William Oliver (1823-1901) was a British painter.
Let \(V\) be a vector space over the scalar field \({\cal F}\), and let \(T\) be a linear operator on \(V\). We shall decompose the vector space into the sum of invariant subspaces of \(V\) under \(T\).
Definition. Let \(V\) be a vector space over the scalar field \({\cal F}\), and let \(W_{1},\cdots ,W_{r}\) be the subspaces of \(V\). We say that $latex W_{1},\cdots ,
W_{r}$ are independent if \(w_{1}+\cdots +w_{r}=\theta\) for \(w_{i}\in W_{i}\), \(i=1,\cdots ,r\), implies \(w_{i}=\theta\) for all \(i=1,\cdots ,r\). \(\sharp\)
Let \(V\) be a vector space over the scalar field \({\cal F}\), and let \(W_{1},\cdots ,W_{r}\) be the subspaces of \(V\). Then, we have that
\[W=W_{1}+\cdots +W_{r}\]
is a subspace spanned by \(\{W_{1},\cdots ,W_{r}\}\), since each vector \(w\in W\) can be expressed as a sum
\begin{equation}{\label{laeq117}}\tag{1}
w=w_{1}+\cdots +w_{r}
\end{equation}
for some \(w_{i}\in W_{i}\), \(i=1,\cdots ,r\). If \(W_{1},\cdots ,W_{r}\) are independent, then the expression in (\ref{laeq117}) is unique. Indeed, if there exist some other \(\widehat{w}_{i}\in W_{i}\) for \(i=1,\cdots ,r\) satisfying
\[w=\widehat{w}_{1}+\cdots +\widehat{w}_{r},\]
then we have
\[\left (w_{1}-\widehat{w}_{1}\right )+\left (w_{2}-\widehat{w}_{2}\right )+\cdots +\left (w_{r}-\widehat{w}_{r}\right )=\theta ,\]
which implies \(w_{i}=\widehat{w}_{i}\) for all \(i=1,\cdots ,r\). If \(r=2\), we see that \(W_{1}\) and \(W_{2}\) are independent if and only if \(W_{1}\cap W_{2}=\{\theta\}\). If \(r>2\), then, for each \(W_{j}\),
\[W_{j}\bigcap\left (\sum_{i\neq j}W_{i}\right )=\{\theta\}.\]
\begin{equation}{\label{lap118}}\tag{2}\mbox{}\end{equation}
Proposition \ref{lap118}. Let \(V\) be a finite-dimensional vector space over the scalar field \({\cal F}\). Let \(W_{1},\cdots ,W_{r}\) be the subspaces of \(V\) and let
\[W=W_{1}+\cdots +W_{r}.\]
Then, the following statements are equivalent.
(a) \(W_{1},\cdots ,W_{r}\) are independent.
(b) For each \(j\) with \(2\leq j\leq r\), we have
\[W_{j}\cap\left (W_{1}+\cdots +W_{j-1}\right )=\{\theta\}.\]
(c) If \(\mathfrak{B}_{i}\) is an ordered basis for \(W_{i}\), \(i=1,\cdots ,r\), then
\[\mathfrak{B}=\mathfrak{B}_{1}\cup\mathfrak{B}_{2}\cup\cdots\cup\mathfrak{B}_{r}\]
is an ordered basis for \(W\) and \(\dim (W)=\sum_{i=1}^{r}\dim (W)_{i}\). \(\sharp\)
\begin{equation}{\label{lat119}}\tag{3}\mbox{}\end{equation}
Lemma \ref{lat119}. Let \(V\) be a vector space over the scalar field \({\cal F}\), and let \(U\) and \(W\) be two subspaces of \(V\). If \(V=U+W\) and \(U\cap W=\{\theta\}\), then \(V=U\oplus W\).
Proof. It is obvious that, given any \(x\in V\), there exist \(u\in U\) and \(w\in W\) satisfying \(x=u+w\). We want to claim that there \(u\) and \(w\) are uniquely determined. Let \(u’\in U\) and \(w’\in W\) such that \(x=u’+w’\). We are going to show \(u=u’\) and \(w=w’\). Since \(x=u+w=u’+w’\), we have \(u-u’=w-w’\). This also says \(u-u’\in U\) and \(w-w’\in W\). Therefore, we obtain
\[u-u’=w-w’\in U\cap W=\{\theta\},\]
which shows \(u=u’\) and \(w=w’\). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{lap120}}\tag{4}\mbox{}\end{equation}
Lemma \ref{lap120}. Let \(V\) be a finite-dimensional vector space over the scalar field \({\cal F}\), and let \(U\) and \(W\) be two subspaces of \(V\). If \(V=U\oplus W\), then
\[\dim V=\dim U+\dim W.\]
Proof. By the definition of direct sum, each element \(x\in V\) has a unique expression \(x=u+w\), where \(u\in U\) and \(w\in W\). Suppose that \(\dim U=r\) and \(\dim W=s\). Let \(\{u_{1},\cdots ,u_{r}\}\) be a basis for \(U\), and let \(\{w_{1},\cdots ,w_{s}\}\) be a basis for \(W\). Therefore, there exists \(\alpha_{1},\cdots ,\alpha_{r},\alpha_{r+1}, \cdots ,\alpha_{r+s}\in {\cal F}\) satisfying
\[u=\alpha_{1}u_{1}+\cdots +\alpha_{r}u_{r}\]
and
\[w=\alpha_{r+1}w_{1}+\cdots +\alpha_{r+s}w_{s},\]
which says that \(x\) can be uniquely expressed as
\[x=u+w=\alpha_{1}u_{1}+\cdots +\alpha_{r}u_{r}+\alpha_{r+1}w_{1}+\cdots +\alpha_{r+s}w_{s}.\]
We conclude that \(\{u_{1},\cdots ,u_{r},w_{1},\cdots ,w_{s}\}\) forms a basis for \(V\), which also says
\[\dim V=r+s=\dim U+\dim W.\]
This completes the proof. \(\blacksquare\)
The above Lemmas \ref{lat119} and \ref{lap120} can refer to the page Vector Spaces. Lemmas \ref{lat119} and \ref{lap120} present the similar properties to Proposition \ref{lap118} for the direct sum of two subspaces of \(V\). Therefore, we can define the direct sum for more than two subspaces of \(V\) according to Proposition \ref{lap118}.
Definition. Let \(V\) be a vector space over the scalar field \({\cal F}\), and let \(W_{1},\cdots ,W_{r}c\) be the subspaces of \(V\). We say that \(W\) is a direct sum of \(W_{1},\cdots ,W_{r}\), denoted by
\[W=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{r}\]
when
\[W=W_{1}+W_{2}+\cdots +W_{r}\]
and \(W_{1},\cdots ,W_{r}\) are independent. \(\sharp\)
Example. Let \(V\) be an \(n\)-dimensional vector space over the scalar field \({\cal F}\), and let \(\{v_{1},\cdots ,v_{n}\}\) be any basis for \(V\). If \(W_{i}\) is a one-dimensional subspace spanned by \(v_{i}\), then we can show \(V=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{r}\). \(\sharp\)
Example. Let \(V\) be the space of all \(n\times n\) matrices over \(\mathbb{C}\). We consider two subspaces
\[W_{1}=\left\{A\in V:A^{\top}=A\right\}\]
and
\[W_{2}=\left\{A\in V:A^{\top}=-A\right\}.\]
Then, we can show that \(V=W_{1}\oplus W_{2}\). Moreover, if \(A\in V\), then the unique expression for \(A=A_{1}+A_{2}\), where \(A_{1}\in W_{1}\) and \(A_{2}\in W_{2}\), can be expressed as
\[A_{1}=\frac{1}{2}\left (A+A^{\top}\right )\]
and
\[A_{2}=\frac{1}{2}\left (A-A^{\top}\right ).\]
\begin{equation}{\label{lat110}}\tag{5}\mbox{}\end{equation}
Lemma \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{6}
P_{T}(\lambda )=(\lambda -\lambda_{1})^{d_{1}}\cdots (\lambda -\lambda_{r})^{d_{r}},
\end{equation}
where \(d_{i}=\dim (W)_{i}\) for \(i=1,\cdots ,r\).
(c) We have
\[\dim (V)=\dim (W_{1})+\dim (W_{2})+\cdots +\dim (W_{r}).\]
The above Lemma \ref{lat110} can refer to the page Eigenvalues and Eigenvectors.
\begin{equation}{\label{lap220}}\tag{7}\mbox{}\end{equation}
Proposition \ref{lap220}. 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 \(V\) is the direct sum of the eigenspaces of \(T\).
Proof. Let \(\lambda_{1},\cdots ,\lambda_{r}\) be the distinct eigenvalues of \(T\), and let \(W_{i}\) be the eigenspaces associated with the eigenvalues \(\lambda_{i}\). Then \(W_{1},\cdots ,W_{r}\) are independent by Proposition \ref{lap109}. In particular, if \(T\) is diagonalizable, then Lemma \ref{lat110} says \(V=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{r}\).
For the converse, suppose that \(V\) is the direct sum of the eigenspaces of \(T\). Let \(\mathfrak{B}_{i}\) be an ordered basis for the eigenspace \(W_{i}\) associated with the eigenvalue \(\lambda_{i}\). From Proposition \ref{lap118}, we see that
\[\mathfrak{B}=\mathfrak{B}_{1}\cup\mathfrak{B}_{2}\cup\cdots\cup\mathfrak{B}_{r}\]
is an ordered basis for \(V\), which consists of eigenvectors of \(T\). Therefore, we conclude that \(T\) is diagonalizable, and the proof is complete. \(\blacksquare\)
Example. Let \(T\) be a linear operator on \(\mathbb{R}^{4}\) defined by
\[T\left (x_{1},x_{2},x_{3},x_{4}\right )=\left (x_{1},x_{2},2x_{3},3x_{4}\right ).\]
It can be shown that \(T\) is diagonalizable with eigenvalues \(1,2,3\). The corresponding eigenspaces are given by
\begin{align*}
W_{1} & =\left\{\left (x_{1},x_{2},0,0\right ):x_{1},x_{2}\in\mathbb{R}\right\}\\
W_{2} & =\left\{\left (0,0,x_{3},0\right ):x_{3}\in\mathbb{R}\right\}\\
W_{3} & =\left\{\left (0,0,0,x_{4}\right ):x_{4}\in\mathbb{R}\right\}
\end{align*}
Proposition \ref{lap220} says
\[\mathbb{R}^{4}=W_{1}\oplus W_{2}\oplus W_{3}.\]
Proposition. Let \(T\) be a linear operator on a finite-dimensional vector space \(V\) over the scalar field \({\cal F}\) satisfying
\[V=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{r},\]
where each \(W_{i}\) is a \(T\)-invariant subspace of \(V\). Suppose that \(f_{i}\) is the characteristic polynomial of \(T_{W_{i}}\) for \(i=1,\cdots ,r\). Then, the product \(f_{1}\cdot f_{2}\cdots f_{r}\) is the characteristic polynomial of \(T\).
Proof. The proof will use the mathematical induction on \(r\). Suppose that \(r=2\). Let \(\mathfrak{B}_{1}\) and \(\mathfrak{B}_{2}\) be the ordered bases for \(W_{1}\) and \(W_{2}\), respectively. Let \(\mathfrak{B}=\mathfrak{B}_{1}\cup\mathfrak{B}_{2}\). since \(V=W_{1}\oplus W_{2}\), from Proposition \ref{lap118}, we see that \(\mathfrak{B}\) is an ordered basis for \(V\). Let \(A_{1}=[T_{W_{1}}]_{\mathfrak{B}_{1}}\) and \(A_{2}=[T_{W_{2}}]_{\mathfrak{B}_{2}}\). Since \(W_{1}\) and \(W_{2}\) are \(T\)-invariant subspaces, we can obtain
\[A=[T]_{\mathfrak{B}}=\left [\begin{array}{cc}
A_{1} & {\bf 0}\\ {\bf 0} & A_{2}
\end{array}\right ].\]
Since \(A\) is in the block form, we have
\begin{align*} P_{T}(\lambda ) & =\det\left (A-\lambda I\right )\\ & =\det\left (A_{1}-\lambda I\right )
\cdot\det\left (A_{2}-\lambda I\right )\\ & =f_{1}(\lambda )\cdot f_{2}(\lambda )\end{align*}
by referring to the arguments in the proof of Proposition\ref{lap224}. Now, we assume that the desired results is valid for \(r-1\) summands, where \(r-1\geq 2\). Suppose that
\[V=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{r}.\]
Let \(W=W_{1}+\cdots +W_{r-1}\). It is not hard to show that \(W\) is \(T\)-invariant and that \(V=W\oplus W_{r}\). Using the case iof \(r=2\), we have
\[P_{T}(\lambda )=P_{T_{W}}(\lambda )\cdot f_{k}(\lambda ),\]
where \(P_{T_{W}}\) is the characteristic polynomial of \(T_{W}\). By the induction hypothesis, we must have
\[P_{T_{W}}(\lambda )=f_{1}(\lambda )\cdot f_{2}(\lambda )\cdots f_{r-1}(\lambda ).\]
Therefore, we obtain
\[P_{T}(\lambda )=f_{1}(\lambda )\cdot f_{2}(\lambda )\cdots f_{r}(\lambda ).\]
This completes the proof. \(\blacksquare\)
Example. Let \(T\) be a linear operator on \(\mathbb{R}^{4}\) defined by
\[T\left (x_{1},x_{2},x_{3},x_{4}\right )=\left (2x_{1}-x_{2},x_{1}+x_{2},x_{3}-x_{4},x_{3}+x_{4}\right ).\]
Let
\[W_{1}=\left\{\left (x_{1},x_{2},0,0\right ):x_{1},x_{2}\in\mathbb{R}\right\}\]
and
\[W_{2}=\left\{\left (0,0,x_{3},x_{4}\right ):x_{3},x_{4}\in\mathbb{R}\right\}.\]
Then, it is not difficult to show that \(W_{1}\) and \(W_{2}\) are \(T\)-invariant subspaces, and that \(\mathbb{R}^{4}=W_{1}\oplus W_{2}\). We can also see that \(\mathfrak{B}_{1}=\{{\bf e}_{1},{\bf e}_{2}\}\) and \(\mathfrak{B}_{2}=\{{\bf e}_{3},{\bf e}_{4}\}\) are the standard ordered bases for \(W_{1}\) and \(W_{2}\), respectively. Therefore, \(\mathfrak{B}=\mathfrak{B}_{1}\cup\mathfrak{B}_{2}\) is an standard ordered basis for \(\mathbb{R}^{4}\). Now, we can obtain
\[A_{1}=[T_{W_{1}}]_{\mathfrak{B}_{1}}=\left [\begin{array}{cc}
2 & -1\\ 1 & 1
\end{array}\right ]\]
and
\[A_{2}=[T_{W_{2}}]_{\mathfrak{B}_{2}}
=\left [\begin{array}{cc}
1 & -1\\ 1 & 1
\end{array}\right ],\]
which imply
\begin{align*} A & =[T]_{\mathfrak{B}}\\ & =\left [\begin{array}{cc}
A_{1} & {\bf 0}\\ {\bf 0} & A_{2}
\end{array}\right ]\\ & =\left [\begin{array}{rrrr}
2 & -1 & 0 & 0\\ 1 & 1 & 0 & 0\\ 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 1
\end{array}\right ].\end{align*}
We can also have
\begin{align*} P_{T}(\lambda ) & =\det\left (A-\lambda I\right )\\ & =\det\left (A_{1}-\lambda I\right )
\cdot\det\left (A_{2}-\lambda I\right )\\ & =f_{1}(\lambda )\cdot f_{2}(\lambda ).\end{align*}
\begin{equation}{\label{lat39}}\tag{8}\mbox{}\end{equation}
Theorem \ref{lat39}. Let \(V\) be a vector space over the scalar field \({\cal F}\), and \(T\) be a linear operator on \(V\). Let \(f\) be a polynomial in \({\cal F}\). Suppose that \(f=f_{1}\cdot f_{2}\), where \(f_{1}\) and \(f_{2}\) are polynomials with \(\deg f_{1},\deg f_{2}\geq 1\), and the greatest common divisor equal to \(1\). We also assume that \(f(T)\) is a zero mapping. Let \(W_{1}\) be the kernel of \(f_{1}(T)\) and \(W_{2}\) be the kernel of \(f_{2}(T)\). Then \(V=W_{1}\oplus W_{2}\) in which \(W_{1}\) and \(W_{2}\) are \(T\)-invariant. \(\sharp\)
Theorem \ref{lat39} is still valid when \(f\) is expressed as a product of several factors. We provide the result over \(\mathbb{C}\).
\begin{equation}{\label{lat234}}\tag{9}\mbox{}\end{equation}
Theorem \ref{lat234}. Let \(V\) be a vector space over \(\mathbb{C}\), and let \(T\) be a linear operator on \(V\). Let \(f\) be a polynomial such that \(f(T)\) is a zero mapping and
\[f(t)=\left (t-\lambda_{1}\right )^{m_{1}}\cdots\left (t-\lambda_{r}\right )^{m_{r}}.\]
where \(\lambda_{1},\cdots ,\lambda_{r}\) are all distinct roots of \(f\). Let \(U_{i}\) be the kernel of \((T-\lambda_{i})^{m_{i}}\) for \(i=1,\cdots ,r\). Then
\[V=U_{1}\oplus U_{2}\oplus\cdots\oplus U_{r}.\]
Definition. Let \(V\) be a vector space over the scalar field \({\cal F}\). A projection of \(V\) is a linear operator \(E\) on \(V\) satisfying \(E^{2}=E\). \(\sharp\)
Suppose that \(E\) is a projection on \(V\). We have the following properties.
- Given any \(v\in V\), we write \(u=E(v)\). Then, we have
\begin{align*} E(u) & =E(E(v))=E^{2}(v)\\ & =E(v)=u.\end{align*}
On the other hand, if \(u=E(u)\), then \(u\) is in the range of \(E\). In other words, \(u\in\mbox{Im}(E)\) if and only if \(E(u)=u\). - Given any \(v\in V\), we have the unique expression
\begin{align*} v & =E(v)+[v-E(v)]\\ & \in\mbox{Im}(E)+\mbox{Ker}(E).\end{align*}
Moreover, we can show
\[V=\mbox{Im}(E)\oplus\mbox{Ker}(E).\]
Let \(R\) and \(N\) be two subspaces of \(V\) such that \(V=R\oplus N\). From the above properties, we can show that there is one and only one projection \(E\) on \(V\) satisfying \(\mbox{Im}(E)=R\) and \(\mbox{Ker}(E)=N\). In this case, the operator \(E\) is also called the projection on \(R\) along \(N\).
Let \(V\) be an \(n\)-dimensional vector space over the scalar field \({\cal F}\). Any projection \(E\) on \(V\) is diagonalizable. Suppose that \(\dim\mbox{Im}(E)=r\) with an ordered basis \(\{v_{1},\cdots ,v_{r}\}\) for \(\mbox{Im}(E)\), and \(\dim\mbox{Ker}(E)=n-r\) with an ordered basis \(\{v_{r+1},\cdots ,v_{n}\}\) for \(\mbox{Ker}(E)\). Then \(\mathfrak{B}=\{v_{1},\cdots ,v_{n}\}\) forms an ordered basis for \(V\) satisfying
\[[E]_{\mathfrak{B}}=\left [\begin{array}{cc}
I_{r\times r} & {\bf 0}\\ {\bf 0} & {\bf 0}
\end{array}\right ],\]
where \(I_{r\times r}\) is an \(r\times r\) identity matrix.
We shall see how the projections can be used to describe the direct-sum decomposition of the vector space \(V\). Suppose that
\[V=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{r}.\]
Any vector \(v\in V\) can be expressed as the following form
\[v=w_{1}+w_{2}+\cdots +w_{r},\]
where \(w_{i}\in W_{i}\) for \(i=1,\cdots ,r\). Now, we define a mapping \(E_{i}:V\rightarrow V\) by \(E_{i}(v)=w_{i}\). Then \(E_{i}\) is well-defined, and we have the following properties.
- The mapping \(E_{i}\) is a projection on \(V\) satisfying \(\mbox{Im}(E_{i})=W_{i}\).
- The null space \(\mbox{Ker}(E_{i})\) of \(E_{i}\) is the subspace
\[\mbox{Ker}(E_{i})=W_{1}+\cdots +W_{i-1}+W_{i+1}+\cdots +W_{n},\]
since \(E_{i}(v)=\theta\) means that \(w_{i}=\theta\). - For each \(v\in V\), we have
\[v=E_{1}(v)+E_{2}(v)+\cdots +E_{n}(v),\]
i.e., \(I=E_{1}+E_{2}+\cdots +E_{n}\). - For \(i\neq j\), we have that \(E_{i}\circ E_{j}\) is a zero mapping. Indeed, we have
\[\mbox{Im}(E_{i})=W_{i}\subseteq W_{1}+\cdots +W_{j-1}+W_{j+1}+\cdots +W_{n}=\mbox{Ker}(E_{j}),\]
which says that \(E_{i}(E_{j}(v))=\theta\) for all \(v\in V\).
The converse of the above properties is given below.
\begin{equation}{\label{lap121}}\tag{10}\mbox{}\end{equation}
Proposition \ref{lap121}. Let \(V\) be a vector space over the scalar field \({\cal F}\). Let \(W_{1},W_{2},\cdots ,W_{r}\) be the subspaces of \(V\) satisfying
\[V=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{r}.\]
Then, there exists \(r\) linear operators \(E_{1},E_{2},\cdots ,E_{r}\) on \(V\) such that the following properties are satisfied.
(a) Each \(E_{i}\) is a projection on \(V\).
(b) \(E_{i}\circ E_{j}\) is a zero mapping for \(i\neq j\).
(c) We have \(I=E_{1}+E_{2}+\cdots +E_{r}\).
(d) We have \(\mbox{Im}(E_{i})=W_{i}\).
Conversely, suppose that there are \(k\) linear operators \(E_{1},E_{2},\cdots ,E_{r}\) on \(V\) such that conditions (a), (b) and (c) are satisfied. Let \(\mbox{Im}(E_{i})=W_{i}\). Then, we have
\[V=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{r}.\]
Theorem. Let \(V\) be a vector space over the scalar field \({\cal F}\). Let \(W_{1},W_{2},\cdots ,W_{r}\) be the subspaces of \(V\) satisfying
\[V=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{r}.\]
Let \(T\) be a linear operator on \(V\), and let \(E_{1},E_{2},\cdots ,E_{r}\) be the projection on \(V\) as in Proposition \ref{lap121}. The, each subspace \(W_{i}\) is \(T\)-invariant if and only if \(T\circ E_{i}=E_{i}\circ T\) for all \(i=1,\cdots ,r\).
\begin{equation}{\label{lat122}}\tag{11}\mbox{}\end{equation}
Theorem \ref{lat122}. Let \(V\) be a finite-dimensional vector space over the scalar field \({\cal F}\), and let \(T\) be a linear operator on \(V\). If \(T\) is diagonalizable and if \(\lambda_{1},\lambda_{2},\cdots ,\lambda_{r}\) are the eigenvalues of \(T\), then there exists \(r\) linear operators \(E_{1},E_{2},\cdots ,E_{r}\) on \(V\) such that the following properties are satisfied.
(a) We have \(T=\lambda_{1}E_{1}+\lambda_{2}E_{2}+\cdots +\lambda_{r}E_{r}\).
(b) We have \(I=E_{1}+E_{2}+\cdots +E_{r}\).
(c)$E_{i}\circ E_{j}$ is a zero mapping for \(i\neq j\).
(d) Each \(E_{i}\) is a projection on \(V\).
(e) The range \(\mbox{Im}(E_{i})\) of \(E_{i}\) is the subspace of eigenvectors associated with eigenvalue \(\lambda_{i}\).
Conversely, suppose that there are \(k\) distinct values \(\lambda_{1},\lambda_{2},\cdots ,\lambda_{r}\) and \(k\) non-zero linear operators \(E_{1},E_{2},\cdots ,E_{r}\) on \(V\) such that conditions (a, (b) and (c) are satisfied. Then \(T\) is diagonalizable and \(\lambda_{1},\lambda_{2},\cdots ,\lambda_{r}\) are the eigenvalues of \(T\) such that conditions (d) and (e) are also satisfied. \(\sharp\)
From the decomposition property of \(T\) as shown in part (a) of Theorem\ref{lat122}, we have the following observations.
- We first compute \(T^{2}\) as follows:
\begin{align*} T^{2} & =\left (\sum_{i=1}^{r}\lambda_{i}E_{i}\right )\circ
\left (\sum_{i=1}^{r}\lambda_{i}E_{i}\right )\\ & =\sum_{i=1}^{r}\sum_{j=1}^{r}
\lambda_{i}\lambda_{j}E_{i}\circ E_{j}\\ & =\sum_{i=1}^{r}\lambda_{i}^{2}E_{i}^{2}
\\ & =\sum_{i=1}^{r}\lambda_{i}^{2}E_{i}.\end{align*}
Therefore, for any positive integer \(k\), we can similarly obtain
\[T^{k}=\sum_{i=1}^{r}\lambda_{i}^{r}E_{i}.\]
In general, if \(g\) is any polynomial over the scalar field \({\cal F}\), then we can show
\begin{equation}{\label{laeq123}}\tag{12}
g(T)=g(\lambda_{1})E_{1}+g(\lambda_{2})E_{2}+\cdots +g(\lambda_{r})E_{r}.
\end{equation} - We consider the Lagrange polynomial corresponding the scalars \(\lambda_{1},\cdots ,\lambda_{r}\) as follows:
\begin{equation}{\label{laeq127}}\tag{13}
p_{j}(x)=\prod_{i\neq j}\frac{x-\lambda_{i}}{\lambda_{j}-\lambda_{i}}.
\end{equation}
Then, we have \(p_{j}(\lambda_{i})=\delta_{ij}\). According to (\ref{laeq123}), we also have
\begin{align*} p_{j}(T) & =\sum_{i=1}^{r}p_{j}(\lambda_{i})E_{i}\\ & =\sum_{i=1}^{r}\delta_{ij}E_{i}=E_{j}.\end{align*}
\begin{equation}{\label{labt125}}\tag{14}\mbox{}\end{equation}
Theorem \ref{labt125}. 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
\begin{equation}{\label{laeq126}}\tag{15}
p(x)=\left (x-\lambda_{1}\right )\cdot\left (x-\lambda_{2}\right )\cdots\left (x-\lambda_{r}\right ),
\end{equation}
where \(\lambda_{1},\cdots ,\lambda_{r}\) are distinct scalars in \({\cal F}\).
Proof. Suppose that \(T\) is diagonalizable and \(g\) is any polynomial. Then, from (\ref{laeq123}), \(g(T)\) is a zero mapping if and only if \(g(\lambda_{i})=0\) for all \(i=1,\cdots ,r\). In particular, the minimal polynomial for \(T\) is (\ref{laeq126}). For the converse, suppose that \(T\) is a linear operator on \(V\) with minimal polynomial \(p\) given in (\ref{laeq126}). We consider the Lagrange polynomial \(p_{j}\) as shown in (\ref{laeq127}). Then, we have
\begin{equation}{\label{laeq128}}\tag{16}
1=p_{1}+\cdots +p_{r}\mbox{ and }x=\lambda_{1}p_{1}+\cdots +\lambda_{r}p_{r}.
\end{equation}
Let \(E_{j}=p_{j}(T)\). Then, from (\ref{laeq128}), it follows
\[I=E_{1}+\cdots +E_{r}\]
and
\[T=\lambda_{1}E_{1}+\lambda_{2}E_{2}+\cdots +\lambda_{r}E_{r}.\]
We also see that if \(i\neq j\), then \(p_{i}p_{j}\) is divisible by the minimal polynomial \(p\), since \(p_{i}p_{j}\) contains every \(x-\lambda_{k}\) as a factor. This shows that \(E_{i}\circ E_{j}\) must be a zero mapping for \(i\neq j\). Since \(\deg p_{i}<\deg p\) for all \(i=1,\cdots ,r\) and \(p\) is a minimal polynomial for \(T\), we must have \(p_{i}(T)=E_{i}\) is not a zero mapping for \(i=1,\cdots ,r\). Therefore, using Theorem \ref{lat122}, we conclude that \(T\) is diagonalizable, and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{lat129}}\tag{17}\mbox{}\end{equation}
Theorem \ref{lat129}. (Primary Decomposition Theorem). Let \(V\) be a finite-dimensional vector space over the scalar field \({\cal F}\), and let \(T\) be a linear operator on \(V\). Let \(p\) be the minimal polynomial for \(T\) given by
\[p(x)=(p_{1}(x))^{d_{1}}\cdot (p_{2}(x))^{d_{2}}\cdots (p_{r}(x))^{d_{r}},\]
where the \(p_{i}\) are the distinct irreducible monic polynomials over \({\cal F}\), and the \(d_{i}\) are the positive integers. Let \(W_{i}=\mbox{Ker}((p_{i}(T))^{d_{i}})\) for \(i=1,\cdots ,r\). Then, we have the following properties.
(i) We have \(V=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{r}\).
(ii) Each \(W_{i}\) is \(T\)-invariant.
(iii) If the operator \(T_{i}=T|_{W_{i}}\) is the restriction of operator \(T\) on \(W_{i}\), then \((p_{i}(x))^{d_{i}}\) is the minimal polynomial for \(T_{i}\). \(\sharp\)
In Theorem \ref{lat129}, suppose that the minimal polynomial \(p\) for \(T\) is given by
\[p(x)=(x-\lambda_{1})^{d_{1}}\cdot (x-\lambda_{2})^{d_{2}}\cdots(x-\lambda_{r})^{d_{r}},\]
where \(\lambda_{1},\cdots ,\lambda_{r}\) are distinct scalars in \({\cal F}\). Let \(W_{i}=\mbox{Ker}((T-\lambda_{i}I)^{d_{i}})\). Then \(V\) is the direct sum of the subspaces \(W_{1},\cdots ,W_{r}\).
Definition. Let \(V\) be a vector space over the scalar field \({\cal F}\), and let \(N\) be a linear operator on \(V\). We say that \(N\) is nilpotent when there exists a positive integer \(d\) such that \(N^{d}\) is a zero mapping.
\begin{equation}{\label{lat192}}\tag{18}\mbox{}\end{equation}
Theorem \ref{lat192}. 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 minimal polynomial for \(T\) has the following form
\[p(x)=(x-\lambda_{1})^{d_{1}}\cdot (x-\lambda_{2})^{d_{2}}\cdots (x-\lambda_{r})^{d_{r}},\]
where \(\lambda_{1},\cdots ,\lambda_{r}\) are distinct scalars in \({\cal F}\), and \(d_{1},\cdots ,d_{r}\) are positive integers. Then,,there exists a diagonalizable operator \(D\) on \(V\) and a nilpotent operator \(N\) on \(V\) satisfying
(a) \(T=D+N\);
(b) \(DN=ND\).
The diagonalizable operator \(D\) and the nilpotent operator \(N\) are uniquely determined by (a) and (b), and each of them is a polynomial in \(T\). \(\sharp\)
If \({\cal F}\) is taken as \(\mathbb{C}\), then the minimal polynomial for \(T\) can definitely be expressed as the following form
\[p(x)=(x-\lambda_{1})^{d_{1}}\cdot (x-\lambda_{2})^{d_{2}}\cdots (x-\lambda_{r})^{d_{r}},\]
where \(\lambda_{1},\cdots ,\lambda_{r}\) are distinct scalars in \(\mathbb{C}\). Therefore, Theorem \ref{lat192} is applicable for the case of \(\mathbb{C}\).
