for $V_{\lambda_{i}}$ for each $i=1,\cdots ,r$, and take the union
\[\mathfrak{B}=\mathfrak{B}_{1}\cup\mathfrak{B}_{2}\cup\cdots\cup\mathfrak{B}_{r}.\]
According to Proposition~\ref{lal219}, it says that $\mathfrak{B}$ is linearly independent. Since $m_{i}=d_{i}$ for all $i$ and $\sum_{i=1}^{r}m_{i}=n$, we have $\sum_{i=1}^{r}d_{i}=n$, which says that $\mathfrak{B}$ contains $n$ vectors. Since $n=\dim (V)$, we conclude that $\mathfrak{B}$ is an ordered basis for $V$ consisting of eigenvectors of $T$, which proves part (ii) and also says that $T$ is diagonalizable. The proof is complete. $\blacksquare$

Let $\mathfrak{B}$ be an ordered basis for an $n$-dimensional vector space $V$ over the scalar field ${\cal F}$. Given any $v\in V$, the coordinate vector $[v]_{\mathfrak{B}}$ is a column vector in ${\cal F}^{n}$. If we define the mapping $\phi_{\mathfrak{B}}:V\rightarrow {\cal F}^{n}$ by $\phi_{\mathfrak{B}}(v)=[v]_{\mathfrak{B}}$, then we can show that $\phi_{\mathfrak{B}}$ is an isomorphism, which is also left as an exercise.

\begin{equation}{\label{lap215}}\tag{18}\mbox{}\end{equation}

Proposition \ref{lap215}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$ with an ordered basis $\mathfrak{B}$, and let $T$ be a linear operator on $V$. Let $A=[T]_{\mathfrak{B}}$, and let Let $\lambda$ be an eigenvalue of $A$, i.e., the eigenvalue of $T$. Then, we have the following properties.

(i) Given $v\in V$, $\phi_{\mathfrak{B}}(v)$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda$ if and only if $v$ is an eigenvector of $T$
corresponding to the eigenvalue $\lambda$.

(ii) A vector $y\in {\cal F}^{n}$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda$ if and only if $\phi_{\mathfrak{B}}^{-1}(y)$ is an eigenvector of $T$ corresponding to $\lambda$.

Proof. Suppose that $v$ is an eigenvector of $T$ corresponding to $\lambda$. Then $T(v)=\lambda v$. Therefore, since $\phi_{\mathfrak{B}}$ is an isomorphism, we have
\begin{align*} A\phi_{\mathfrak{B}}(v) & =\phi_{\mathfrak{B}}(T(v))\\ & =\phi_{\mathfrak{B}}(\lambda v)\\ & =\lambda\phi_{\mathfrak{B}}(v).\end{align*}
Since $v\neq\theta$, we also have $\phi_{\mathfrak{B}}(v)\neq {\bf 0}$, which says that $\phi_{\mathfrak{B}}(v)$ is an eigenvector of $A$. Conversely, suppose that $\phi_{\mathfrak{B}}(v)$ is an eigenvector of $A$ corresponding to $\lambda$. Then, we have
\begin{align*} \phi_{\mathfrak{B}}(T(v)) & =A\phi_{\mathfrak{B}}(v)\\ & =\lambda\phi_{\mathfrak{B}}(v)\\ & =\phi_{\mathfrak{B}}(\lambda v).\end{align*}
Since $\phi_{\mathfrak{B}}$ is injective, we have $T(v)=\lambda v$, which shows that $v$ is an eigenvector of $T$ corresponding to $\lambda$. Since $\phi_{\mathfrak{B}}$ is bijective, part (ii) is the equivalent formulation of part (i), and the proof is complete. $\blacksquare$

Example. Let $V$ be the vector space consisting of all polynomials with degree less than or equal to $2$ and coefficients in $\mathbb{R}$, and let $T$ be a linear operator on $V$ defined by
\[T(f(x))=f(x)+(x+1)f'(x).\]
Let $\mathfrak{B}=\{1,x,x^{2}\}$ be the standard ordered basis for $V$. Then
\[[T]_{\mathfrak{B}}=A=\left [\begin{array}{rrr}
1 & 1 & 0\\ 0 & 2 & 2\\ 0 & 0 & 3
\end{array}\right ].\]
The characteristic polynomial $P_{T}$ for $T$ (or $P_{A}$ for $A$) is given by
\begin{align*} \det (\lambda I-A) & =\left |\begin{array}{ccc}
\lambda -1 & 1 & 0\\ 0 & \lambda -2 & 2\\ 0 & 0 & \lambda -3
\end{array}\right |\\ & =(\lambda -1)(\lambda -2)(\lambda -3).\end{align*}
Therefore, the eigenvalues of $T$ (or $A$) are $1$, $2$ and $3$. We can apply Proposition \ref{lap215} to find all eigenvectors of $T$.

The eigenvectors of $A$ corresponding to the eigenvalue $1$ have the form of
\[\left\{t\left [\begin{array}{c}
1\\ 0\\ 0
\end{array}\right ]:t\in\mathbb{R}\mbox{ and }t\neq 0\right\}.\]
According to Proposition \ref{lap215}, the eigenvector of $T$ corresponding to the eigenvalue $1$ have the form of
\[\phi_{\mathfrak{B}}^{-1}\left (t\left [\begin{array}{c}
1\\ 0\\ 0
\end{array}\right ]\right )=t\phi_{\mathfrak{B}}^{-1}\left (\left [\begin{array}{c}
1\\ 0\\ 0
\end{array}\right ]\right )=t\cdot 1=t.\]
This says that the non-zero constant polynomial are the eigenvectors of $T$ corresponding to the eigenvalue $1$.
The eigenvectors of $A$ corresponding to the eigenvalue $2$ have the form of
\[\left\{t\left [\begin{array}{c}
1\\ 1\\ 0
\end{array}\right ]:t\in\mathbb{R}\mbox{ and }t\neq 0\right\}.\]
Since $\phi_{\mathfrak{B}}^{-1}$ is an isomorphism, the eigenvector of $T$ corresponding to the eigenvalue $2$ have the form of
\begin{align*}
\phi_{\mathfrak{B}}^{-1}\left (t\left [\begin{array}{c}
1\\ 1\\ 0
\end{array}\right ]\right ) & =t\phi_{\mathfrak{B}}^{-1}\left (\left [\begin{array}{c}
1\\ 0\\ 0
\end{array}\right ]+\left [\begin{array}{c}
0\\ 1\\ 0
\end{array}\right ]\right )\\
& =t\cdot\left (\phi_{\mathfrak{B}}^{-1}\left (
\left [\begin{array}{c}
1\\ 0\\ 0
\end{array}\right ]\right )+\phi_{\mathfrak{B}}^{-1}\left (\left [\begin{array}{c}
0\\ 1\\ 0
\end{array}\right ]\right )\right )\\
& =t\cdot (1+x).
\end{align*}
This says that the polynomials of the form $t\cdot (1+x)$ are the eigenvectors of $T$ corresponding to the eigenvalue $2$.
The eigenvectors of $A$ corresponding to the eigenvalue $3$ have the form of
\[\left\{t\left [\begin{array}{c}
1\\ 2\\ 1
\end{array}\right ]:t\in\mathbb{R}\mbox{ and }t\neq 0\right\}.\]
Therefore, the eigenvector of $T$ corresponding to the eigenvalue $3$ have the form of
\begin{align*}
\phi_{\mathfrak{B}}^{-1}\left (t\left [\begin{array}{c}
1\\ 2\\ 1
\end{array}\right ]\right ) & =t\phi_{\mathfrak{B}}^{-1}\left (\left [\begin{array}{c}
1\\ 0\\ 0
\end{array}\right ]+\left [\begin{array}{c}
0\\ 2\\ 0
\end{array}\right ]+\left [\begin{array}{c}
0\\ 0\\ 1
\end{array}\right ]\right )\\
& =t\cdot\left (\phi_{\mathfrak{B}}^{-1}\left (
\left [\begin{array}{c}
1\\ 0\\ 0
\end{array}\right ]\right )+\phi_{\mathfrak{B}}^{-1}\left (\left [\begin{array}{c}
0\\ 2\\ 0
\end{array}\right ]\right )+\phi_{\mathfrak{B}}^{-1}\left (\left [\begin{array}{c}
0\\ 0\\ 1
\end{array}\right ]\right )\right )\\
& =t\cdot (1+2x+x^{2}).
\end{align*}
This says that the polynomials of the form $t\cdot (1+2x+x^{2})$ are the eigenvectors of $T$ corresponding to the eigenvalue $3$.

By taking $t=1$ in the above cases, it is easy to see that
\[\widehat{\mathfrak{B}}=\left\{1,1+x,1+2x+x^{2}\right\}\]
is an ordered basis for $V$. Therefore, $T$ is diagonalizable with
\[[T]_{\widehat{\mathfrak{B}}}=\left [\begin{array}{rrr}
1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3
\end{array}\right ].\]

Example. Let $V$ be the vector space consisting of all polynomials with degree less than or equal to $2$ and coefficients in $\mathbb{R}$, and let $T$ be a linear operator on $V$ defined by
\[T(f(x))=f(1)+f'(0)x+\left [f'(0)+f”(0)\right ]x^{2}.\]
Let $\mathfrak{B}=\{1,x,x^{2}\}$ be the standard ordered basis for $V$. Then
\[[T]_{\mathfrak{B}}=A=\left [\begin{array}{rrr}
1 & 1 & 1\\ 0 & 1 & 0\\ 0 & 1 & 2
\end{array}\right ].\]
The characteristic polynomial $P_{T}$ for $T$ (or $P_{A}$ for $A$) is given by
\begin{align*} \det (\lambda I-A) & =\left |\begin{array}{ccc}
\lambda -1 & 1 & 1\\ 0 & \lambda -1 & 0\\ 0 & 1 & \lambda -2
\end{array}\right |\\ & =(\lambda -1)^{2}(\lambda -2).\end{align*}
Therefore, the eigenvalues of $T$ (or $A$) are $1$ and $2$ with multiplicities $2$ and $1$, respectively.

For eigenvalue $1$, we consider
\begin{align*}
\left [\begin{array}{c}
0\\ 0 \\ 0
\end{array}\right ] & =(I_{3}-A)X\\ & =\left (\left [\begin{array}{rrr}
1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1
\end{array}\right ]-\left [\begin{array}{rrr}
1 & 1 & 1\\ 0 & 1 & 0\\ 0 & 1 & 2
\end{array}\right ]\right )\left [\begin{array}{c}
x\\ y\\ z
\end{array}\right ]\\
& =\left [\begin{array}{rrr}
0 & -1 & -1\\ 0 & 0 & 0\\ 0 & -1 & -1
\end{array}\right ]\left [\begin{array}{c}
x\\ y\\ z
\end{array}\right ]\\ & =\left [\begin{array}{c}
-y-z\\ 0\\ -y-z
\end{array}\right ].
\end{align*}
The set of all solutions is given by
\[V_{1}=\left\{s\left [\begin{array}{c}
1\\ 0\\ 0
\end{array}\right ]+t\left [\begin{array}{r}
0\\ -1\\ 1
\end{array}\right ]:s,t\in\mathbb{R}\right\},\]
which is the eigenspace with eigenvalue $1$ and basis
\[\left\{\left [\begin{array}{c}
1\\ 0\\ 0
\end{array}\right ],\left [\begin{array}{r}
0\\ -1\\ 1
\end{array}\right ]\right\}.\]
For eigenvalue $2$, we consider
\begin{align*}
\left [\begin{array}{c}
0\\ 0 \\ 0
\end{array}\right ] & =(2I_{3}-A)X\\ & =\left (\left [\begin{array}{rrr}
2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2
\end{array}\right ]-\left [\begin{array}{rrr}
1 & 1 & 1\\ 0 & 1 & 0\\ 0 & 1 & 2
\end{array}\right ]\right )\left [\begin{array}{c}
x\\ y\\ z
\end{array}\right ]\\
& =\left [\begin{array}{rrr}
1 & -1 & -1\\ 0 & 1 & 0\\ 0 & -1 & 0
\end{array}\right ]\left [\begin{array}{c}
x\\ y\\ z
\end{array}\right ]\\ & =\left [\begin{array}{c}
x-y-z\\ y\\ -y
\end{array}\right ].
\end{align*}
The set of all solutions is given by
\[V_{2}=\left\{t\left [\begin{array}{r}
1\\ 0\\ 1
\end{array}\right ]:t\in\mathbb{R}\right\},\]
which is the eigenspace with eigenvalue $2$ and basis
\[\left\{\left [\begin{array}{c}
1\\ 0\\ 1
\end{array}\right ]\right\}.\]

Therefore, the union of these two bases given by
\[\left\{\left [\begin{array}{c}
1\\ 0\\ 0
\end{array}\right ],\left [\begin{array}{r}
0\\ -1\\ 1
\end{array}\right ],\left [\begin{array}{c}
1\\ 0\\ 1
\end{array}\right ]\right\}\]

is an ordered basis for $\mathbb{R}^{3}$ consisting of eigenvectors of $A$, which are also the coordinate vectors relative to standard basis $\mathfrak{B}=\{1,x,x^{2}\}$ of the vectors in the ordered basis
\[\widehat{\mathfrak{B}}=\left\{1,-x+x^{2},1+x^{2}\right\}\]
for $V$ consisting of eigenvectors of $T$. Therefore, $T$ is diagonalizable with
\[[T]_{\widehat{\mathfrak{B}}}=\left [\begin{array}{rrr}
1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 2
\end{array}\right ].\]

Definition. Let $V$ be a vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Let $W$ be a subspace of $V$. We say that $W$ is invariant under $T$ or $T$-invariant when $T(W)\subseteq W$.

Example. Let $V$ be a vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Then we have the following trivial $T$-invariant subspaces.

The zero subspace and the whole space $V$ are $T$-invariant.
$\mbox{Im}(T)$ and $\mbox{Ker}(T)$ are $T$-invariant.
The eigenspace $V_{\lambda}$ for any eigenvalue of $T$ is $T$-invariant. $\sharp$

Proposition. Let $V$ be a vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Let $f$ be a polynomial in ${\cal F}$. If $W$ is the kernel of $f(T)$, then $W$ is an invariant subspace of $V$ under $T$.

Proof. Given $w\in W$, i.e., $(f(T))(w)=\theta$, we want to show that $(f(T))(A(w))=\theta$. Since $tf(t)=f(t)t$, i.e., $Tf(T)=f(T)T$, we have
\begin{align*} (f(T))(A(w)) & =(f(T)T)(w)=(Tf(T))(w)\\ & =T((f(T))(w))=T(\theta )=\theta .\end{align*}
This completes the proof. $\blacksquare$

\begin{equation}{\label{lad227}}\tag{19}\mbox{}\end{equation}

Definition \ref{lad227}. Let $V$ be a vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Given any $v\in V$, the subspace
\[W=\mbox{span}\left (\left\{v,T(v),T^{2}(v),\cdots\right\}\right )\]
is called the $T$-cyclic subspace of $V$ generated by $v$

\begin{equation}{\label{laex139}}\tag{20}\mbox{}\end{equation}

Example. \ref{laex139}. We provide some simple $T$-cyclic subspaces.

For any linear operator $T$ on $V$, we have $\langle\theta\rangle_{T}=\{\theta\}$ the zero subspace of $V$.
We have that $\dim\langle v\rangle_{T}=1$ if and only if $v$ is an eigenvector for $T$.
We have $\dim\langle v\rangle_{I}=1$, where $I$ is an identity operator and $v\neq\theta$. This also says that if $\dim (V)>1$, then the identity operator $I$ has no cyclic vector.

\begin{equation}{\label{lap223}}\tag{21}\mbox{}\end{equation}

Remark \ref{lap223}. Let $V$ be a vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Let $W$ be the $T$-cyclic subspace of $V$ generated by $v\in V$. Then, we have the following observations.

$W$ is $T$-invariant.
Any $T$-invariant subspace of $V$ containing $v$ also contains $W$. In other words, $W$ is the smallest $T$-invariant subspace of $V$ containing $v$.

\begin{equation}{\label{lae221}}\tag{22}\mbox{}\end{equation}

Example \ref{lae221}. Let $T$ be a linear operator on $\mathbb{R}^{3}$ defined by
\[T(x,y,z)=\left (-y+z,x+z,3z\right ).\]
We shall find the $T$-cyclic subspace generated by ${\bf e}_{1}=(1,0,0)$. Now, we have
\begin{align*} T({\bf e}_{1}) & =T(1,0,0)\\ & =(0,1,0)={\bf e}_{2}\end{align*}
and
\begin{align*} T^{2}({\bf e}_{1}) & =T({\bf e}_{2})\\ & =(-1,0,0)=-{\bf e}_{1}.\end{align*}
Therefore, we can also obtain
\[T^{3}({\bf e}_{1})=T(-{\bf e}_{1})=-{\bf e}^{2}\]
and
\[T^{4}({\bf e}_{1})=T(-{\bf e}_{2})={\bf e}^{1}.\]
This says
\begin{align*}
\mbox{span}\left (\left\{{\bf e}_{1},T({\bf e}_{1}),T^{2}({\bf e}_{1}),
\cdots\right\}\right ) & =\mbox{span}\left\{\left ({\bf e}_{1},{\bf e}_{2},
-{\bf e}_{1},-{\bf e}_{2}\right\}\right )\\
& =\mbox{span}\left\{\left ({\bf e}_{1},{\bf e}_{2}\right\}\right )\\
& =\left\{(x,y,0):x,y\in\mathbb{R}\right\}.
\end{align*}

Example. Let $V$ be the space of all polynomial functions from $\mathbb{R}$ into $\mathbb{R}$ which have degree less than or equal to $2$. Let $T$ be a linear operator on $V$ defined by $T(f)=f’$. Then, the $T$-cyclic subspace generated by $x^{2}\in V$ is
\[\mbox{span}\left (\left\{x^{2},2x,2\right\}\right )=V.\]

\begin{equation}{\label{lap222}}\tag{23}\mbox{}\end{equation}

Proposition \ref{lap222}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Let $W$ be a $T$-cyclic subspace of $V$ generated by a non-zero vector $v\in V$ such that $\dim (W)=k$. Then, we have the following properties.

(i) The set $\{v,T(v),T^{2}(v),T^{k-1}(v)\}$ is a basis for $W$.

(ii) For $a_{0},a_{1},a_{2},\cdots ,a_{k-1}\in {\cal F}$, if
\[a_{0}v+a_{1}T(v_{1})+a_{2}T^{2}(v)+\cdots +a_{k-1}T^{k-1}(v)+T^{k}(v)=\theta ,\]
then the characteristic polynomial of $T_{W}$ is
\[P_{T_{W}}(\lambda )=(-1)^{k}\cdot\left (a_{0}+a_{1}t+a_{2}\lambda^{2}+\cdots
+a_{k-1}\lambda^{k-1}+\lambda^{k}\right ).\]

Proof. To prove part (i), since $v\neq\theta$, the set $\{v\}$ is linearly independent. Let $r$ be the largest positive integer such that
\[\mathfrak{B}=\left\{v,T(v),T^{2}(v),\cdots ,T^{r-1}(v)\right\}\]
is linearly independent. Such integer $r$ must exists, since $V$ is finite-dimensional. Let $Z=\mbox{span}(\mathfrak{B})$. Then $\mathfrak{B}$ is a basis for $Z$. We are going to claim that $T^{r}(v)\in Z$. Suppose that $T^{r}(v)\not\in Z$. Then, $T^{r}(v)\not\in\mbox{span}(\mathfrak{B})$. According to Proposition \ref{lap156}, the set $\mbox{span}(\mathfrak{B})\cup\{T^{r}(v)\}$ must be independent, which contradicts the fact that $r$ is the largest positive integer such that $\mbox{span}(\mathfrak{B})$ is linearly independent. This shows that $T^{r}(v)\in Z$, which will be used to claim that $Z$ is a $T$-invariant subspace of $V$. Let $w\in Z$. Then there exists $b_{0},b_{1},\cdots , b_{r-1}\in {\cal F}$ satisfying
\[w=b_{0}v+b_{1}T(v)+b_{2}T^{2}(v)+\cdots +b^{r-1}T^{r-1}(v).\]
Therefore, we have
\[T(w)=b_{0}T(v)+b_{1}T^{2}(v)+b_{2}T^{3}(v)+\cdots +b^{r-1}T^{r}(v),\]
which says that $T(w)$ is a linea combinations of vectors in $Z$. This shows $T(w)\in Z$, i.e., $Z$ is $T$-invariant. Since $v\in Z$, using part (ii) of Remark \ref{lap223}, we see that $W$ is the smallest $T$-invariant subspace of $V$ containing $v$. Therefore, we must have $W\subseteq Z$. Since $Z\subseteq W$ is obvious, we conclude that $Z=W$. This shows that $\mathfrak{B}$ is a basis for $W$ such that $\dim (W)=r$. Therefore, we must $r=k$.

To prove part (ii), we can regard $\mathfrak{B}$ as an ordered basis for $W$. Since
\[a_{0}v+a_{1}T(v_{1})+a_{2}T^{2}(v)+\cdots +a_{k-1}T^{k-1}(v)+T^{k}(v)=\theta ,\]
we can have
\[[T_{W}]_{\mathfrak{B}}=\left [\begin{array}{ccccc}
0 & 0 & \cdots & 0 & -a_{0}\\
1 & 0 & \cdots & 0 & -a_{1}\\
\vdots & \vdots & \cdots & \vdots & \vdots\\
0 & 0 & \cdots & 1 & -a_{k-1}
\end{array}\right ],\]
which has the characteristic polynomial
\[f(\lambda )=(-1)^{k}\cdot\left (a_{0}+a_{1}t+a_{2}\lambda^{2}+\cdots
+a_{k-1}\lambda^{k-1}+\lambda^{k}\right ).\]
Therefore, we mush have $P_{T_{W}}(\lambda )=f(\lambda )$. This completes the proof. $\blacksquare$

Example. Continued from Example \ref{lae221}, let $W$ be the $T$-cyclic subspace generated by ${\bf e}_{1}$. Then, we have obtained
\begin{align*} W & =\mbox{span}\left\{\left ({\bf e}_{1},{\bf e}_{2}\right\}\right )\\ & =\left\{(x,y,0):x,y\in\mathbb{R}\right\}.\end{align*}

Let $\mathfrak{B}_{W}=\{{\bf e}_{1},{\bf e}_{2}\}$ be an ordered basis for $W$. Then, we have
\[[T_{w}]_{\mathfrak{B}_{W}}=\left [\begin{array}{rr}
0 & -1\\ 1 & 0
\end{array}\right ].\]
Therefore, the characteristic polynomial of $T_{W}$ is given by
\[P_{T_{W}}(\lambda )=\det\left [\begin{array}{rr}
-\lambda & -1\\ 1 & -\lambda
\end{array}\right ]=\lambda^{2}+1.\] We have that
\[\{{\bf e}_{1},T({\bf e}_{1})\}=\{{\bf e}_{1},{\bf e}_{2}\}\]
is a basis for $W$. Since $T^{2}({\bf e}_{1})=-{\bf e}_{1}$, we have
\[1{\bf e}_{1}+0T({\bf e}_{1})+T^{2}({\bf e}_{1})=\theta\]
By part (ii) of Proposition \ref{lap222}, we obtain
\begin{align*} P_{T_{W}}(\lambda ) & =(-1)^{2}\cdot\left (1+0\lambda+\lambda^{2}\right )\\ & =\lambda^{2}+1.\end{align*}

If the subspace $W$ is $T$-invariant, then $T$ can induce a liner operator $T_{W}$ on $W$ defined by $T_{W}(v)=T(v)$ for $v\in W$. Suppose that $\dim (V)=n$ and $\dim (W)=r$. Now, we choose an ordered basis $\mathfrak{B}_{V}=\{v_{1},\cdots ,v_{n}\}$ for $V$ such that $\mathfrak{B}_{W}=\{v_{1},\cdots ,v_{r}\}$ is an ordered basis for $W$. Let $A=[T]_{\mathfrak{B}_{V}}$ and $B=[T_{W}]_{\mathfrak{B}_{W}}$. By definition, we have
\[T(v_{j})=\sum_{i=1}^{n}a_{ij}v_{i}.\]
Since $W$ is invariant under $T$, it follows that $T(v_{j})\in W$ for $j=1,\cdots ,r$, which also means
\[T(v_{j})=\sum_{i=1}^{r}a_{ij}v_{i}.\]
In other words, we have $a_{ij}=0$ for $j=1,\cdots ,r$ and $i=r+1,\cdots ,n$. Therefore, the matrix $A$ has the following block form
\begin{equation}{\label{laeq114}}\tag{24}
A=\left [\begin{array}{cc}
B & C\\ {\bf 0} & D
\end{array}\right ],
\end{equation}
where $B$ is an $r\times r$ matrix, $C$ is an $r\times (n-r)$ matrix, and $D$ is an $(n-r)\times (n-r)$ matrix.

\begin{equation}{\label{lap224}}\tag{25}\mbox{}\end{equation}

Proposition \ref{lap224}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Let $W$ be a $T$-invariant subspace of $V$. The characteristic polynomial for the restriction operator $T_{W}$ divides the characteristic polynomial for $T$.

Proof. Let $\dim (V)=n$ and $\dim (W)=r$. We have $A=[T]_{\mathfrak{B}_{V}}$ and $B=[T_{W}]_{\mathfrak{B}_{W}}$ as shown in (\ref{laeq114}). Since $A$ is in the block form, we have
\[\mbox{det}(A-\lambda I_{n})=\mbox{det}(B-\lambda I_{r})\cdot\mbox{det}(D-\lambda I_{n-r}).\]
This completes the proof. $\blacksquare$

Suppose that $T$ is a diagonalizable operator with distinct eigenvalues $\lambda_{1},\cdots ,\lambda_{r}$, where $r<n$. Since $\lambda_{i}$ is a root of the characteristic polynomial $P_{T}$, we also assume that the root $\lambda_{i}$ has multiplicity $d_{i}$. Then $d_{1}+\cdots +d_{r}=n$ and
\[P_{T}(\lambda )=(\lambda -\lambda_{1})^{d_{1}}\cdots (\lambda -\lambda_{r})^{d_{r}}.\]
Let $I_{i}$ be the $d_{i}\times d_{i}$ identity matrix. Then, there is an ordered basis $\mathfrak{B}$ for $V$ such that $[T]_{\mathfrak{B}}$ is the block form shown below
\[[T]_{\mathfrak{B}}=\left [\begin{array}{cccc}
\lambda_{1}I_{1} & {\bf 0} & \cdots & {\bf 0}\\
{\bf 0} & \lambda_{2}I_{2} & \cdots & {\bf 0}\\
\vdots & \vdots & \cdots & \vdots\\
{\bf 0} & {\bf 0} & \cdots & \lambda_{r}I_{r}
\end{array}\right ].\]

\begin{equation}{\label{lap107}}\tag{26}\mbox{}\end{equation}

Proposition \ref{lap107}. Let $V$ be vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Suppose that $T(v)=\lambda v$. If $f$ is any polynomial, then $(f(T))(v)=f(c)v$.

Proof. We have
\begin{align*} T^{2}(v) & =T(T(v))=T(\lambda v)\\ & =\lambda T(v)=\lambda^{2}v.\end{align*}
In general, we can obtain $T^{n}(v)=\lambda^{n}v$. Using the linearity of $T$, we can show that $(f(T))(v)=f(c)v$, and the proof is complete. $\blacksquare$

For example, if $f(x)=x^{2}+2x+3$, then $f(T)=T^{2}+2T+I$ is a linear operator on $V$. Therefore, we have
\begin{align*}
(f(T))(v) & =(T^{2}+2T+I)(v)=T^{2}(v)+2T(v)+I(v)\\
& =\lambda^{2}v+2\lambda v+v=(\lambda^{2}+2\lambda +1)v=f(\lambda )v.
\end{align*}

Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Suppose that $V$ has a basis $\{v_{1},\cdots ,v_{n}\}$ consisting of eigenvectors of $T$ with the corresponding eigenvalues $\{\lambda_{1},\cdots ,\lambda_{n}\}$. Then, the characteristic polynomial of $T$ is
\[P_{T}(\lambda )=(\lambda -\lambda_{1})\cdots (\lambda -\lambda_{n}).\]
In this case, we define
\[P_{T}(T)=(T-\lambda_{1} I)\cdots (T-\lambda_{n}I).\]
Then, we have $(P_{T}(T))(v_{i})=0$ for all $i=1,\cdots ,n$. This also says that $P_{T}(T)$ is a zero mapping, i.e., $(P_{T}(T))(v)=\theta$ for all $v\in V$.

\begin{equation}{\label{lat38}}\tag{27}\mbox{}\end{equation}

Theorem \ref{lat38}. (Cayley-Hamilton) Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. If $P_{T}$ is the characteristic polynomial for $T$, then $P_{T}(T)$ is a zero mapping, i.e., $(P_{T}(T))(v)=\theta$ for all $v\in V$.

Proof. If $v=\theta$, then it is obvious that $(P_{T}(T))(v)=\theta$, since $P_{T}(T)$ is linear. Therefore, we assume that $v\neq\theta$. Let $W$ be the $T$-cyclic subspace generated by $v$ with $\dim (W)=r$. From part (i) of Proposition \ref{lap222}, the set $\{v,T(v),T^{2}(v),T^{k-1}(v)\}$ is a basis for $W$. Therefore, there exists $a_{0},a_{1},a_{2},\cdots ,a_{k-1}\in {\cal F}$ satisfying
\begin{equation}{\label{laeq225}}\tag{28}
a_{0}v+a_{1}T(v_{1})+a_{2}T^{2}(v)+\cdots +a_{k-1}T^{k-1}(v)+T^{k}(v)=\theta .
\end{equation}
By part (ii) of Proposition \ref{lap222}, the characteristic polynomial of $T_{W}$ is
\begin{equation}{\label{laeq226}}\tag{29}
P_{T_{W}}(\lambda )=(-1)^{k}\cdot\left (a_{0}+a_{1}t+a_{2}\lambda^{2}+\cdots+a_{k-1}\lambda^{k-1}+\lambda^{k}\right ).
\end{equation}
Combining (\ref{laeq225}) and (\ref{laeq226}), we obtain
\[(P_{T_{W}}(T))(v)=(-1)^{k}\cdot\left (a_{0}I+a_{1}T+a_{2}T^{2}+\cdots+a_{k-1}T^{k-1}+T^{k}\right )(v)=\theta .\]
Using Proposition \ref{lap224}, we see that $P_{T_{W}}$ divides $P_{T}$, which says that there exists a polynomial $q$ such that $P_{T}=q\cdot P_{T_{W}}$. Therefore, we have
\begin{align*} (P_{T}(T))(v) & =(q(T)\circ P_{T_{W}}(T))(v)\\ & =(q(T))((P_{T_{W}}(T))(v))\\ & =(q(T))(\theta )=\theta .\end{align*}
This completes the proof. $\blacksquare$

Let $T$ be a linear operator on $\mathbb{R}^{2}$ defined by
\[T(x,y)=(x+2y,-2x+y)\]
and let $\mathfrak{B}$ be the standard basis for $\mathbb{R}^{2}$. Then, we have
\[[T]_{\mathfrak{B}}=A=\left [\begin{array}{rr}
1 & 2\\ -2 & 1
\end{array}\right ].\]
The characteristic polynomial is given by
\begin{align*} P_{T}(\lambda ) & =\det\left (A-\lambda I_{2}\right )\\ & =\lambda^{2}-2\lambda +5.\end{align*}
The Cayley-Hamilton theorem says that
\[P_{T}(T)=T^{2}-2T+5I\]
is a zero mapping. We can also show
\[P_{A}(A)=P_{T}(A)=A^{2}-2A+5I_{2}\]
is a zero matrix. $\sharp$

Corollary. Let $A$ be an $n\times n$ matrix of complex numbers and let $P_{A}$ be its characteristic polynomial. Then $P_{A}(A)$ is a zero matrix.

Proof. We can regard $A$ as a linear mapping from $\mathbb{C}^{n}$ into itself. Then, the results follows immediately from Theorem \ref{lat38}. $\blacksquare$

\begin{equation}{\label{lap109}}\tag{30}\mbox{}\end{equation}

Proposition \ref{lap109}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Let $\lambda_{1},\cdots ,\lambda_{r}$ be the distinct eigenvalues of $T$, and let $W_{i}$ be the eigenspaces associated with the eigenvalues $\lambda_{i}$. If $W=W_{1}+W_{2}+\cdots +W_{r}$, then
\[\dim (W)=\dim (W_{1})+\dim (W_{2})+\cdots +\dim (W_{r}).\]
Moreover, if $\mathfrak{B}_{i}$ is an ordered basis for $W_{i}$, then $\mathfrak{B}=\bigcup_{i=1}^{r}\mathfrak{B}_{i}$ arranged in the order $\mathfrak{B}_{1},\cdots ,\mathfrak{B}_{r}$ is an ordered basis for $W$.

Proof. For any $w_{i}\in W_{i}$, we have $T(w_{i})=\lambda_{i}w_{i}$. This says that $W$ is the space spanned by all of the eigenvectors of $T$, and $\mathfrak{B}$ spans $W$. Therefore, we remain to prove that $\mathfrak{B}$ is linearly independent. Suppose that $w_{1}+\cdots +w_{r}=\theta$. We shall show that $w_{i}=\theta$. Let $f$ be any polynomial. Proposition \ref{lap107} says
\begin{align*}
\theta & =(f(T))(\theta )=(f(T))\left (w_{1}+\cdots +w_{r}\right )\\
& =(f(T))(w_{1})+\cdots +(f(T))(w_{r})=f(\lambda_{1})w_{1}+\cdots +
f(\lambda_{r})w_{r}.
\end{align*}
In particular, we take polynomials $f_{1},\cdots ,f_{r}$ by
\[f_{i}(\lambda_{j})=\delta_{ij}=\left\{\begin{array}{ll}
1 & i=j\\ 0 & i\neq j.
\end{array}\right .\]
Then, we obtain
\begin{align*} \theta & =(f_{i}(T))(\theta )\\ & =f_{i}(\lambda_{1})w_{1}+\cdots +f_{i}(\lambda_{r})w_{r}
\\ & =\sum_{j=1}^{r}\delta_{ij}w_{j}=w_{i}\end{align*}
for each $i$. The linear combination of vectors in $\mathfrak{B}$ has the form of $b_{1}+\cdots +b_{r}$, where $b_{i}$ is also the linear combination of the vectors in $\mathfrak{B}_{i}$. Therefore, if $b_{1}+\cdots +b_{r}=\theta$, then $b_{i}=\theta$ for all $i$. Since $\mathfrak{B}_{i}$ is independent, the scalars (coefficients) presented in the linear combination $b_{i}$ must be zero. Therefore, the scalars (coefficients) presented in the linear combination $b_{1}+\cdots +b_{r}$ must also be zero. This shows the independence, and the proof is complete. $\blacksquare$

\begin{equation}{\label{lat110}}\tag{31}\mbox{}\end{equation}

Theorem \ref{lat110}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Let $\lambda_{1},\cdots ,\lambda_{r}$ be the distinct eigenvalues of $T$, and let $W_{i}=\mbox{Ker}(T-\lambda_{i}I)$. Then, the following statements are equivalent.

(a) $T$ is diagonalizable.

(b) The characteristic polynomial of $T$ is
\begin{equation}{\label{laeq108}}\tag{32}
P_{T}(\lambda )=(\lambda -\lambda_{1})^{d_{1}}\cdots (\lambda -\lambda_{r})^{d_{r}},
\end{equation}
where $d_{i}=\dim (W)_{i}$ for $i=1,\cdots ,r$.

Proof. We have observed that (a) implies (b). If (\ref{laeq108}) holds, then $\dim (V)=d_{1}+\cdots +d_{r}$, since the degree of characteristic polynomial $P_{T}$ is the dimension of $V$. This shows that (b) implies (c). Suppose that (c) holds. Proposition \ref{lap109} says that $V=W_{1}+\cdots +W_{r}$, i.e., the eigenvectors of $T$ spans $V$. This completes the proof. $\blacksquare$

\begin{equation}{\label{laex137}}\tag{33}\mbox{}\end{equation}

Example \ref{laex137}. Let $T$ be a linear operator on $\mathbb{R}^{3}$ which is represented in the standard ordered basis by the matrix
\[A=\left [\begin{array}{rrr}
5 & -6 & -6\\ -1 & 4 & 2\\ 3 & -6 & -4
\end{array}\right ].\]
The characteristic polynomial is given by
\[P_{T}(\lambda )=\left |\begin{array}{ccc}
\lambda -5 & 6 & 6\\ 1 & \lambda -4 & -2\\ -3 & 6 & \lambda +4
\end{array}\right |=(\lambda -2)^{2}(\lambda -1).\]
This says that the eigenvalues are $\lambda_{1}=1$ and $\lambda_{2}=2$. Then, we have
\[\lambda_{1}I_{3}-A=I_{3}-A=\left [\begin{array}{rrr}
-4 & 6 & 6\\ 1 & -3 & -2\\ -3 & 6 & 5
\end{array}\right ]\]
and
\[\lambda_{2}I_{3}-A=2I_{3}-A=\left [\begin{array}{rrr}
-3 & 6 & 6\\ 1 & -2 & -2\\ -3 & 6 & 6
\end{array}\right ].\]
It is not difficult to show $\mbox{rank}(\lambda_{1}I_{3}-A)=2$ and $\mbox{rank}(\lambda_{2}I_{3}-A)=1$, i.e., $\mbox{rank}(\lambda_{1}I-T)=2$ and $\mbox{rank}(\lambda_{2}I-T)=1$. Since
\[3=\dim\mathbb{R}^{3}=\mbox{rank}(\lambda I-T)+\mbox{nullity}(\lambda I-T),\]
we have $\dim (W_{1})=1$ and $\dim (W_{2})=2$. Using Theorem \ref{lat110}, it follows that $T$ is diagonalizable. By solving
\begin{align*} (\lambda_{1}I_{3})X-A & =\left [\begin{array}{rrr}
-4 & 6 & 6\\ 1 & -3 & -2\\ -3 & 6 & 5
\end{array}\right ]\left [\begin{array}{c}
x_{1}\\ x_{2}\\ x_{3}
\end{array}\right ]\\ & =\left [\begin{array}{c}
0\\ 0\\ 0
\end{array}\right ],\end{align*}
we can obtain the eigenvector $v_{1}=(3,-1,3)$.
\begin{align*} (\lambda_{2}I_{3})X-A & =\left [\begin{array}{rrr}
-3 & 6 & 6\\ 1 & -2 & -2\\ -3 & 6 & 6
\end{array}\right ]\left [\begin{array}{c}
x_{1}\\ x_{2}\\ x_{3}
\end{array}\right ]\\ & =\left [\begin{array}{c}
0\\ 0\\ 0
\end{array}\right ],\end{align*}
we can obtain the eigenvectors $v_{2}=(2,1,0)$ and $v_{3}=(2,0,1)$. If we take $\mathfrak{B}=\{v_{1},v_{2},v_{3}\}$ as the ordered basis for $\mathbb{R}^{3}$, then $[T]_{\mathfrak{B}}$ is a diagonal matrix
\[D=\left [\begin{array}{rrr}
1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2
\end{array}\right ].\]
If we take the eigenvectors as the column vector to form the matrix
\[P=\left [\begin{array}{rrr}
3 & 2 & 2\\ -1 & 1 & 0\\ 3 & 0 & 1
\end{array}\right ],\]
then $P^{-1}AP=D$. $\sharp$

Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$ with $\dim (V)=n$, and let $T$ be a linear operator on $V$. Let $W$ be the subspace spanned by all of the eigenvectors of $T$. Let $\lambda_{1},\cdots ,\lambda_{s}$ be the distinct eigenvalues of $T$. For each $i$, let $W_{i}$ be the subspace of all eigenvectors associated with the eigenvalue $\lambda_{i}$, and let $\mathfrak{B}_{W_{i}}$ be an ordered basis for $W_{i}$. Proposition \ref{lap109} says that $\mathfrak{B}_{W}=\bigcup_{i=1}^{r}\mathfrak{B}_{W_{i}}$ is an ordered basis for $W$ and
\[r=\dim (W)=\dim (W_{1})+\dim (W_{2})+\cdots +\dim (W_{s}).\]
Let $\mathfrak{B}_{W}=\{v_{1},\cdots ,v_{r}\}$ such that the first few $v$’s form the basis $\mathfrak{B}_{W_{1}}$, the next few $v$’s form the basis $\mathfrak{B}_{W_{2}}$, and so on. Let $r_{i}=\dim (W_{i})$. We also have
\begin{align}
T(v_{j}) & =\lambda_{1}v_{j}\mbox{ for }j=1,\cdots ,r_{1}\\
T(v_{j+r_{1}}) & =\lambda_{2}v_{j+r_{1}}\mbox{ for }j=1,\cdots ,r_{2}\nonumber\\
T(v_{j+r_{2}}) & =\lambda_{3}v_{j+r_{2}}\mbox{ for }j=1,\cdots ,r_{3}\nonumber\\
\vdots & \vdots\vdots\nonumber\\
T(v_{j}) & =\lambda_{s}v_{j}\mbox{ for }j=r_{s}+1,\cdots ,n\label{laeq116}\tag{34}.
\end{align}
For any $w\in W$, we have
\[w=\alpha_{1}v_{1}+\cdots +\alpha_{r}v_{r}, \]
which implies
\begin{align*} T(w) & =\sum_{j=1}^{r}\alpha_{j}T(v_{j})\\ & =\beta_{1}v_{1}+\cdots +\beta_{r}v_{r}\in W\end{align*}
for some $\beta_{i}\in {\cal F}$ by (\ref{laeq115}). This says that the subspace $W$ is $T$-invariant. Now, we take any other vectors $v_{r+1},\cdots ,v_{n}$ such that $\mathfrak{B}_{V}=\{v_{1},\cdots ,v_{n}\}$ is an ordered basis for $V$. Then $[T]_{\mathfrak{B}_{V}}$ has the block from shown in (\ref{laeq114}) and $B=[T]_{\mathfrak{B}_{W}}$ is a diagonal matrix given by
\begin{align*} B & =[T]_{\mathfrak{B}_{W}}\\ & =\left [\begin{array}{cccc}
B_{1} & {\bf 0} & \cdots & {\bf 0}\\
{\bf 0} & B_{2} & \cdots & {\bf 0}\\
\vdots & \vdots & \vdots & \vdots\\
{\bf 0} & {\bf 0} & \cdots & B_{s}
\end{array}\right ],\end{align*}
where $B_{i}$ is an $r_{i}\times r_{i}$ diagonal matrix such that all the entries in the diagonal are the same value $\lambda_{i}$. The characteristic polynomial for $T_{W}$ is given by
\[P_{T_{W}}(\lambda )=(\lambda -\lambda_{1})^{r_{1}}\cdot (\lambda -\lambda_{2})^{r_{2}}\cdots (\lambda -\lambda_{s})^{r_{s}}.\]
From Theorem \ref{lat110}, it follows that $T$ is diagonolizable if and only if $r=n$, i.e., $r_{1}+\cdots +r_{s}=n$.

\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}

Minimal Polynomials.

The Cayley-Hamilton Theorem \ref{lat38} says that, for any linear operator $T$ on a finite-dimensional vector space $V$ over the scalar field ${\cal F}$, its characteristic polynomial $P_{T}$ satisfies that $P_{T}(T)$ is a zero mapping. It is possible that there exists another polynomial $p$ with $\deg p\leq\deg P_{T}$ such that $p(T)$ is also a zero mapping. Therefore, we propose the following definition.

\begin{equation}{\label{lad111}}\tag{35}\mbox{}\end{equation}

Definition \ref{lad111}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. We say that $p$ is a minimal polynomial for $T$ when the following conditions are satisfied.

$p$ is a monic polynomial (i.e., the leading coefficient is $1$) over the scalar field ${\cal F}$.
$p(T)$ is a zero mapping.
If $f$ is another polynomial such that $f(T)$ is a zero mapping, then $\deg f\geq\deg p$. $\sharp$

We can show that if there is another monic polynomial $\widehat{p}$ satisfies the above three conditions, then $\widehat{p}=p$. This shows that uniqueness of minimal polynomial, and says that Definition \ref{lad111} is well-defined. The minimal polynomial of a given matrix $A$ is defined to be the minimal polynomial of its corresponding linear operator $T_{A}$.

\begin{equation}{\label{lar113}}\tag{36}\mbox{}\end{equation}

Remark \ref{lar113}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. The Cayley-Hamilton Theorem \ref{lat38} says that the minimal polynomial $p$ divides the characteristic polynomial $P_{T}$ for $T$, since $p(T)$ and $P_{T}(T)$ are zero mappings and $\deg p\leq\deg P_{T}$.

\begin{equation}{\label{lap112}}\tag{37}\mbox{}\end{equation}

Proposition \ref{lap112}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. The characteristic and minimal polynomials for $T$ have the same roots except for multiplicities.

Proof. Let $p$ be a minimal polynomial for $T$. We shall show that $p(\lambda )=0$ if and only if $\lambda$ is an eigenvalue of $T$. Suppose that $p(\lambda )=0$. Then $p(x)=(x-\lambda )q(x)$, where $q(x)$ is a polynomial with $\deg q<\deg p$. The definition of minimal polynomial says that $q(T)$ cannot be a zero mapping. Therefore there exists $v_{0}\in V$ such that $(q(T))(v_{0})\neq\theta$. We write $v_{1}=(q(T))(v_{0})$. Then, we have
\begin{align*} \theta & =(p(T))(v_{0})\\ & =((T-\lambda I)q(T))(v_{0})\\ & =(T-\lambda I)(v_{1}),\end{align*}
which says that $\lambda$ is an eigenvalue of $T$. For the converse, suppose that $T(v)=\lambda v$ for some $v\neq\theta$ and $\lambda\in {\cal F}$. We have $(p(T))(v)=p(\lambda )v$. Since $p(T)$ is a zero mapping and $v\neq\theta$, we must have $p(c)=0$. This completes the proof. $\blacksquare$

Example. We consider the following matrix
\[A=\left [\begin{array}{rrr}
3 & -1 & 0\\ 0 & 2 & 0\\ 1 & -1 & 2
\end{array}\right ].\]
Its characteristic polynomial is given by
\begin{align*} P_{A}(\lambda ) & =\det (\lambda I_{3}-A)\\ & =\det\left [\begin{array}{rrr}
\lambda -3 & 1 & 0\\ 0 & \lambda -2 & 0\\ -1 & 1 & \lambda -2
\end{array}\right ]\\ & =(\lambda -2)^{2}\cdot (\lambda -3).\end{align*}
Using Proposition \ref{lap112}, the minimal polynomial must be either $(\lambda -2)\cdot (\lambda -3)$ or $(\lambda -2)^{2}\cdot (\lambda -3)$. Since $(A-2)(A-3)$ is a zero matrix, we conclude that the minimal polynomial is $p(\lambda )=(\lambda -2)\cdot (\lambda -3)$. $\sharp$

Example. Let $T$ be a linear operator on $\mathbb{R}^{2}$ defined by
\[T(x,y)=(2x+5y,6x+y).\]
Let $\mathfrak{B}$ be the standard basis for $\mathbb{R}^{2}$. Then, we have
\[A=[T]_{\mathfrak{B}}=\left [\begin{array}{cc}
2 & 5\\ 6 & 1\end{array}\right ].\]
Its characteristic polynomial is given by
\begin{align*} P_{T}(\lambda ) & =\det (\lambda I_{2}-A)\\ & =\det\left [\begin{array}{cc}
\lambda -2 & -5\\ -6 & \lambda -1\end{array}\right ]\\ & =(\lambda -7)\cdot (\lambda +4).\end{align*}
Using Proposition \ref{lap112}, the minimal polynomial must be $(\lambda -7)\cdot (\lambda +4)$. $\sharp$

\begin{equation}{\label{laex246}}\tag{38}\mbox{}\end{equation}

Example \ref{laex246}. Let $V$ be the vector space consisting of all polynomials with degree less than or equal to $2$ and coefficients in $\mathbb{R}$, and let $T$ be a linear operator on $V$ defined by $T(f(x))=f'(x)$. Let $\mathfrak{B}$ be the standard basis for $V$. Then, we have
\[[T]_{\mathfrak{B}}=\left [\begin{array}{rrr}
0 & 1 & 0\\ 0 & 0 & 2\\ 0 & 0 & 0
\end{array}\right ].\]
Its characteristic polynomial is given by $P_{T}(\lambda )=\lambda^{3}$. Therefore, the minimal polynomial may be $t$, $t^{2}$ or $t^{3}$. Since $T^{2}(x^{2})=2\neq 0$, it follows that the minimal polynomial of $T$ must be $t^{3}$. $\sharp$

Proposition. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Let $W$ be a $T$-invariant subspace of $V$. The minimal polynomial for the restriction operator $T_{W}$ divides the minimal polynomial for $T$.

Proof. Let $\dim (V)=n$ and $\dim (W)=r$. We consider the $k$th power of $A$ as given in (\ref{laeq114}), which can be obtained below
\[A^{k}=\left [\begin{array}{cc}
B^{k} & E_{k}\\ {\bf 0} & D^{k}
\end{array}\right ],\]
where $E_{k}$ is some $r\times (n-r)$ matrix depending on $k$. Therefore, if the polynomial $p$ satisfies $p(A)={\bf 0}$ the zero matrix, then $p(B)$ is also a zero matrix. This completes the proof. $\blacksquare$

Example. Let $T$ be a linear operator on $\mathbb{R}^{4}$ defined by
\[T\left (x_{1},x_{2},x_{3},x_{4}\right )=\left (x_{1}+x_{2}+2x_{3}-x_{4},x_{2}+x_{4},2x_{3}-x_{4},x_{3}+x_{4}\right ).\]
Let $W$ be a subspace of $\mathbb{R}^{4}$ defined by
\[W=\left\{\left (x_{1},x_{2},0,0\right ):x_{1},x_{2}\in\mathbb{R}\right\}.\]
Since
\[T\left (x_{1},x_{2},0,0\right )=\left (x_{1}+x_{2},x_{2},0,0\right )\in W,\]
it follows that $W$ is $T$-invariant. We also see that
\[\mathfrak{B}_{W}=\left\{{\bf e}_{1},{\bf e}_{2}\right\}\]
is an ordered basis for $W$. Let
\[\mathfrak{B}_{V}=\left\{{\bf e}_{1},{\bf e}_{2},{\bf e}_{3},{\bf e}_{4}\right\}\]
be an ordered basis for $V$. Then, we have
\[B=[T_{W}]_{\mathfrak{B}_{W}}=\left [\begin{array}{cc}
1 & 1\\ 0 & 1
\end{array}\right ]\]

and
\[A=[T]_{\mathfrak{B}_{V}}=\left [\begin{array}{rrrr}
1 & 1 & 2 & -1\\ 0 & 1 & 0 & 1\\ 0 & 0 & 2 & -1\\ 0 & 0 & 1 & 1
\end{array}\right ].\]
Finally, we can obtain
\begin{align*}
P_{T}(\lambda ) & =\det\left (A-\lambda I_{4}\right )\\
& =\det\left [\begin{array}{cccc}
1-\lambda & 1 & 2 & -1\\ 0 & 1-\lambda & 0 & 1\\ 0 & 0 & 2-\lambda & -1\\
0 & 0 & 1 & 1-\lambda
\end{array}\right ]\\
& =\det\left [\begin{array}{cc}
1-\lambda & 1\\ 0 & 1-\lambda
\end{array}\right ]\cdot\det\left [\begin{array}{cc}
2-\lambda & -1\\ 1 & 1-\lambda
\end{array}\right ]\\
& =P_{T_{W}}(\lambda )\cdot\det\left [\begin{array}{cc}
2-\lambda & -1\\ 1 & 1-\lambda
\end{array}\right ].
\end{align*}

\begin{equation}{\label{lat245}}\tag{39}\mbox{}\end{equation}

Theorem \ref{lat245}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$ with $\dim (V)=n$, and let $T$ be a linear operator on $V$. Then $T$ is diagonalizable if and only if the minimal polynomial of $T$ is of the form
\[p(t)=\left (t-\lambda_{1}\right )\left (t-\lambda_{2}\right )\cdots\left (t-\lambda_{r}\right ),\]
where $r\leq n$ and $\lambda_{1},\cdots ,\lambda_{r}$ are distinct eigenvalues of $T$.

Proof. Suppose that $T$ is diagonalizable. Then, there is a basis $\mathfrak{B}=\{v_{1},\cdots ,v_{n}\}$ for $V$ which consists of eigenvectors of $T$. Let $\lambda_{1},\cdots ,\lambda_{r}$ be the distinct eigenvalues of $T$, and define
\[p(t)=\left (t-\lambda_{1}\right )\left (t-\lambda_{2}\right )\cdots\left (t-\lambda_{r}\right ).\]
From Proposition \ref{lap112}, we see that $p$ and the minimal polynomial have also the same roots, which means that $p$ divides the minimal polynomial. In order to show that $p$ is the minimal polynomial, by definition, we remain to claim that $p(T)$ is a zero mapping. For any $v_{i}\in\mathfrak{B}$, we have $(T-\lambda_{j}I)(v_{i})=\theta$ for some $\lambda_{j}$. Since $p$ can be written as $p(t)=q_{j}(t)\cdot (t-\lambda_{j})$ for some polynomial $q_{j}$, we have
\begin{align*} (p(T))(v_{i}) & =(q_{j}(T))((T-\lambda_{j}I)(v_{i}))\\ & =(q_{j}(T))(\theta )=\theta .\end{align*}
This shows that $p(T)$ is indeed a zero mapping. Therefore, $p$ is a minimal polynomial of $T$.

Conversely, suppose that there exist distinct scalars $\lambda_{1},\cdots ,\lambda_{r}$ such that the minimal polynomial of $T$ is
\[p(t)=\left (t-\lambda_{1}\right )\left (t-\lambda_{2}\right )\cdots\left (t-\lambda_{r}\right ).\]
Proposition \ref{lap112} says that these $\lambda_{1},\cdots ,\lambda_{r}$ are the eigenvalues of $T$. We are going to prove it by mathematical induction on $n=\dim (V)$. It is obvious that $T$ is diagonalizable for $n=1$. We assume that $T$ is diagonalizable for $\dim V<n$ and $n>1$. Now, we consider $\dim V=n$ and $W=\mbox{Im}(T-\lambda_{r}I)$. Since $\lambda_{r}$ is an eigenvalue of $T$, it says $W\neq V$ (check it !). If $W=\{\theta\}$, then $T=\lambda_{r}I$, which says that $T$ is diagonalizable. Therefore, we assume $0<\dim (W)<n$. It is obvious that $W$ is $T$-invariant. Let $T_{W}$ be the linear operator that is the restriction of $T$ on $W$. Since $p$ is a minimal polynomial, by definition, we have
\begin{align*}
\theta & =(p(T))(v)\\ & =((T-\lambda_{1}I)(T-\lambda_{2}I)\cdots (T-\lambda_{r-1}I)(T-\lambda_{r}I))(v)\\
& =((T-\lambda_{1}I)(T-\lambda_{2}I)\cdots (T-\lambda_{r-1}I))(x),
\end{align*}
where $x=(T-\lambda_{r}I)(v)\in W$. This shows that, fro any $x\in W$, we have
\[((T-\lambda_{1}I)(T-\lambda_{2}I)\cdots (T-\lambda_{r-1}I))(x)=\theta ,\]
which says that the minimal polynomial of $T_{W}$ divides the following polynomial
\begin{equation}{\label{laeq258}}\tag{40}
\left (t-\lambda_{1}\right )\left (t-\lambda_{2}\right )\cdots\left (t-\lambda_{r-1}\right ).
\end{equation}
Therefore, the minimal polynomial of $T_{W}$ must be the form of (\ref{laeq258}). By the induction hypothesis, the linear operator $T_{W}$ must be diagonalizable. Proposition \ref{lap112} also says that $\lambda_{r}$ cannot be the eigenvalue of $T_{W}$. Therefore, we obtain
\begin{equation}{\label{laeq259}}\tag{41}
W\cap\mbox{Ker}(T-\lambda_{r}I)=\{\theta\}.
\end{equation}
Let
\[\mathfrak{B}_{1}=\left\{w_{1},\cdots ,w_{m}\right\}\]
be a basis for $W$ that consists of eigenvectors of $T_{W}$, which are also eigenvector of $T$. Let
\[\mathfrak{B}_{2}=\left\{v_{1},\cdots ,v_{p}\right\}\]
be a basis for $\mbox{Ker}(T-\lambda_{r}I)$. Then $\mathfrak{B}_{1}$ and $\mathfrak{B}_{2}$ are disjoint by (\ref{laeq259}). By considering $T-\lambda_{r}I$ and using Theorem \ref{lat71}, we also have $n=m+p$. Next, we shall claim that $\mathfrak{B}=\mathfrak{B}_{1}\cup\mathfrak{B}_{2}$ is linearly independent. For any scalars $\alpha_{1},\cdots ,\alpha_{m}$ and $\beta_{1},\cdots ,\beta_{r}$ satisfying
\[\alpha_{1}w_{1}+\cdots +\alpha_{m}w_{m}+\beta_{1}v_{1}+\cdots +\beta_{p}v_{p}=\theta .\]
Let
\[x=\alpha_{1}w_{1}+\cdots +\alpha_{m}w_{m}\]

and
\[y=\beta_{1}v_{1}+\cdots +\beta_{p}v_{p}.\]
Then $x\in W$ and $y\in \mbox{Ker}(T-\lambda_{r}I)$ satisfy $x+y=\theta$. It says that $x=-y\in W\cap\mbox{Ker}(T-\lambda_{r}I)$. Therefore, we have $x=\theta$, which implies $\alpha_{i}=0$ for $i=1,\cdots ,m$, since $\mathfrak{B}_{1}$ is linearly independent. Similarly, we can show that $y=\theta$, which also implies $\beta_{j}=0$ for $j=1,\cdots ,p$, since $\mathfrak{B}_{2}$ is linearly independent. Therefore, we conclude that $\mathfrak{B}$ is linearly independent subset of $V$ consisting of $n=\dim (V)$ eigenvectors of $T$. It means that $\mathfrak{B}$ is a basis for $V$ consisting of eigenvectors of $T$. Therefore, the linear operator $T$ is diagonalizable. This completes the proof. $\blacksquare$

Example. Continued from Example \ref{laex246}, since the minimal polynomial is $p(t)=t^{3}$, Theorem \ref{lat245} says that the differential operator $T$ is not diagonalizable. $\sharp$

Example. Let $V$ be the vector space consisting of all $2\times 2$ matrices over $\mathbb{R}$. We shall determine all matrices $A$ that are satisfying $A^{2}-3A+2I_{2}={\bf 0}$. Let
\[f(t)=t^{2}-3t+2=(t-1)(t-2).\]
Since $f(A)={\bf 0}$, the minimal polynomial $p$ of $A$ divides $f$. Therefore, the possible minimal polynomials of $A$ are $t-1$, $t-2$ or $(t-1)(t-2)$. If $p(t)=t-1$ or $p(t)=t-2$, then $A=I_{2}$ or $A=2I_{2}$, respectively. If $p(t)=(t-1)(t-2)$, then $A$ is diagonalizable by Theorem \ref{lat245} with eigenvalues $1$ and $2$. Therefore, the matrix $A$ must be similar to
\[D=\left [\begin{array}{cc}
1 & 0\\ 0 & 2
\end{array}\right ]\]
In other words, we have $A=PDP^{-1}$ for some invertible $2\times 2$ matrix. $\sharp$

Example. Let $A$ be an $n\times n$ matrix over $\mathbb{R}$ satisfying $A^{3}=A$. We are going to show that $A$ is diagonalizable. Let
\[f(t)=t^{3}-t=t(t+1)(t-1).\]
Then $f(A)={\bf 0}$. Therefore, the minimal polynomial $p$ divides $f$. Since $p$ and $f$ have the same roots, it follows that $p(t)=t(t+1)(t-1)$. This shows that $A$ is diagonalizable by Theorem \ref{lat245}. $\sharp$

\begin{equation}{\label{e}}\tag{E}\mbox{}\end{equation}

Triangulation.

Definition. Let $V$ be an $n$-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. A finite sequence $\{V_{1},\cdots ,V_{n}\}$ of subspaces of $V$ is called a fan of $T$ when $V_{i}\subseteq V_{i+1}$ for $i=1,\cdots ,n-1$, $\dim (V_{i})=i$ for $i=1,\cdots ,n$ and each $V_{i}$ is $T$-invariant for $i=1,\cdots ,n$. A finite sequence $\{v_{1},\cdots ,v_{n}\}$ in $V$ is called a fan basis of $T$ if and only if $\{v_{1},\cdots ,v_{i}\}$ is a basis of $V_{i}$ for $i=1,\cdots ,n$. $\sharp$

We see that the fan basis exists. For example, let $v_{1}$ be a basis of $V_{1}$. We can extend $v_{1}$ to a basis $\{v_{1},v_{2}\}$ of $V_{2}$, and also to a basis $\{v_{1},v_{2},v_{3}\}$ of $V_{3}$. Inductively, we can finally obtain a basis $\{v_{1},\cdots ,v_{i}\}$ of $V_{i}$.

Proposition. Let $\{v_{1},\cdots ,v_{n}\}$ be a fan basis of a linear mapping. Then, the matrix associated with $T$ with respect to this fan basis is an upper triangular matrix.

Proof. Since $T(V_{i})\subseteq V_{i}$ for each $i=1,\cdots ,n$, there exists $a_{ij}\in {\cal F}$ satisfying
\begin{align*}
T(v_{1}) & =a_{11}v_{1}\\
T(v_{2}) & =a_{21}v_{1}+a_{22}v_{2}\\
& \vdots\\
T(v_{i}) & =a_{1i}v_{1}+a_{2i}v_{2}+\cdots +a_{ii}v_{i}\\
& \vdots\\
T(v_{n}) & =a_{1n}v_{1}+a_{2n}v_{2}+\cdots +a_{nn}v_{i},
\end{align*}
which means that the matrix associated with $T$ with respect to this fan basis is the triangular matrix
\[\left [\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n}\\
0 & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \cdots & \vdots\\
0 & 0 & \cdots & a_{nn}
\end{array}\right ].\]
This completes the proof. $\blacksquare$

Definition. The triangulation of linear mappings and matrices is defined below.

Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. We say that $T$ is triangulable when there exists an ordered basis $\mathfrak{B}$ of $V$ such that the associated matrix $[T]_{\mathfrak{B}}$ of $T$ is a triangular matrix.
Let $A$ be an $n\times n$ matrix over the scalar field ${\cal F}$. We say that $A$ is triangulable when there exists a non-singular matrix $B$ in ${\cal F}$ such that $B^{-1}AB$ is a triangular matrix. $\sharp$

We see that if the matrix $A$ is triangulable, then its corresponding linear mapping $T_{A}:{\cal F}^{n}\rightarrow {\cal F}^{n}$ is triangulable.

Theorem. Let $V$ be a finite-dimensional vector space over the complex numbers with $\dim (V)\geq 1$. If $T:V\rightarrow V$ is a linear mapping, then there exists a fan of $T$ in $V$.

Corollary. Let $V$ be a finite-dimensional vector space over the complex numbers with $\dim (V)\geq 1$. If $T:V\rightarrow V$ is a linear mapping, then there exists a basis of $V$ such that the associated matrix of $T$ is a triangular matrix with respect to this basis.

Corollary. Let $A$ be a matrix of complex numbers. Then there exists a non-singular matrix $B$ such that $B^{-1}AB$ is a triangular matrix.

Theorem. Let $V$ be a finite-dimensional vector space over the complex numbers with $\dim (V)\geq 1$. Suppose that $V$ is endowed with a positive Hermitian product. Let $T:V\rightarrow V$ is a unitary mapping. Then there exists an orthonormal basis of $V$ consisting of eigenvectors of $T$.

Corollary. Let $A$ be a complex unitary matrix. Then there exists a unitary matrix $U$ such that $U^{-1}AU$ is a diagonal matrix.

Proposition. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$ such that the minimal polynomial for $T$ has the form
\[p(x)=\left (x-c_{1}\right )^{d_{1}}\cdot\left (x-c_{2}\right )^{d_{2}}\cdots\left (x-c_{r}\right )^{d_{r}},\]
where $c_{i}\in {\cal F}$ for $i=1,\cdots ,r$. Let $W$ be a $T$-invariant proper subspace of $V$ $(W\neq V)$. Then, there exists a vector $v\in V$ satisfying $v\not\in W$ and $(T-\lambda I)(v)\in W$ for some eigenvalue $\lambda$ of $T$.

Theorem. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Then $T$ is triangulable if and only if the minimal polynomial for $T$ has the form
\[p(x)=\left (x-c_{1}\right )^{d_{1}}\cdot\left (x-c_{2}\right )^{d_{2}}\cdots\left (x-c_{r}\right )^{d_{r}},\]
where $c_{i}\in {\cal F}$ for $i=1,\cdots ,r$.

\begin{equation}}{\label{lat124}}\tag{42}\mbox{}\end{equation}
Theorem \ref{lat124}. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Then $T$ is diagonalizable if and only if the minimal polynomial for $T$ has the form
\[p(x)=\left (x-c_{1}\right )\cdot\left (x-c_{2}\right )\cdots\left (x-c_{r}\right ),\]
where $c_{1},\cdots ,c_{r}$ are distinct elements in ${\cal F}$.

Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let $T$ be a linear operator on $V$. Suppose that the characteristic polynomial for $T$ can be factored as
\[P_{T}(\lambda )=\left (\lambda -\lambda_{1}\right )^{d_{1}}\cdot\left (\lambda -\lambda_{2}\right )^{d_{2}}\cdots
\left (\lambda -\lambda_{r}\right )^{d_{r}}.\]
We have two different methods for determining whether or not $T$ is diagonalizable.

(Method 1). For each $i$, we need to see whether we can find $d_{i}$ independent eigenvectors associated with the eigenvalue $\lambda_{i}$. If this is true, then these eigenvectors can form an ordered basis $\mathfrak{B}$ such that $[T]_{\mathfrak{B}}$ is a diagonal matrix and $\lambda_{i}$ will repeat $d_{i}$ times in the diagonal of $[T]_{\mathfrak{B}}$.
(Method 2). We need to check whether or not
\[(T-\lambda_{1}I)(T-\lambda_{2}I)\cdots (T-\lambda_{r}I)\]
is a zero mapping.

Definition. Let $V$ be a vector space over a scalar field ${\cal F}$, and let ${\cal T}$ be a family of all linear operators on $V$. A subspace $W$ of $V$ is called ${\cal T}$-invariant when $T(W)\subseteq W$ for all $T\in {\cal T}$. We say that $V$ is a simple ${\cal T}$-space when $V\neq\{\theta\}$ and the ${\cal T}$-subspaces of $V$ are only $V$ and $\{\theta\}$. $\sharp$

We see that the subspace $W$ of $V$ is ${\cal T}$-invariant if and only if $W$ is $T$-invariant for all $T\in {\cal T}$.

\begin{equation}{\label{lap40}}\tag{43}\mbox{}\end{equation}
Proposition \ref{lap40}. Let $T\in {\cal T}$ such that $ST=TS$ for all $S\in {\cal T}$. Then, the image and kernel of $T$ are ${\cal T}$-invariant subspaces.

Proof. Let $w$ be in the image of $T$. This says that there exists $v\in V$ satisfying $w=T(v)$. Then, we have
\begin{align*} S(w) & =S(T(v))=(ST)(v)\\ & =(TS)(v)=T(S(v)),\end{align*}
which says that $S(w)$ is also in the image of $T$. Therefore, the image of $T$ is ${\cal T}$-invariant. Let $u$ be in the kernel of $T$, i.e., $T(u)=\theta$. Then, we have
\begin{align*} (T(S(u)) & =(TS)(u)=(ST)(u)\\ & =S(T(u))=S(\theta )=\theta ,\end{align*}
which says that $S(u)$ is in the kernel of $T$. Therefore, the kernel of $T$ is ${\cal T}$-invariant. This completes the proof. $\blacksquare$

Theorem. Let $V$ be a vector space over a scalar field ${\cal F}$, and let ${\cal T}$ be a family of all linear operators on $V$. Assume that $V$ is a simple ${\cal T}$-space. If $T\in {\cal T}$ satisfying $ST=TS$ for all $S\in {\cal T}$, then either $T$ is invertible, or $T$ is a zero mapping.

Proof. If $T$ is not a zero mapping, Proposition \ref{lap40} says $\ker T=\{\theta\}$, which also means that the image of $T$ is all of $V$. Therefore, $T$ is invertible. This completes the proof. $\blacksquare$

Theorem. Let $V$ be a finite-dimensional vector space over $\mathbb{C}$, and let ${\cal T}$ be a family of all linear operators on $V$. Assume that $V$ is a simple ${\cal T}$-space. If $T\in {\cal T}$ satisfying $ST=TS$ for all $S\in {\cal T}$, then there exists $\lambda$ satisfying $T=\lambda I$. $\sharp$

Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let ${\cal T}$ be a family of linear operators on $V$. We may ask whether or not we can simultaneously triangulate or diagonalize the linear operators in ${\cal T}$. In other words, we try to find an ordered basis $\mathfrak{B}$ for $V$ such that the matrices $[T]_{\mathfrak{B}}$ are triangular (or diagonal) for all $T\in {\cal T}$. The family ${\cal T}$ is said to be a commuting family when $TU=UT$ for any $T,U\in {\cal T}$.

Proposition. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let ${\cal T}$ be a commuting family of triangulable linear operators on $V$. Let $W$ be a proper subspace of $V$ which is invariant under ${\cal T}$. Then, there exists a vector $v\in V$ such that $v\not\in W$ and the vector $T(v)$ is in the subspace spanned by $v$ and $W$ for each $T\in {\cal T}$.

Theorem. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let ${\cal T}$ be a commuting family of triangulable linear operators on $V$. Then there exists an ordered basis $\mathfrak{B}$ for $V$ such that the matrices $[T]_{\mathfrak{B}}$ are triangular for all $T\in {\cal T}$.

Corollary. Let ${\cal A}$ be a commuting family of $n\times n$ matrices over $\mathbb{C}$. Then, there exists a non-singular $n\times n$ matrix $P$ over $\mathbb{C}$ such that $P^{-1}AP$ is upper-triangular for every $A\in {\cal A}$.

Theorem. Let $V$ be a finite-dimensional vector space over the scalar field ${\cal F}$, and let ${\cal T}$ be a commuting family of diagonalizable linear operators on $V$. Then, there exists an ordered basis $\mathfrak{B}$ for $V$ such that the matrices $[T]_{\mathfrak{B}}$ are diagonal for all $T\in {\cal T}$.

Eigenspaces.

Diagonalization.

Characteristic Polynomials.

Minimal Polynomials.

Triangulation.

Hsien-Chung Wu

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Eigenspaces.

Diagonalization.

Characteristic Polynomials.

Minimal Polynomials.

Triangulation.

Hsien-Chung Wu

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