4.2 Jordan canonical form

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Block diagonal form

Let $S$ be a linear operator over a complex vector space $V$, with characteristic polynomial and minimal annihilating polynomial given by Eq. (4.13) and Eq. (4.14), respectively. The restriction to complex vector spaces ensures that all roots of the polynomials $f(z)$ and $\Delta (z)$ are eigenvalues. A canonical form can also be derived for operators on real vector spaces, but it relies heavily on the complex version.

If $(z-{\lambda }_i{)}^{k_i}$ is the $i$th elementary divisor of $S$, define the subspace

$$V_i=\{u\mid (S-{\lambda }_i{)}^{k_i}u=0\}.$$

This subspace is invariant under $S$, for

$$\begin{array}{rl}u\in V_i& ⟹(S-{\lambda }_i{)}^{k_i}u=0\\ & ⟹(S-{\lambda }_i{)}^{k_i}Su=S(S-{\lambda }_i{)}^{k_i}u=0\\ & ⟹Su\in V_i.\end{array}$$

Our first task will be to show that $V$ is a direct sum of these invariant subspaces,

$$V=V_1\oplus V_2\oplus \dots$$

Lemma 4.4 · Annihilation order of generalized eigenvectors

If $(z-{\lambda }_i{)}^{k_i}$ is an elementary divisor of the operator $S$ and $(S-{\lambda }_i{)}^{k_i+r}u=0$ for some $r>0,$ , then $(S-{\lambda }_i{)}^{k_i}u=0$

Proof

There is clearly no loss of generality in setting $i=1$ in the proof of this result. The proof proceeds by induction on $r$.

Case $r=1$ : Let $u$ be any vector such that $(S-{\lambda }_1{)}^{k_1+1}u=0$ and set

$$v=(S-{\lambda }_1{)}^{k_1}u.$$

Since $(S-{\lambda }_1)v=0$ , this vector satisfies the eigenvector equation, $Sv={\lambda }_{1}v$ . If $\Delta (S)$ is the minimal annihilating polynomial of $S$, then

$$\begin{array}{c}0=\Delta (S)u=(S-{\lambda }_2{)}^{k_2}\dots (S-{\lambda }_m{)}^{k_m}v\\ =({\lambda }_1-{\lambda }_2{)}^{k_2}\dots ({\lambda }_1-{\lambda }_m{)}^{k_m}v.\end{array}$$

As all ${\lambda }_i\ne {\lambda }_j$ for $i\ne j$ it follows that $v=0$ , which proves the case $r=1$

Case $r>1$ : Suppose the lemma has been proved for $r-1$ . Then

$$\begin{array}{rl}(S-{\lambda }_1{)}^{k_1+r}u=0& ⟹(S-{\lambda }_1{)}^{k_1+r-1}(S-{\lambda }_1)u=0\\ & ⟹(S-{\lambda }_1{)}^{k_1}(S-{\lambda }_1)u=0\text{ by induction hypothesis }\\ & ⟹(S-{\lambda }_1{)}^{k_1}u=0\text{ by the case }r=1,\end{array}$$

which concludes the proof of the lemma.

If $\dim V_1=p$, let $h_1,\dots ,h_p$ be a basis of $V_1$ and extend to a basis of $V$ using Theorem 3.7:

$$h_1,h_2,\dots ,h_p,h_{p+1},\dots ,h_n.\text{\hspace{1em}(4.16)}$$

Of course if $(z-{\lambda }_1{)}^{k_1}$ is the only elementary divisor of $S$ then $p=n$ , since $V_1=V$ as every vector is annihilated by $(S-{\lambda }_1{)}^{k_1}$ . If, however, $p<n$ we will show that the vectors

$$h_1,h_2,\dots ,h_p,{\hat{h}}_1,{\hat{h}}_2,\dots ,{\hat{h}}_{n-p}\text{\hspace{1em}(4.17)}$$

form a basis of $V$, where

$${\hat{h}}_a=(S-{\lambda }_1{)}^{k_1}{h}_{p+a}(a=1,\dots ,n-p).\text{\hspace{1em}(4.18)}$$

Since the vectors listed in Eq. (4.17) are $n$ in number it is only necessary to show that they are linearly independent. Suppose that for some constants $c^1,\dots ,c^p,{\hat{c}}^1,\dots ,{\hat{c}}^{n-p}$

$$\sum _{i=1}^p{c}^i{h}_i+\sum _{a=1}^{n-p}{\hat{c}}^a{\hat{h}}_a=0.\text{\hspace{1em}(4.19)}$$

Apply $(S-{\lambda }_1{)}^{k_1}$ to this equation. The first sum on the left is annihilated since it belongs to $V_1$ , resulting in

$$(S-{\lambda }_1{)}^{k_1}\sum _{a=1}^{n-p}{\hat{c}}^a{\hat{h}}_a=(S-{\lambda }_1{)}^{2k_1}\sum _{a=1}^{n-p}{\hat{c}}^a{h}_{p+a}=0,$$

and since $k_1>0$ we conclude from Lemma 4.4 that

$$(S-{\lambda }_1{)}^{k_1}\sum _{a=1}^{n-p}{\hat{c}}^a{h}_{p+a}=0.$$

Hence $\begin{array}{r}{\sum }_{a=1}^{n-p}{\hat{c}}^a{h}_{p+a}\in V_1\end{array}$ , and there exist constants $d^1,d^2,\dots ,d^p$ such that

$$\sum _{a=1}^{n-p}{\hat{c}}^a{h}_{p+a}=\sum _{i=1}^p{d}^i{h}_i.$$

As the set $\{h_1,\dots ,h_n\}$ is by definition a basis of $V$, these constants must vanish: ${\hat{c}}^a=d^i=0$ for all $a=1,\dots ,n-p$ and $i=1,\dots ,p.$ . Substituting into Eq. (4.19), it fol lows from the linear independence of $h_1,\dots ,h_p$ that $c^1,\dots ,c^p$ all vanish as well. This proves the linear independence of the vectors in Eq. (4.17).

Let $W_1=L({\hat{h}}_1,{\hat{h}}_2,\dots ,{\hat{h}}_{n-p})$ . By Eq. (4.18) every vector $x\in W_1$ is of the form

$$x=(S-{\lambda }_1{)}^{k_1}y$$

since this is true of each of the vectors spanning $W_1$ . Conversely, suppose $x=(S-{\lambda }_1{)}^{k_1}y$ and let $\{y^1,\dots ,y^n\}$ be the components of $y$ with respect to the original basis Eq. (4.16); then

$$x=(S-{\lambda }_1{)}^{k_1}\left(\sum _i{y}^i{h}_i+\sum _a{y}^{p+a}{h}_{p+a}\right)=\sum _a{y}^{p+a}{\hat{h}}_a\in W_1.$$

Hence $W_1$ consists precisely of all vectors of the form $x=(S-{\lambda }_1{)}^{k_1}y$ where $y$ is an arbitrary vector of $V.$ Furthermore $W_1$ is an invariant subspace of $V$, for if $x\in W_1$ then

$$x=(S-{\lambda }_1{)}^{k_1}y⟹Sx=(S-{\lambda }_1{)}^{k_1}Sy⟹Sx\in W_1.$$

Hence $W_1$ and $V_1$ are complementary invariant subspaces, $V=V_1\oplus W_1$ , and the matrix of $S$ with respect to the basis Eq. (4.17) has block diagonal form

$$S=\left(\begin{array}{cc}{S}_1& O\\ O& T_1\end{array}\right)$$

where ${\mathsf{S}}_1$ is the matrix of $S_1=S{|}_{V_1}$ and ${\mathsf{T}}_1$ is the matrix of $T_1=\mathcal{S}{|}_{{\mathcal{W}}_1}$

Now on the subspace $V_1$ we have, by definition,

$$(S_1-{\lambda }_1{)}^{k_1}=(S-{\lambda }_1{)}^{k_1}{|}_{V_1}=0.$$

Hence ${\lambda }_1$ is the only eigenvalue of $S_1$ , for if $u\in V_1$ is an eigenvector of $S_1$ corresponding to an eigenvalue $\sigma$ ,

$$S_{1}u=\sigma u,$$

then

$$(S_1-{\lambda }_1{)}^{k_1}u=(\sigma -{\lambda }_1{)}^{k_1}u=0,$$

from which it follows that $\sigma ={\lambda }_1$ since $u\ne 0$ . The characteristic equation $o$ ${\cdot }_{S_1}$ is therefore

$$\det ({\mathsf{S}}_1-z\mathsf{I})=(-1{)}^p({\lambda }_1-z{)}^p.\text{\hspace{1em}(4.20)}$$

Furthermore, the operator

$$(T_1-{\lambda }_1{)}^{k_1}=(S-{\lambda }_1{)}^{k_1}{|}_{W_1}:W_1\to W_1$$

is invertible. For, let $x$ be an arbitrary vector in $W_1$ and set $x=(S-{\lambda }_1{)}^{k_1}y$ where $y\in V$ Let $y=y_1+y_2$ be the unique decomposition such that $y_1\in V_1$ and $y_2\in W_1$ ; then

$$x=(S-{\lambda }_1{)}^{k_1}{y}_2=(T_1-{\lambda }_1{)}^{k_1}{y}_2,$$

and since any surjective (onto) linear operator on a finite dimensional vector space is bijective (one-to-one), the map $(T_1-{\lambda }_1{)}^{k_1}$ must be invertible on $W_1$ . Hence

$$\det (T_1-{\lambda }_{1}I)\ne 0\text{\hspace{1em}(4.21)}$$

and ${\lambda }_1$ cannot be an eigenvalue of $T_1$ .

The characteristic equation of $S$ is

$$\det (\mathsf{S}-z\mathsf{I})=\det ({\mathsf{S}}_1-z\mathsf{I})\det ({\mathsf{T}}_1-z\mathsf{I})$$

and from Eq. (4.20) and Eq. (4.21) the only way the right-hand side can equal the expression in Eq. (4.13) is if

$$p_1=p\text{ and }\det ({\mathsf{T}}_1-z\mathsf{I})=(-1{)}^{n-p}(z-{\lambda }_2{)}^{p_2}\dots (z-{\lambda }_m{)}^{p_m}.$$

Hence, the dimension $p_i$ of each space $V_i$ is equal to the multiplicity of the eigenvalue ${\lambda }_i$ and from the Cayley–Hamilton Theorem 4.3 it follows that $p_i\ge k_i$

Repeating this process on $T_1$ , and proceeding inductively, it follows that

$$V=V_1\oplus V_2\oplus \cdots \oplus V_m,$$

and setting

$$h_{11},h_{12},\dots ,h_{1p_1},h_{21},\dots ,h_{2p_2},\dots ,h_{m1},\dots ,h_{m{p}_m}$$

to be a basis adapted to this decomposition, the matrix of $S$ has block diagonal form

$$\mathsf{S}=\left(\begin{array}{cccc}{\mathsf{S}}_1& 0& \dots & 0\\ 0& {\mathsf{S}}_2& \dots & 0\\ \dots & \dots & \dots & \dots \\ 0& 0& \dots & {\mathsf{S}}_m\end{array}\right).\text{\hspace{1em}(4.22)}$$

The restricted operators $S_i=\mathbf{S}{|}_{V_i}$ each have a minimal polynomial equation of the form

$$(S_i-{\lambda }_i{)}^{k_i}=0,$$

so that

$$S_i={\lambda }_i{\mathrm{id}}_i+N_i\text{ where }{N}_i^{k_i}=0.\text{\hspace{1em}(4.23)}$$

Nilpotent operators

Any operator $N$ satisfying an equation of the form $N^k=0$ is called a nilpotent operator. The matrix $N$ of any nilpotent operator satisfies ${\mathsf{N}}^k=0$ , and is called a nilpotent matrix. From Eq. (4.23), each $p_i\times p_i$ matrix ${\mathsf{S}}_i$ in the decomposition Eq. (4.22) is a multiple of the unit matrix plus a nilpotent matrix ${\mathsf{N}}_i$ ,

$${\mathsf{S}}_i={\lambda }_i\mathsf{I}+{\mathsf{N}}_i\text{ where }({\mathsf{N}}_i{)}^{k_i}=0.$$

We next find a basis that expresses the matrix of any nilpotent operator in a standard (canonical) form.

Let $U$ be a finite dimensional space, not necessarily complex, $\dim$ $U=p$ , and $N$ a nilpotent operator on $U$. Set $k$ to be the smallest positive integer such that $N^k=0$ . Evidently $k=1$ if $N=0$ , while if $N\ne 0$ then $k>1$ and $N^{k-1}\ne 0$ . Define the subspaces $X_i$ of $U$ by

$$X_i=\{x\in U|N^{i}x=0\}(i=0,1,\dots ,k).$$

These subspaces form an increasing sequence,

$$\{0\}=X_0\subset X_1\subset X_2\subset \cdots \subset X_k=U$$

and all are invariant under $N$, for if $u\in X_i$ then $Nu$ also belongs to $X_i$ since $N^{i}Nu=N{N}^{i}u=N0=0$ . The set inclusions are strict inclusion in every case, for suppose that $X_i=X_{i+1}$ for some $i\le k-1$ . Then for any vector $x\in X_{i+2}$ we have

$$N^{i+2}x=N^{i+1}Nx=0⟹N^{i}Nx=0⟹x\in X_{i+1}.$$

Hence $X_{i+1}=X_{i+2}$ , and continuing inductively we find that $X_i=\cdot \cdot \cdot =X_{k-1}=X_k=U$ This leads to the conclusion that $N^{k-1}x=0$ for all $x\in U$ , which contradicts the assumption that $N^{k-1}\ne 0$ . Hence none of the subspaces $X_i$ can be equal to each other.

We call a set of vectors $v_1,v_2,\dots ,v_s$ belonging to $X_i$ linearly independent with respect to $X_{i-1}$ if

$$a^1{v}_1+\cdots +a^s{v}_s\in X_{i-1}⟹a^1=\cdots =a^s=0.$$

Lemma 4.5 · Quotient dimensions and relative linear independence

Set $r_i=\dim X_i-\dim X_{i-1}>0$ so that $p=r_1+r_2+\cdot \cdot \cdot +r_k$ . Then $r_i$ is the maximum number of vectors in $X_i$ that canform a set that is linearly independent with respect to $X_{i-1}$

Proof

Let $\dim$ $X_{i-1}=q$ , $\dim$ $X_i=q^{\prime }>q$ and let $u_1,\dots ,u_q$ be a basis of $X_{i-1}$. Suppose $\{v_1,\dots ,v_r\}$ is a maximal set of vectors l.i. with respect to $X_{i-1}$ ; that is, a set that cannot be extended to a larger such set. Such a maximal set must exist since any set of vectors that is l.i. with respect to $X_{i-1}$ is linearly independent and therefore cannot exceed $q$ in number. We show that $\mathcal{S}=\{u_1,\dots ,u_q,v_1,\dots ,v_r\}$ is a basis of $X_i$ :

(a) $S$ is a l.i. set since

$$\sum _{i=1}^q{a}^i{u}_i+\sum _{a=1}^r{b}^a{v}_a=0$$

implies firstly that all $b^a$ vanish by the requirement that the vectors $v_a$ are l.i. with respect to $X_{i-1}$ , and secondly all the ${\mathbf{a}}^i=0$ because the $u_i$ are l.i.

(b) $S$ spans $X_i$ else there would exist a vector $x$ that cannot be expressed as a linear combination of vectors of $S$, and $S\cup \{x\}$ would be linearly independent. In that case, the set of vectors $\{v_1,\dots ,v_r,x\}$ would be l.i. with respect to $X_{i-1}$ , for $\mathrm{if}{\sum }_{a=1}^r{b}^a{v}_a+bx\in X_{i-1}$ then from the linear independence of $S\cup \{x\}$ all $b^a=0$ and $b=0$ . This contradicts the maximality of $v_1,\dots ,v_r$ , and $S$ must span the whole of $X_i$

This proves the lemma.

Let $\{h_1,\dots ,h_{r_k}\}$ be a maximal set of vectors in $X_k$ that is l.i. with respect to $X_{k-1}$ . From Lemma 4.5 we have $r_k=\dim X_k-\dim X_{k-1}$ . The vectors

$$h_1^{\prime }=N{h}_1,h_2^{\prime }=N{h}_2,\dots ,h_{r_k}^{\prime }=N{h}_{r_k}$$

all belong to $X_{k-1}$ and are l.i. with respect to $X_{k-2}$ , for if

$$a^1{h}_1^{\prime }+a^2{h}_2^{\prime }+\cdots +a^{r_k}{h}_{r_k}^{\prime }\in X_{k-2}$$

then

$$N^{k-1}(a^1{h}_1+a^2{h}_2+\cdots +a^{r_k}{h}_{r_k})=0,$$

from which it follows that

$$a^1{h}_1+a^2{h}_2+\cdots +a^{r_k}{h}_{r_k}\in X_{k-1}.$$

Since $\{h_1,\dots ,h_{r_k}\}$ are l.i. with respect to $X_{k-1}$ we must have

$$a^1=a^2=\cdots =a^{r_k}=0.$$

Hence $r_{k-1}\ge r_k$ . Applying the same argument to all other $X_i$ gives

$$r_k\le r_{k-1}\le \cdots \le r_2\le r_1.\text{\hspace{1em}(4.24)}$$

Now complete the set $\{h_1^{\prime },\dots ,h_{r_k}^{\prime }\}$ to a maximal system of vectors in $X_{k-1}$ that is l.i. with respect to $X_{k-2}$

$$h_1^{\prime },\dots ,h_{r_k}^{\prime },h_{r_{k+1}}^{\prime },\dots ,h_{r_{k-1}}^{\prime }.$$

Similarly, define the vectors $h_i^{\prime \prime }=N{h}_i^{\prime }(i=1,\dots ,r_{k-1})$ and extend to a maximal system $\{h_1^{\prime \prime },h_2^{\prime \prime },\dots ,h_{r_{k-2}}^{\prime \prime }\}$ in $X_{k-2}$ . Continuing in this way, form a series of $r_1+r_2+\cdots +r_k=p=\dim U$ vectors that are linearly independent and form a basis of $U$ and may be displayed in the following scheme:

$$\begin{array}{ccccccccc}{h}_1& \dots & h_{r_k}& & & & & & \\ h_1^{\prime }& \dots & h_{r_k}^{\prime }& \dots & h_{r_{k-1}}^{\prime }& & & & \\ h_1^{\prime \prime }& \dots & h_{r_k}^{\prime \prime }& \dots & h_{r_{k-1}}^{\prime \prime }& \dots & h_{r_{k-2}}^{\prime \prime }& & \\ \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \\ h_1^{(k-1)}& \dots & h_{r_k}^{(k-1)}& \dots & h_{r_{k-1}}^{(k-1)}& \dots & \dots & \dots & h_{r_1}^{(k-1)}.\end{array}$$

Let $U_a$ be the subspace generated by the $a$th column where $a=1,\dots ,r_1$ . These subspaces are all invariant under $N$ , since $\dot{N}{h}_a^{(j)}=h_a^{(j+1)}$ , and the bottom elements $h_a^{(k-1)}\in X_1$ are annihilated by $N$,

$$N{h}_a^{(k-1)}\in X_0=\{0\}.$$

Since the vectors $h_a^{(i)}$ are linearly independent and form a basis for $U$, the subspaces $U_a$ are non-intersecting, and

$$U=U_1\oplus U_2\oplus \cdots \oplus U_{r_1},$$

where the dimension $d(a)=\dim U_a$ of the $a$th subspace is given by height of the $a$th column. In particular, $d(1)=k$ and

$$\sum _{a=1}^{r_1}d(a)=p.$$

If a basis is chosen in $U_a$ by proceeding up the $a$th column starting from the vector in the bottom row,

$$f_{a1}=h_a^{(k-1)},f_{a2}=h_a^{(k-2)},\dots ,f_{ad(a)}=h_a^{(k-d(a))},$$

then the matrix of $N_a=N{|}_{U_a}$ has all components zero except for 1’s in the superdiagonal,

$${\mathsf{N}}_a=\left(\begin{array}{cccc}0& 1& 0& \dots \\ 0& 0& 1& \dots \\ & & \ddots & \\ 0& 0& \dots & 1\\ 0& 0& \dots & 0\end{array}\right).\text{\hspace{1em}(4.25)}$$

Exercise

Check this matrix representation by remembering that the components of the matrix of an operator $M$ with respect to a basis $\{u_i\}$ are given by

$$M{u}_i=M_i^j{u}_j.$$

Now set $M=N_a,u_1=f_{a1},u_2=f_{a2},\dots ,u_{d(a)}=f_{ad(a)}$ and note that $M{u}_1=0,M{u}_2=u_1$ , etc.

Selecting a basis for $U$ that runs through the subspaces $U_1,\dots ,U_{r_1}$ in order,

$$e_1=f_{11},e_2=f_{12},\dots ,e_k=f_{1k},e_{k+1}=f_{21},\dots ,e_p=f_{r_{1}d(r_1)},$$

the matrix of $N$ appears in block diagonal form

$$N=\left(\begin{array}{cccc}{N}_1& & & \\ & N_2& & \\ & & \ddots & \\ & & & N_{r_1}\end{array}\right)\text{\hspace{1em}(4.26)}$$

where each submatrix ${\mathsf{N}}_a$ has the form Eq. (4.25).

Jordan canonical form

Let $V$ be a complex vector space and $S:V\to V$ a linear operator on $V$. To summarize the above conclusions: there exists a basis of $V$ such that the operator $S$ has matrix $S$ in block diagonal form Eq. (4.22), and each ${\mathsf{S}}_i$ has the form ${\mathsf{S}}_i={\lambda }_i\mathsf{I}+{\mathsf{N}}_i$ , where ${\mathsf{N}}_i$ is a nilpotent matrix. The basis can then be further specialized such that each nilpotent matrix is in turn decomposed into a block diagonal form Eq. (4.26) such that the submatrices along the diagonal all have the form Eq. (4.25). This is called the Jordan canonical form of the matrix $S$.

In other words, if $S$ is an arbitrary $n\times n$ complex matrix, then there exists a non-singular complex matrix $A$ such that ${\mathsf{ASA}}^{-1}$ is in Jordan form. The essential features of the matrix $S$ can be summarized by the following Segré characteristics:

Eigenvalues	${\lambda }_1$	…	${\lambda }_m$
Multiplicities	$p_1$	…	$p_m$
	$(d_{11}\dots d_{1r_1})$	…	$(d_{m1}\dots d_{m{r}_m})$

where $r_i$ is the number of eigenvectors corresponding to the eigenvalue ${\lambda }_i$ and

$$\sum _{a=1}^{r_i}{d}_{ia}=p_i,\sum _{i=1}^m{p}_i=n=\dim V.$$

The Segré characteristics are determined entirely by properties of the operator $S$ such as its eigenvalues and its elementary divisors. It is important, however, to realize that the Jordan canonical form only applies in the context of a complex vector space since it depends critically on the fundamental theorem of algebra. For a real matrix there is no guarantee that a real similarity transformation will convert it to the Jordan form.

Example 4.6

Let $S$ be a transformation on a four-dimensional vector space having matrix with respect to a basis $\{e_1,e_2,e_3,e_4\}$ whose components are

$$\mathsf{S}=\left(\begin{array}{cccc}1& 1& -1& 0\\ 0& 1& 0& -1\\ 1& 0& 1& 1\\ 0& 1& 0& 1\end{array}\right).$$

The characteristic equation can be written in the form

$$\det (\mathsf{S}-\lambda \mathsf{I})=((\lambda -1{)}^2+1{)}^2=0,$$

which has two roots $\lambda =1\pm i$ , both of which are repeated roots. Each root corresponds to just a single eigenvector, written in column vector form as

$${\mathbf{f}}_1=\left(\begin{array}{c}1\\ 0\\ -i\\ 0\end{array}\right)\text{ and }{\mathbf{f}}_3=\left(\begin{array}{c}1\\ 0\\ i\\ 0\end{array}\right),$$

satisfying

$$\mathbf{S}{\mathbf{f}}_1=(1+i){\mathbf{f}}_1\text{ and }\mathbf{S}{\mathbf{f}}_3=(1-i){\mathbf{f}}_3.$$

Let ${\mathbf{f}}_2$ and ${\mathbf{f}}_4$ be the vectors

$${\mathbf{f}}_2=\left(\begin{array}{c}0\\ 1\\ 0\\ -i\end{array}\right)\text{ and }{\mathbf{f}}_4=\left(\begin{array}{c}0\\ 1\\ 0\\ i\end{array}\right),$$

and we find that

$$\mathbf{S}{\mathbf{f}}_2={\mathbf{f}}_1+(1+i){\mathbf{f}}_2\text{ and }\mathbf{S}{\mathbf{f}}_4={\mathbf{f}}_3+(1-i){\mathbf{f}}_4.$$

Expressing these column vectors in terms of the original basis

$$f_1=e_1-i{e}_3,f_2=e_2-i{e}_4,f_3=e_1+i{e}_3,f_4=e_2+i{e}_4$$

provides a new basis with respect to which the matrix of operator $S$ has block diagona Jordan form

$${\mathsf{S}}^{\prime }=\left(\begin{array}{cccc}1+i& 1& 0& 0\\ 0& 1+i& 0& 0\\ 0& 0& 1-i& 1\\ 0& 0& 0& 1-i\end{array}\right).$$

The matrix $A$ needed to accomplish this form by the similarity transformation ${\mathsf{S}}^{\prime }=\mathsf{AS}{\mathsf{A}}^{-1}$ is found by solving for the $e_i$ in terms of the $f_j$

$$e_1=\frac{1}{2}(f_1+f_3),e_2=\frac{1}{2}(f_2+f_4),e_3=\frac{1}{2}i(f_1-f_3),e_4=\frac{1}{2}i(f_2-f_4),$$

which can be written

$$e_j=A_j^i{f}_i\text{ where }\mathsf{A}=[A_j^i]=\frac{1}{2}\left(\begin{array}{cccc}1& 0& i& 0\\ 0& 1& 0& i\\ 1& 0& -i& 0\\ 0& 1& 0& -i\end{array}\right).$$

The matrix ${\mathsf{S}}^{\prime }$ is summarized by the Segré characteristics:

$1+i$	$1-i$
2	2
(2)	(2)

Exercise

Verify that ${\mathsf{S}}^{\prime }=\mathsf{AS}{\mathsf{A}}^{-1}$ in Example 4.6.

Problems

Problem 4.2

On a vector space $V$ let $S$ and $T$ be two commuting operators, $ST=TS$

(a) Show that if $v$ is an eigenvector of $T$ then so is $Sv$

(b) Show that a basis for $V$ can be found such that the matrices of both $S$ and $T$ with respect to thi basis are in upper triangular form.

Problem 4.3

For the operator $T:V\to V$ on a four-dimensional vector space given in Problem 3.10, show that no basis exists such that the matrix of $T$ is diagonal. Find a basis in which the matrix

of $T$ has the Jordan form

$$\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& \lambda & 1\\ 0& 0& 0& \lambda \end{array}\right)$$

for some $\lambda$, and calculate the value of $\lambda$.

Problem 4.4

Let $S$ be the matrix

$$\mathsf{S}=\left(\begin{array}{cccc}i-1& 1& 0& 0\\ -1& 1+i& 0& 0\\ -1-2i& 2i& -i& 1\\ 2i-1& 1& 0& -i\end{array}\right).$$

Find the minimal annihilating polynomial and the characteristic polynomial of this matrix, its eigenvalues and eigenvectors, and find a basis that reduces it to its Jordan canonical form