William Clarke Wontner (1857-1930) was an English portrait painter.
Definition. Let \(V\) be a vector space over the scalar field \({\cal F}\) and let \(T\) be a linear operator on \(V\). Given \(\lambda\in {\cal F}\), a non-zero vector \(v\in V\) is called a generalized eigenvector of \(T\) corresponding to \(\lambda\) when there exists an integer \(r\geq 1\) satisfying \((T-\lambda I)^{r}(v)=\theta\). This non-zero vector \(v\) is also called \((T-\lambda I)\)-cyclic. The smallest positive integer \(r\) having this property is called a period of \(v\) relative to \(T-\lambda I\). If the non-zero vector \(v\) is also \((T-\lambda I)\)-cyclic with period \(r\), then the following ordered set
\[\left\{v,(T-\lambda I)(v),\cdots ,(T-\lambda I)^{r-1}(v)\right\}\]
is called a cycle of \(T\) corresponding to \(\lambda\) with length \(r\). \(\sharp\)
It is obvious that if \(r\) is a period of \(v\) relative to \(T-\lambda I\), then we have \((T-\lambda I)^{k}(v)\neq\theta\) for \(0\leq k<r\). Let \(\widehat{v}=(T-\lambda I)^{r-1}(v)\). We obtain
\begin{align*} \theta & =(T-\lambda I)^{r}(v)\\ & =(T-\lambda )\left ((T-\lambda I)^{r-1}(v)\right )\\ & =(T-\lambda )(\widehat{v}),\end{align*}
which says \(T(\widehat{v})=\lambda\widehat{v}\). It means that \((T-\lambda I)^{r-1}(v)\) is an eigenvector of \(T\) with the corresponding eigenvalue \(\lambda\).
\begin{equation}{\label{lap41}}\tag{1}\mbox{}\end{equation}
Proposition \ref{lap41}. Let \(V\) be a vector space over the scalar field \({\cal F}\) and let \(T\) be a linear operator on \(V\). If the non-zero vector \(v\) is \((T-\lambda I)\)-cyclic with period \(r\), then the vectors \(v,(T-\lambda I)(v),\cdots ,(T-\lambda I)^{r-1}(v)\) are linearly independent. \(\sharp\)
Proposition. Let \(V\) be a vector space over the scalar field \({\cal F}\) and let \(T\) be a linear operator on \(V\). Let \(\lambda\) be an eigenvalue of \(T\), and let
\[\mathfrak{B}_{i}=\left\{v_{i},(T-\lambda I)(v_{i}),\cdots ,(T-\lambda I)^{r_{i}-1}(v_{i})\right\}\]
be cycles of \(T\) corresponding to \(\lambda\) with length \(r_{i}\) for \(i=1,\cdots ,m\). If the eigenvectors \((T-\lambda I)^{r_{i}-1}(v_{i})\) of \(T\) are all distinct for \(i=1,\cdots ,m\), then the \(m\) cycles are disjoint. \(\sharp\)
\begin{equation}{\label{lap241}}\tag{2}\mbox{}\end{equation}
Proposition \ref{lap241}. Let \(V\) be a vector space over the scalar field \({\cal F}\) and let \(T\) be a linear operator on \(V\). Let \(\lambda\) be an eigenvalue of \(T\) and let
\[\mathfrak{B}_{i}=\left\{v_{i},(T-\lambda I)(v_{i}),\cdots ,(T-\lambda I)^{r_{i}-1}(v_{i})\right\}\]
be cycles of \(T\) corresponding to \(\lambda\) with length \(r_{i}\) for \(i=1,\cdots ,m\). If the eigenvectors \((T-\lambda I)^{r_{i}-1}(v_{i})\) of \(T\) for \(i=1,\cdots ,m\) are all distinct and linearly independent, then the \(m\) cycles are disjoint, and their union given below
\[\mathfrak{B}=\bigcup_{i=1}^{m}\mathfrak{B}_{i}\]
is linearly independent. \(\sharp\)
Let \(T\) be a linear operator on a finite-dimensional vector space \(V\) over the scalar field \({\cal F}\). Suppose that there exists an ordered basis \(\mathfrak{B}\) for \(T\) such that the matrix \([T]_{\mathfrak{T}}\) for \(T\) has the following block form
\[[T]_{\mathfrak{B}}=\left [\begin{array}{cccc}
A_{1} & {\bf 0} & \cdots & {\bf 0}\\
{\bf 0} & A_{2} & \cdots & {\bf 0}\\
\vdots & \vdots & \cdots & \vdots\\
{\bf 0} & {\bf 0} & \cdots & J_{m}
\end{array}\right ],\]
where each \(A_{i}\) has the following form
\begin{equation}{\label{laeq252}}\tag{3}
A_{i}=\left [\begin{array}{cccccc}
\lambda_{i} & 1 & 0 & \cdots & 0 & 0\\
0 & \lambda_{i} & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \cdots & \cdots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & \lambda_{i} & 1\\
0 & 0 & 0 & \cdots & 0 & \lambda_{i}
\end{array}\right ]
\end{equation}
for some eigenvalue \(\lambda_{i}\) for \(T\). Then, this matrix \([T]_{\mathfrak{T}}\) is called the Jordan normal form for \(T\), and \(\mathfrak{B}\) is called a Jordan basis for \(T\). Each block \(A_{i}\) is called a Jordan block corresponding to \(\lambda_{i}\).
Definition. Let \(V\) be a \(r\)-dimensional vector space over the scalar field \(\mathbb{F}\). We say that the vector space \(V\) is cyclic when there exist a non-zero vector \(v\in V\) and a \(\lambda\in\mathbb{F}\) such that \(v\) is \((T-\lambda I)\)-cyclic with period \(r\). \(\sharp\)
According to the above discussion, we see that if \(V\) is cyclic with the corresponding non-zero vector \(v\) and scalar \(\lambda\), then \(\lambda\) is an eigenvalue of \(T\) with the corresponding eigenvector \((T-\lambda I)^{r-1}(v)\). Proposition \ref{lap41} says that the following cycle of \(T\) corresponding to \(\lambda\) with length \(r\)
\[\mathfrak{B}=\left\{v,(T-\lambda I)(v),\cdots ,(T-\lambda I)^{r-1}(v)\right\}\]
is a basis of \(V\). Now, for each \(k=0,1,\cdots ,r-1\), we have
\begin{align*}
T((T-\lambda I)^{k}(v)) & =(T-\lambda I)(T-\lambda I)^{k}(v)+\lambda I(T-\lambda I)^{k}(v)\\
& =(T-\lambda I)^{k+1}(v)+\lambda (T-\lambda I)^{k}(v).
\end{align*}
Therefore, the associated matrix for \(T\) with respect to this basis \(\mathfrak{B}\) is given by
\[[T]_{\mathfrak{B}}=\left [\begin{array}{cccccc}
\lambda & 1 & 0 & \cdots & 0 & 0\\
0 & \lambda & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \cdots & \cdots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & \lambda & 1\\
0 & 0 & 0 & \cdots & 0 & \lambda
\end{array}\right ].\]
This matrix has \(\lambda\) on diagonal, \(1\) above the diagonal, and \(0\) everywhere else.
Suppose that the vector space \(V\) can be expressed as a direct sum of \(T\)-invariant subspaces as given by
\begin{equation}{\label{laeq232}}\tag{4}
V=V_{1}\oplus V_{2}\oplus\cdots\oplus V_{m}
\end{equation}
such that each \(V_{i}\) is cyclic with the corresponding eigenvalue \(\lambda\) for \(i=1,\cdots ,m\). If we select a cycle of \(T\) for each \(V_{i}\), then the union of these cycles forms a basis \(\mathfrak{B}\) for \(V\) such that the matrix \([T]_{\mathfrak{B}}\) for \(T\) has the following block form
\begin{equation}{\label{laeq233}}\tag{5}
[T]_{\mathfrak{B}}=\left [\begin{array}{cccc}
A_{1} & {\bf 0} & \cdots & {\bf 0}\\
{\bf 0} & A_{2} & \cdots & {\bf 0}\\
\vdots & \vdots & \cdots & \vdots\\
{\bf 0} & {\bf 0} & \cdots & A_{m}
\end{array}\right ],
\end{equation}
where each \(A_{i}\) has the following form
\[A_{i}=\left [\begin{array}{cccccc}
\lambda_{i} & 1 & 0 & \cdots & 0 & 0\\
0 & \lambda_{i} & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \cdots & \cdots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & \lambda_{i} & 1\\
0 & 0 & 0 & \cdots & 0 & \lambda_{i}
\end{array}\right ].\]
This shows that \(\mathfrak{B}\) is a Jordan basis for \(V\), and the matrix \([T]_{\mathfrak{B}}\) for \(T\) is a Jordan normal form.
Theorem. Let \(V\) be a finite-dimensional vector space over the scalar field \(\mathbb{F}\) with \(V\neq\{\theta\}\), and let \(T\) be a linear operator on \(V\). Then \(V\) can be expressed as a direct sum of \(T\)-invariant cyclic subspaces. In other words, there exists a Jordan basis for \(V\) such that the matrix for \(T\) is a Jordan normal form. \(\sharp\)
Theorem. (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 \(v\in V\).
\begin{equation}{\label{lat234}}\tag{6}\mbox{}\end{equation}
Lemma \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
\[V=U_{1}\oplus U_{2}\oplus\cdots\oplus U_{r}.\]
The above Lemma \ref{lat234} can refer to the page Decomposition of Vector Spaces.
\begin{equation}{\label{lap236}}\tag{7}\mbox{}\end{equation}
Proposition \ref{lap236}. 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 \(P_{T}\) is given by
\[P_{T}(\lambda )=(\lambda -\lambda_{1})^{d_{1}}\cdots (\lambda -\lambda_{r})^{d_{m}}.\]
Suppose that \(\mathfrak{B}\) is a basis for \(V\) such that it is a union of \(m\) cycles of \(T\) given by
\[\mathfrak{B}=\bigcup_{i=1}^{m}\mathfrak{B}_{i}.\]
Then, we have the following properties.
(i) For each cycle \(\mathfrak{B}_{i}\), the subspace \(W_{i}=\mbox{span}(\mathfrak{B}_{i})\) is \(T\)-invariant, and the matrix \([T_{W}]_{\mathfrak{B}_{i}}\) for the restriction \(T_{W}\) is a Jordan block.
(ii) The basis \(\mathfrak{B}\) is also a Jordan basis for \(V\).
Proof. By the Cayley-Hamilton Theorem, we see that \((P_{T}(T))(v)=\theta\) for all \(v\in V\). By referring to (\ref{laeq232}) and (\ref{laeq233}), the results follow immediately from Lemma \ref{lat234}. \(\blacksquare\)
Definition. Let \(V\) be a vector space over the scalar field \({\cal F}\), and let \(T\) be a linear operator on \(V\). Let \(\lambda\) be an eigenvalue of \(T\). The generalized eigenspace of \(T\) corresponding to \(\lambda\) is denoted by \(\bar{V}_{\lambda}\) and is defined by
\[\bar{V}_{\lambda}=\left\{v\in V:(T-\lambda I)^{r}(v)=\theta\mbox{ for some positive integer }r\right\}.\]
We can see that \(\bar{V}_{\lambda}\) is a subspace of \(V\), and consists of the zero vector and all generalized eigenvectors corresponding to \(\lambda\).
Proposition. Let \(V\) be a vector space over the scalar field \({\cal F}\) and let \(T\) be a linear operator on \(V\). Let \(\lambda\) be an eigenvalue of \(T\). Then, we have the following properties.
(i) The generalized eigenspace \(\bar{V}_{\lambda}\) is a \(T\)-invariant subspace of \(V\), and contains the eigenspace \(V_{\lambda}\) of \(T\).
(ii) For any \(\mu\in {\cal F}\) with \(\mu\neq\lambda\), the restriction of \(T-\mu I\) to \(\bar{V}_{\lambda}\) is one-to-one.
\begin{equation}{\label{lap237}}\tag{8}\mbox{}\end{equation}
Proposition \ref{lap237}. 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 \(P_{T}\) is given by
\[P_{T}(\lambda )=(\lambda -\lambda_{1})^{d_{1}}\cdots (\lambda -\lambda_{m})^{d_{m}}.\]
Then, we have \(\dim (\bar{V}_{\lambda_{i}})\leq d_{i}\) and
\[\bar{V}_{\lambda_{i}}=\mbox{Ker}\left ((T-\lambda_{i}I)^{d_{i}}\right )\]
for \(i=1,\cdots ,m\).
Proposition. 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 \(P_{T}\) is given by
\[P_{T}(\lambda )=(\lambda -\lambda_{1})^{d_{1}}\cdots (\lambda -\lambda_{m})^{d_{m}}.\]
Then, for every \(v\in V\), there exist \(v_{i}\in\bar{V}_{\lambda_{i}}\) for \(i=1,\cdots ,m\) satisfying
\[v=v_{1}+v_{2}+\cdots +v_{m}.\]
\begin{equation}{\label{lat229}}\tag{9}\mbox{}\end{equation}
Theorem \ref{lat229}. 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 \(P_{T}\) is given by
\[P_{T}(\lambda )=(\lambda -\lambda_{1})^{d_{1}}\cdots (\lambda -\lambda_{m})^{d_{m}}.\]
Let \(\mathfrak{B}_{i}\) be an ordered basis for the subspace \(\bar{V}_{\lambda_{i}}\) for \(i=1,\cdots ,m\). Then, we have the following properties.
(i) The sets \(\mathfrak{B}_{i}\) and \(\mathfrak{B}_{j}\) are disjoint for \(i\neq j\), i.e., \(\mathfrak{B}_{i}\cap\mathfrak{B}_{j}=\emptyset\) for \(i\neq j\).
(ii) The union
\[\mathfrak{B}=\mathfrak{B}_{1}\cup\mathfrak{B}_{2}\cup\cdots\cup\mathfrak{B}_{m}\]
is an ordered basis for \(V\).
(iii) For each \(i=1,\cdots ,m\), we have \(\dim (\bar{V}_{\lambda_{i}})=d_{i}\).
\begin{equation}{\label{lat230}}\tag{10}\mbox{}\end{equation}
Lemma \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. \(\blacksquare\)
The above Lemma \ref{lat230} can refer to the page Eigenvalues and Eigenvectors.
Theorem. 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 \(P_{T}\) is given by
\[P_{T}(\lambda )=(\lambda -\lambda_{1})^{d_{1}} \cdots (\lambda -\lambda_{m})^{d_{m}}.\]
Then \(T\) is diagonalizable if and only if \(V_{\lambda_{i}}=\bar{V}_{\lambda_{i}}\) for all \(i=1,\cdots ,m\).
Proof. Using part (iii) of Theorem \ref{lat229} and part (i) of Lemma \ref{lat230}, it follows that \(T\) is diagonalizable if and only if \(\dim (V_{\lambda_{i}})=\dim (\bar{V}_{\lambda_{i}})\) for all \(i=1,\cdots ,m\). Since \(V_{\lambda_{i}}\subseteq\bar{V}_{\lambda_{i}}\), we must have that \(\dim (V_{\lambda_{i}})=\dim (\bar{V}_{\lambda_{i}})\) if and only if \(V_{\lambda_{i}}=\bar{V}_{\lambda_{i}}\) for all \(i=1,\cdots ,m\). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{lat235}}\tag{11}\mbox{}\end{equation}
Theorem \ref{lat235}. Let \(V\) be a finite-dimensional vector space over the scalar field \({\cal F}\), and let \(T\) be a linear operator on \(V\) with an eigenvalue \(\lambda\). Then \(\bar{V}_{\lambda}\) has an ordered basis consisting of a union of disjoint cycles of \(T\) corresponding to \(\lambda\).
Theorem. 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 \(P_{T}\) is given by
\[P_{T}(\lambda )=(\lambda -\lambda_{1})^{d_{1}}\cdots (\lambda -\lambda_{m})^{d_{m}}.\]
Then \(T\) has a Jordan normal form.
Proof. Since \(\lambda_{1},\cdots ,\lambda_{m}\) are the distinct eigenvalues of \(T\), for each \(\lambda_{i}\), Theorem \ref{lat235} says that there is an ordered basis \(\mathfrak{B}_{i}\) consisting of a union of disjoint cycles of \(T\) corresponding to \(\lambda_{i}\). Let
\[\mathfrak{B}=\bigcup_{i=1}^{m}\mathfrak{B}_{i}.\]
Then part (ii) of Proposition \ref{lap236} says that \(\mathfrak{B}\) is a Jordan basis for \(V\). This completes the proof. \(\blacksquare\)
Example. We consider the following \(3\times 3\) matrix
\[A=\left [\begin{array}{rrr}
3 & 1 & -2\\ -1 & 0 & 5\\ -1 & -1 & 4
\end{array}\right ].\]
In order to find a Jordan normal form. we need to find a Jordan basis for the linear operator \(T_{A}\). The characteristic polynomial of \(A\) is given by
\begin{align*} P_{T_{A}}(\lambda ) & =P_{A}(\lambda )\\ & =\det\left (\lambda I-A\right )\\ & =(\lambda -3)(\lambda -2)^{2}.\end{align*}
Therefore, \(\lambda_{1}=3\) and \(\lambda_{2}=2\) are the eigenvalues of \(A\) with multiplicities \(1\) and \(2\), respectively. By part (iii) of Theorem \ref{lat229}, we have \(\dim(\bar{V}_{\lambda_{1}})=1\) and \(\dim (\bar{V}_{\lambda_{2}})=2\). Using Proposition \ref{lap237}, we also have
\[\bar{V}_{\lambda_{1}}=\mbox{Ker}\left (T-3I\right )\]
and
\[\bar{V}_{\lambda_{2}}=\mbox{Ker}\left ((T-2I)^{2}\right ).\]
Since \(V_{\lambda_{1}}=\mbox{Ker}\left (T-3I\right )\), we obtain \(\bar{V}_{\lambda_{1}}=V_{\lambda_{1}}\). It is not difficult to observe that \((-1,2,1)\) is an eigenvector of \(T\) corresponding to the eigenvalue \(\lambda_{1}=3\). Therefore,
\[\mathfrak{B}_{1}=\left\{\left [\begin{array}{r}
-1\\ 2\\ 1\end{array}\right ]\right\}.\]
is a basis for \(\bar{V}_{\lambda_{1}}\). Since \(\dim (\bar{V}_{\lambda_{2}})=2\) and \(\bar{V}_{\lambda_{2}}\) has a basis consisting of a union of cycles, this basis is either a union of two cycles with length \(1\) or a cycle with length \(2\). Since \((T-\lambda )^{r-1}(v)\) is an eigenvector for each cycle, if this basis is a union of two cycles with length \(1\), then the two vectors in the basis would be eigenvectors of \(T\), which contradicts the fact of
\begin{align*} \dim (V_{\lambda_{2}}) & =3-\mbox{rank}(A-2I)\\ & =3-\mbox{rank}
\left [\begin{array}{rrr}
1 & 1 & -2\\ -1 & -2 & 5\\ -1 & -1 & 2
\end{array}\right ]\\ & =3-2=1.\end{align*}
Therefore, the basis for \(\bar{V}_{\lambda_{2}}\) must be a single cycle with length \(2\). Therefore, we need to find a non-zero vector \(v\) such that \((A-2I)v\neq {\bf 0}\) and \((A-2I)^{2}v={\bf 0}\). We can show that the basis for the solution space of the homogeneous system \((A-2I)^{2}X={\bf 0}\) is given by
\[\left\{\left [\begin{array}{r}
1\\ -3\\ -1\end{array}\right ],\left [\begin{array}{r}
-1\\ 2\\ 0\end{array}\right ]\right\}.\]
Now, we take \(v=(-1,2,0)\). Then \((A-2I)v=(1,-3,-1)\neq {\bf 0}\). Therefore, we obtain the cycle
\[\mathfrak{B}_{2}=\left\{v,(A-2I)v\right\}=\left\{\left [\begin{array}{r}
1\\ -3\\ -1\end{array}\right ],\left [\begin{array}{r}
-1\\ 2\\ 0\end{array}\right ]\right\}\]
as a basis for \(\bar{V}_{\lambda_{2}}\). Therefore, the union of \(\mathfrak{B}_{1}\) and \(\mathfrak{B}_{1}\) given by
\begin{align*} \mathfrak{B} & =\mathfrak{B}_{1}\cup\mathfrak{B}_{2}
\\ & =\left\{\left [\begin{array}{r}
-1\\ 2\\ 1\end{array}\right ],\left [\begin{array}{r}
1\\ -3\\ -1\end{array}\right ],\left [\begin{array}{r}
-1\\ 2\\ 0\end{array}\right ]\right\}\end{align*}
is a Jordan basis for \(A\). The Jordan normal form for \(A\) is given by
\[J=[T_{A}]_{\mathfrak{B}}=\left [\begin{array}{r|rr}
3 & 0 & 0\\ \hline
0 & 2 & 1\\ 0 & 0 & 0
\end{array}\right ].\]
If we take the matrix \(P\) whose columns are the vectors in \(\mathbb{B}\), then we have \(J=P^{-1}AP\). \(\sharp\)
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 \(T(f(x))=-f(x)-f'(x)\). Let \(\mathfrak{B}=\{1,x,x^{2}\}\) be the standard ordered basis for \(V\). Then, we have
\[[T]_{\mathfrak{B}}=A=\left [\begin{array}{rrr}
-1 & -1 & 0\\ 0 & -1 & -2\\ 0 & 0 & -1
\end{array}\right ].\]
The characteristic polynomial \(P_{T}\) for \(T\) (or \(P_{A}\) for \(A\)) is given by
\[P_{T}(\lambda )=\det\left (\lambda I-A\right )=(\lambda +1)^{3}.\]
Therefore, \(\lambda =-1\) is the only eigenvalues of \(T\). By Theorem \ref{lat229}, We have \(\bar{V}_{\lambda}=V\), i.e., \(\mathfrak{B}\) is also a basis for
$\bar{V}_{\lambda}$. Now
\begin{align*} \dim\left (V_{\lambda}\right ) & =3-\mbox{rank}(A+I)\\ & =3-\mbox{rank}
\left [\begin{array}{rrr}
0 & -1 & 0\\ 0 & 0 & -2\\ 0 & 0 & 0
\end{array}\right ]\\ & =3-2=1.\end{align*}
Since \((T-\lambda )^{r-1}(v)\) is an eigenvector for each cycle, the basis for \(\bar{V}_{\lambda}\) cannot be a union of two or three cycles, which also says that there do not exist two or three linearly independent eigenvectors. This says that the desired basis for \(\bar{V}_{\lambda}\) is a single cycle with
length \(3\). If
\[\mathfrak{B}=\left\{v,(T+I)(v),(T+I)^{2}(v)\right\}\]
is a cycle with length \(3\), then it determines a single Jordan block given by
\[[T]_{\mathfrak{B}}=A=\left [\begin{array}{rrr}
-1 & 1 & 0\\ 0 & -1 & 1\\ 0 & 0 & -1
\end{array}\right ],\]
which is also a Jordan normal form of \(T\). Since \((T+I)^{2}(v)\neq\theta\) in the cycle \(\mathfrak{B}\), for any basis for \(\bar{V}_{\lambda}\), there must be a vector satisfies the condition \((T+I)^{2}(v)\neq\theta\). By testing the vector \(x^{2}\) in the standard basis \(\{1,x,x^{2}\}\), we see that \((T+I)^{2}(x^{2})\neq\theta\). Therefore, the Jordan basis for \(T\) is given by
\[\left\{x^{2},(T+I)\left (x^{2}\right ),(T+I)^{2}\left (x^{2}\right )\right\}=\left\{x^{2},-2x,2\right\}.\]
Theorem. 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 \(P_{T}\) is given by
\[P_{T}(\lambda )=(\lambda -\lambda_{1})^{d_{1}}\cdots (\lambda -\lambda_{m})^{d_{m}}.\]
Then \(V\) is the direct sum of the generalized eigenspaces of \(T\). \(\sharp\)
According to the above results, each generalized eigenspace \(\bar{V}_{\lambda_{i}}\) contains an ordered basis \(\mathfrak{B}_{i}\) which consists of a union of disjoint cycles of \(T\) corresponding to \(\lambda_{i}\). Moreover, the union \(\mathfrak{B}=\bigcup_{i=1}^{m}\mathfrak{B}_{i}\) is a Jordan basis for \(T\). Let \(T_{i}\) be the restriction of \(T\) to \(\bar{V}_{\lambda_{i}}\). Then \(J_{i}=[T_{i}]_{\mathfrak{B}_{i}}\) is the Jordan normal form of \(T_{i}\). We also have
\[J=[T]_{\mathfrak{B}}=\left [\begin{array}{cccc}
J_{1} & {\bf 0} & \cdots & {\bf 0}\\
{\bf 0} & J_{2} & \cdots & {\bf 0}\\
\vdots & \vdots & \cdots & \vdots\\
{\bf 0} & {\bf 0} & \cdots & J_{m}
\end{array}\right ].\]
In order to efficiently find the Jordan normal form, we adopt the following convention. The basis \(\mathfrak{B}_{i}\) for \(\bar{V}_{\lambda_{i}}\) will be ordered such that the cycles appear in order of decreasing length. In other words, if \(\mathfrak{B}_{i}\) is a disjoint union of cycles \(\mathfrak{C}_{1},\mathfrak{C}_{2},\cdots ,\mathfrak{C}_{d_{i}}\) with the corresponding lengths \(r_{1},r_{2},\cdots ,r_{d_{i}}\), then we arrange the cycles satisfying \(r_{1}\geq r_{2}\geq\cdots\geq r_{d_{i}}\).
\begin{equation}{\label{laex239}}\tag{12}\mbox{}\end{equation}
Example \ref{laex239}. If the ordered basis \(\mathfrak{B}_{i}\) for \(\bar{V}_{\lambda_{i}}\) is the union of four cycles \(\mathfrak{C}_{1},\mathfrak{C}_{2},\mathfrak{C}_{3},\mathfrak{C}_{4}\) with respective lengths \(r_{1}=3\), \(r_{2}=3\), \(r_{3}=2\) and \(r_{4}=1\). Then, we have
\[J_{i}=[T_{i}]_{\mathfrak{B}_{i}}=\left [\begin{array}{ccccccccc}
\lambda_{i} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & \lambda_{i} & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & \lambda_{i} & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & \lambda_{i} & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & \lambda_{i} & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & \lambda_{i} & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & \lambda_{i} & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \lambda_{i} & 0\\ \
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \lambda_{i}
\end{array}\right ].\]
We shall introduce the {\bf dot diagram} of \(T_{i}\) to visualize the Jordan normal form \(J_{i}\). The dot diagram of \(T_{i}\) is shown below
\[\begin{array}{llll}
\bullet (T-\lambda_{i}I)^{r_{1}-1}(v_{1}) & \bullet (T-\lambda_{i}I)^{r_{2}-1}(v_{2})
& \cdots & \bullet (T-\lambda_{i}I)^{r_{d_{i}}-1}(v_{d_{i}})\\
\bullet (T-\lambda_{i}I)^{r_{1}-2}(v_{1}) & \bullet (T-\lambda_{i}I)^{r_{2}-2}(v_{2})
& \cdots & \bullet (T-\lambda_{i}I)^{r_{d_{i}}-2}(v_{d_{i}})\\
\vdots & \vdots & \cdots & \vdots\\
\vdots & \vdots & \cdots & \bullet (T-\lambda_{i}I)(v_{d_{i}})\\
\vdots & \vdots & \cdots & \bullet v_{d_{i}}\\
\vdots & \bullet (T-\lambda_{i}I)(v_{2}) &&\\
\vdots & \bullet v_{2} &&\\
\bullet (T-\lambda_{i}I)(v_{1}) &&&\\
\bullet v_{1} &&&
\end{array}\]
The dot diagram of \(T_{i}\) consists of \(d_{i}\) columns, where one column for each cycle. The \(j\)th column consists of \(r_{j}\) dots that correspond to the vectors in the cycle \(\mathfrak{C}_{j}\).
Example. Continued from Example \ref{laex239}, we have \(n_{i}=4\). The dot diagram of \(T_{i}\) is given by
\[\begin{array}{llll}
\bullet & \bullet & \bullet & \bullet\\
\bullet & \bullet & \bullet &\\
\bullet & \bullet &&
\end{array}\]
In the sequel, we shall devise a method to compute the dot diagram of \(T_{i}\). The basis \(\mathfrak{B}_{i}\) for \(T_{i}\) is fixed such that \(\mathfrak{B}_{i}\) is a disjoint union of cycles \(\mathfrak{C}_{1},\mathfrak{C}_{2},\cdots ,\mathfrak{C}_{d_{i}}\) with the corresponding lengths \(r_{1},r_{2},\cdots ,r_{d_{i}}\) satisfying \(r_{1}\geq r_{2}\geq\cdots\geq r_{d_{i}}\).
Proposition. For any positive integer \(r\), the vectors corresponding to the dots in the first \(r\) rows of the dot diagram of \(T_{i}\) constitute a basis for the null space \(\mbox{Ker}((T-\lambda_{i}I)^{r})\). In other words, the number of dots in the first \(r\) rows of the dot diagram of \(T_{i}\) is equal to the nullity of \((T-\lambda_{i}I)^{r}\). \(\sharp\)
If \(r=1\) in the above proposition, we have the following interesting result.
Corollary. The dimension of \(V_{\lambda_{i}}\) is \(n_{i}\). In other words, in a Jordan normal form of \(T_{i}\), the number of Jordan blocks corresponding to \(\lambda_{i}\) is equal to \(\dim (V_{\lambda_{i}})\). \(\sharp\)
\begin{equation}{\label{lap240}}\tag{13}\mbox{}\end{equation}
Proposition \ref{lap240}. Let \(r_{j}\) denote the number of dots in the \(j\)th rows of the dot diagram of \(T_{i}\). Then, we have the following formulas.
\[r_{1}=\dim (V)-\mbox{rank}(T-\lambda_{i}I)\]
and
\[r_{j}=\mbox{rank}((T-\lambda_{i}I)^{j-1})-\mbox{rank}((T-\lambda_{i}I)^{j})\mbox{ for }j>1.\]
We are going to use the above results to find a Jordan normal forms.
\begin{equation}{\label{laex242}}\tag{14}\mbox{}\end{equation}
Example \ref{laex242}. For the matrix given below
\[A=\left [\begin{array}{rrrr}
2 & -1 & 0 & 1\\ 0 & 3 & -1 & 0\\ 0 & 1 & 1 & 0\\ 0 & -1 & 0 & 3
\end{array}\right ],\]
we shall find the Jordan normal form of \(A\) and a Jordan basis for the linear operator \(T_{A}\). The characteristic polynomial of \(A\) is
\[\mbox{det}\left (A-\lambda I_{4}\right )=(\lambda -2)^{3}(\lambda -3),\]
which says that the matrix \(A\) has two distinct eigenvalues \(\lambda_{1}=2\) and \(\lambda_{2}=3\) with multiplicities \(3\) and \(1\), respectively. Let \(T_{1}\) and \(T_{2}\) be the restriction of \(T_{A}\) to the generalized eigenspaces \(\bar{V}_{\lambda_{1}}\) and \(\bar{V}_{\lambda_{2}}\), respectively. Suppose that \(\mathfrak{B}_{1}\) is a Jordan basis for \(T_{1}\). Since \(\lambda_{1}\) has multiplicity \(3\), it follows that \(\dim (\bar{V}_{\lambda_{1}})=3\) by part (iii) of Theorem \ref{lat229}. This also says that the dot diagram of \(T_{1}\) has \(3\) dots. According to Proposition \ref{lap240}, we have
\begin{align*} r_{1} & =4-\mbox{rank}(A-2I_{4})\\ & =4-\mbox{rank}\left (\left [\begin{array}{rrrr}
0 & -1 & 0 & 1\\ 0 & 1 & -1 & 0\\ 0 & 1 & -1 & 0\\ 0 & -1 & 0 & 1
\end{array}\right ]\right )\\ & =4-2=2\end{align*}
and
\begin{align*} r_{2} & =\mbox{rank}(A-2I_{4})-\mbox{rank}((A-2I_{4})^{2})\\ & =2-1=1.\end{align*}
As a matter of fact, \(r_{2}\) can be easily realized, since the dot diagram contains \(3\) dots, i.e., \(r_{2}=3-r_{1}\). Therefore, the dot diagram of \(T_{1}\) is given by
\[\begin{array}{cc}
\bullet & \bullet\\ \bullet &
\end{array}\]
which says
\[J_{1}=[T_{1}]_{\mathfrak{B}_{1}}=\left [\begin{array}{ccc}
2 & 1 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2
\end{array}\right ].\]
On the other hand, since \(\lambda_{2}=3\) with multiplicity \(1\), it follows \(\dim (\bar{V}_{\lambda_{2}}=1\). Therefore, any basis \(\mathfrak{B}_{2}\) for \(\bar{V}_{\lambda_{2}}\) consists of a single eigenvector corresponding to \(\lambda_{2}=3\), which also says
\[J_{2}=[T_{2}]_{\mathfrak{B}_{2}}=[3].\]
Taking \(\mathfrak{B}=\mathfrak{B}_{1}\cup\mathfrak{B}_{2}\), we obtain
\[J=[T_{A}]_{\mathfrak{B}}=\left [\begin{array}{cc}
J_{1} & {\bf 0}\\ {\bf 0} & J_{2}\end{array}\right ]
=\left [\begin{array}{cccc}
2 & 1 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 3
\end{array}\right ],\]
which is the Jordan normal form of \(A\). Now, we want to find a Jordan basis for \(T_{A}\). We first determine the Jordan basis for \(T_{1}\). It can be realized that the dot diagram of \(T_{1}\) is given by
\[\begin{array}{ll}
\bullet (T_{A}-2I)(v_{1}) & \bullet v_{2}\\ \bullet v_{1} &
\end{array},\]
which also says \(v_{1}\in\mbox{Ker}((T_{A}-2I)^{2})\) and \(v_{1}\not\in\mbox{Ker}(T_{A}-2I)\). Since
\[A-2I_{4}=\left [\begin{array}{rrrr}
0 & -1 & 0 & 1\\ 0 & 1 & -1 & 0\\ 0 & 1 & -1 & 0\\ 0 & -1 & 0 & 1
\end{array}\right ]\]
and
\[(A-2I_{4})^{2}=\left [\begin{array}{rrrr}
0 & -2 & 1 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & -2 & 1 & 1
\end{array}\right ],\]
we see that
\[\left\{\left [\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right ],
\left [\begin{array}{c}0\\ 1\\ 2\\ 0\end{array}\right ],
\left [\begin{array}{c}0\\ 1\\ 0\\ 2\end{array}\right ]\right\}\]
is a basis for \(\mbox{Ker}((T_{A}-2I)^{2})=\bar{V}_{\lambda_{2}}\). We can see that the last two vectors do not in \(\mbox{Ker}(T_{A}-2I)\). Therefore, we choose one of these as \(v_{1}\). Now, we pick
\[v_{1}=\left [\begin{array}{c}0\\ 1\\ 0\\ 2\end{array}\right ].\]
Then, we also obtain
\begin{align*} (T_{A}-2I)(v_{1}) & =(A-2I_{4}(v_{1})\\ & =\left [\begin{array}{rrrr}
0 & -1 & 0 & 1\\ 0 & 1 & -1 & 0\\ 0 & 1 & -1 & 0\\ 0 & -1 & 0 & 1
\end{array}\right ]\end{align*}
and
\[(A-2I_{4})^{2}=\left [\begin{array}{rrrr}
0 & -2 & 1 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & -2 & 1 & 1
\end{array}\right ]\left [\begin{array}{c}0\\ 1\\ 0\\ 2\end{array}\right ]
=\left [\begin{array}{c}-1\\ -1\\ -1\\ -1\end{array}\right ].\]
Now, we can choose \(v_{2}\) to be a vector in the eigenspace \(V_{\lambda_{1}}\) that is linearly independent of \((T_{A}-2I)(v_{1})\). For example, we pick
\[v_{2}=\left [\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right ].\]
Therefore, we can form a Jordan basis
\[\mathfrak{B}_{1}=\left\{\left [\begin{array}{c}-1\\ -1\\ -1\\ -1\end{array}\right ],
\left [\begin{array}{c}0\\ 1\\ 0\\ 2\end{array}\right ],
\left [\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right ]\right\}.\]
The linear independence of \(\mathfrak{B}_{1}\) can be guaranteed by Proposition \ref{lap241}, since \(v_{2}\) was chosen to be linearly independent of \((T_{A}-2I)(v_{1})\). Since \(\lambda_{2}=3\) has multiplicity \(1\), we have
\[\dim (\bar{V}_{\lambda_{2}})=\dim (V_{\lambda_{2}})=1.\]
This says that any eigenvector of \(T_{A}\) corresponding to \(\lambda_{2}=3\) constitutes an appropriate basis \(\mathfrak{B}_{2}\) for \(T_{2}\). For example, we can take
\[\mathfrak{B}_{2}=\left\{\left [\begin{array}{c}1\\ 0\\ 0\\ 1
\end{array}\right ]\right\}.\]
Therefore, we obtain
\begin{align*} \mathfrak{B} & =\mathfrak{B}_{1}\cup\mathfrak{B}_{2}
\\ & =\left\{\left [\begin{array}{c}-1\\ -1\\ -1\\ -1\end{array}\right ],
\left [\begin{array}{c}0\\ 1\\ 0\\ 2\end{array}\right ],
\left [\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right ],
\left [\begin{array}{c}1\\ 0\\ 0\\ 1
\end{array}\right ]\right\}.\end{align*}
is a Jordan basis for \(T_{A}\). We also note that if
\[P=\left [\begin{array}{rrrr}
-1 & 0 & 1 & 1\\ -1 & 1 & 0 & 0\\ -1 & 2 & 0 & 0\\ -1 & 0 & 0 & 1
\end{array}\right ],\]
where the column vectors consist of vectors in the Jordan basis \(\mathfrak{B}\), then \(J=P^{-1}AP\). \(\sharp\)
Example. For the matrix given below
\[A=\left [\begin{array}{rrrr}
2 & -4 & 2 & 2\\ -2 & 0 & 1 & 3\\ -2 & -2 & 3 & 3\\ -2 & -6 & 3 & 7
\end{array}\right ],\]
we shall find the Jordan normal form of \(A\) and a Jordan basis for the linear operator \(T_{A}\). The characteristic polynomial of \(A\) is
\[\mbox{det}\left (\lambda I_{4}-A\right )=(\lambda -2)^{2}(\lambda -4)^{2},\]
which says that the matrix \(A\) has two distinct eigenvalues \(\lambda_{1}=2\) and \(\lambda_{2}=4\) with the same multiplicities \(2\). Let \(T_{1}\) and \(T_{2}\) be the restriction of \(T_{A}\) to the generalized eigenspaces \(\bar{V}_{\lambda_{1}}\) and \(\bar{V}_{\lambda_{2}}\), respectively. Suppose that \(\mathfrak{B}_{1}\) is a Jordan basis for \(T_{1}\). Since \(\lambda_{1}\) has multiplicity \(2\), it follows that \(\dim (\bar{V}_{\lambda_{1}})=2\) by part (iii) of Theorem \ref{lat229}. This also says that the dot diagram of \(T_{1}\) has \(2\) dots. According to Proposition \ref{lap240}, we have
\begin{align*} r_{1} & =4-\mbox{rank}(2I_{4}-A)\\ & =4-2=2.\end{align*}
Therefore, the dot diagram of \(T_{1}\) is given by
\[\begin{array}{cc}
\bullet & \bullet
\end{array}\]
which says
\[J_{1}=[T_{1}]_{\mathfrak{B}_{1}}=\left [\begin{array}{cc}
2 & 0\\ 0 & 2
\end{array}\right ].\]
On the other hand, since \(\lambda_{2}=4\) with multiplicity \(2\), it follows \(\dim (\bar{V}_{\lambda_{2}}=2\). This also says that the dot diagram of \(T_{2}\) has \(2\) dots. According to Proposition \ref{lap240}, we have
\begin{align*} r_{1} & =4-\mbox{rank}(4I_{4}-A)\\ & =4-3=1.\end{align*}
Therefore, we also have \(r_{2}=2-r_{1}=1\), and the dot diagram of \(T_{1}\) is given by
\[\begin{array}{c}
\bullet\\ \bullet
\end{array}\]
which says
\[J_{2}=[T_{2}]_{\mathfrak{B}_{2}}=\left [\begin{array}{cc}
4 & 1\\ 0 & 4
\end{array}\right ].\]
Taking \(\mathfrak{B}=\mathfrak{B}_{1}\cup\mathfrak{B}_{2}\), we obtain
\begin{align*} J & =[T_{A}]_{\mathfrak{B}}=\left [\begin{array}{cc}
J_{1} & {\bf 0}\\ {\bf 0} & J_{2}\end{array}\right ]
\\ & =\left [\begin{array}{cc|cc}
2 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ \hline
0 & 0 & 4 & 1\\ 0 & 0 & 0 & 4
\end{array}\right ],\end{align*}
which is the Jordan normal form of \(A\). Now, we want to find a Jordan basis for \(T_{A}\). We first determine the Jordan basis for \(T_{1}\). It can be realized that the dot diagram of \(T_{1}\) is given by
\[\begin{array}{ll}
\bullet (T_{A}-2I)(v_{1}) & \bullet v_{2}
\end{array}.\]
In this case, \(\mathfrak{B}_{1}\) can be any basis for \(V_{\lambda_{1}}=\mbox{Ker}(T_{A}-2I)\). For example, we can take
\[\mathfrak{B}_{1}=\left\{\left [\begin{array}{c}2\\ 1\\ 0\\ 2\end{array}\right ],
\left [\begin{array}{c}0\\ 1\\ 2\\ 0\end{array}\right ]\right\}.\]
On the other hand, it can be realized that the dot diagram of \(T_{2}\) is given by
\[\begin{array}{l}
\bullet (T_{A}-4I)(v)\\ \bullet v
\end{array}.\]
In this case, \(\mathfrak{B}_{2}\) is a cycle of length \(2\). Therefore, we must choose \(v\in V\) such that \(v\in\bar{V}_{\lambda_{2}}=\mbox{Ker}((T_{A}-4I)^{2})\) and \(v\not\in\mbox{Ker}(T_{A}-4I)\). We can use the method of finding \(v_{1}\) in Example \ref{laex242} to choose this vector \(v\). Here, we illustrate another method. A simple calculation shows that a basis for the null space \(\mbox{Ker}(T_{A}-4I)\) is
\[\left\{\left [\begin{array}{c}0\\ 1\\ 1\\ 1
\end{array}\right ]\right\}.\]
Then, we can choose \(v\) to be any solution of the following system of linear equations
\[(A-4I_{4})X=\left [\begin{array}{c}0\\ 1\\ 1\\ 1
\end{array}\right ].\]
Now, we take
\[v=\left [\begin{array}{r}1\\ -1\\ -1\\ 0
\end{array}\right ].\]
Therefore, we obtain
\begin{align*} \mathfrak{B}_{2} & =\left\{v,(T_{A}-4I)(v)\right\}
\\ & =\left\{\left [\begin{array}{c}0\\ 1\\ 1\\ 1\end{array}\right ],
\left [\begin{array}{r}1\\ -1\\ -1\\ 0\end{array}\right ]\right\}.\end{align*}
The Jordan basis for \(T_{A}\) is given by
\begin{align*} \mathfrak{B} & =\mathfrak{B}_{1}\cup\mathfrak{B}_{2}
\\ & =\left\{\left [\begin{array}{c}2\\ 1\\ 0\\ 2\end{array}\right ],
\left [\begin{array}{c}0\\ 1\\ 2\\ 0\end{array}\right ],
\left [\begin{array}{c}0\\ 1\\ 1\\ 1\end{array}\right ],
\left [\begin{array}{r}1\\ -1\\ -1\\ 0\end{array}\right ]\right\}.\end{align*}
If we take
\[P=\left [\begin{array}{rrrr}
2 & 1 & 0 & 2\\ 0 & 1 & 2 & 0\\ 0 & 1 & 1 & 1\\ 1 & -1 & -1 & 0
\end{array}\right ],\]
where the column vectors consist of vectors in the Jordan basis \(\mathfrak{B}\), then \(J=P^{-1}AP\). \(\sharp\)
Example. Let \(V\) be the real vector space of polynomials in two real variables \(x\) and \(y\) with degree at most \(2\). Then
\begin{equation}{\label{laeq243}}\tag{15}
\mathfrak{B}=\left\{1,x,y,x^{2},y^{2},xy\right\}
\end{equation}
is an ordered basis for \(V\). Let \(T\) be a linear operator on \(V\) defined by
\[T(f(x,y))=\frac{\partial}{\partial x}f(x,y).\]
We shall find a Jordan normal form and Jordan basis for \(T\). We immediately have
\[A=[T]_{\mathfrak{B}}=\left [\begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 2 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0
\end{array}\right ].\]
The characteristic polynomial of \(T\) is given by
\[\det\left (\lambda I_{6}-A\right )
=\left [\begin{array}{cccccc}
\lambda & 1 & 0 & 0 & 0 & 0\\
0 & \lambda & 0 & 2 & 0 & 0\\
0 & 0 & \lambda & 0 & 0 & 1\\
0 & 0 & 0 & \lambda & 0 & 0\\
0 & 0 & 0 & 0 & \lambda & 0\\
0 & 0 & 0 & 0 & 0 & \lambda
\end{array}\right ]=\lambda^{6}.\]
This says that \(\lambda =0\) is the only eigenvalue of \(T\) with multiplicity \(6\). We also see that \(\bar{V}_{\lambda}=V\). According to Proposition \ref{lap240}, we have
\begin{align*} r_{1} & =6-\mbox{rank}(A)\\ & =6-3=3.\end{align*}
Since
\[A^{2}=\left [\begin{array}{cccccc}
0 & 0 & 0 & 2 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0
\end{array}\right ],\]
we have
\begin{align*} r_{2} & =\mbox{rank}(A)-\mbox{rank}(A^{2})\\ & =3-1=2.\end{align*}
Since there are \(6\) dots in the dot diagram, we must have \(r_{3}=1\). Therefore, the dot diagram of \(T\) is given by
\[\begin{array}{lll}
\bullet & \bullet & \bullet\\
\bullet & \bullet &\\
\bullet &&
\end{array}\]
This also says that the Jordan normal form is
\[J=\left [\begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0
\end{array}\right ].\]
Now, we want to find the Jordan basis for \(T\). Since the first column of the dot diagram of \(T\) consists of three dots, i.e., the first cycle has length \(3\), we need to find a polynomial \(f_{1}(x,y)\) satisfying
\[\frac{\partial}{\partial x}f_{1}(x,y)\neq 0, \quad
\frac{\partial^{2}}{\partial x^{2}}f_{1}(x,y)\neq 0\mbox{ and }
\frac{\partial^{3}}{\partial x^{3}}f_{1}(x,y)=0.\]
By checking the basis \(\mathfrak{B}\) given in (\ref{laeq243}), since \(\bar{V}_{\lambda}=V\), we can take \(f_{1}(x,y)=x^{2}\). In this case, we have
\begin{align*} (T-\lambda I)(f_{1}(x,y)) & =T(f_{1}(x,y))\\ & =\frac{\partial}{\partial x}x^{2}=2x\end{align*}
and
\begin{align*} (T-\lambda I)^{2}(f_{1}(x,y)) & =T^{2}(f_{1}(x,y))\\ & =\frac{\partial^{2}}{\partial x^{2}}x^{2}=2.\end{align*}
Similarly, since the second column of the dot diagram of \(T\) consists of two dots, , i.e., the second cycle has length \(2\), we need to find a polynomial \(f_{2}(x,y)\) satisfying
\[\frac{\partial}{\partial x}f_{2}(x,y)\neq 0\mbox{ and }\frac{\partial^{2}}{\partial x^{2}}f_{2}(x,y)=0.\]
Since the choice must be linearly independent of the polynomials that are already chosen in the first cycle, the only choice in (\ref{laeq243}) must be \(f_{2}(x,y)=xy\). In this case, we have
\begin{align*} (T-\lambda I)(f_{2}(x,y)) & =T(f_{2}(x,y))\\ & =\frac{\partial}{\partial x}xy=y.\end{align*}
Finally, the third column consists of a single dot, which says that we should pick a single polynomial that lies in the null space \(\mbox{Ker}(T)\). The suitable choice in (\ref{laeq243}) is \(f_{3}(x,y)=y^{2}\). Therefore, the corresponding polynomials for the dot diagram is given by
\[\begin{array}{lll}
\bullet 2 & \bullet y & \bullet y^{2}\\
\bullet 2x & \bullet xy &\\
\bullet x^{2} &&
\end{array}\]
This shows that
\[\left\{2,2x,x^{2},y,xy,y^{2}\right\}\]
is a Jordan basis for \(T\). \(\sharp\)
