4.1 Eigenspaces and characteristic equations

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Invariant subspaces

A subspace $U$ of $V$ is said to be invariant under a linear operator $S:V\to V$ if

$$SU=\{Su\mid u\in U\}\subseteq U.$$

In this case, the action of $S$ restricted to the subspace $U,S{|}_U,$ gives rise to a linear operator on $U$.

Example 4.1

Let $V$ be a three-dimensional vector space with basis $\{e_1,e_2,e_3\}$ , and $S$ the operator defined by

$$\begin{array}{l}S{e}_1=e_2+e_3,\\ S{e}_2=e_1+e_3,\\ S{e}_3=e_1+e_2.\end{array}$$

Let $U$ be the subspace of all vectors of the form $(a+b)e_1+b{e}_2+(-a+b)e_3$ , where $a$ and $b$ are arbitrary scalars. This subspace is spanned by $f_1=e_1-e_3$ and $f_2=e_1+e_2+e_3$ and is invariant under $S$, since

$$S{f}_1=-f_1,S{f}_2=2f_2⟹S(a{f}_1+b{f}_2)=-a{f}_1+2b{f}_2\in U.$$

Exercise

Show that if both $U$ and $W$ are invariant subspaces of $V$ under an operator $S$ then so is their intersection $U\cap W$ and their sum $U+W=\{v=u+w\mid u\in U,w\in W\}$

Suppose $\dim$ $U=m<n=\dim V$ and let $\{e_1,\dots ,e_m\}$ be a basis of $U$. By Theorem 3.7 this basis can be extended to a basis $\{e_1,\dots ,e_n\}$ spanning all of $V$. The invariance of $U$ under $S$ implies that the first $m$ basis vectors are transformed among themselves,

$$S{e}_a=\sum _{b=1}^m{S}_a^b{e}_b(a\le m).$$

In such a basis, the components $S_i^k$ of the operator $S$ vanish for $i\le m,k>m$ , and the $n\times n$ matrix $\mathsf{S}=[S_i^k]$ has the upper block diagonal form

$$\mathsf{S}=\left(\begin{array}{cc}{\mathsf{S}}_1& {\mathsf{S}}_3\\ \mathsf{O}& {\mathsf{S}}_2\end{array}\right).$$

The submatrix ${\mathsf{S}}_1$ is the $m\times m$ matrix of components of $S{|}_U$ expressed in the basis $\{e_1,\dots ,e_m\}$ , while ${\mathsf{S}}_3$ and ${\mathsf{S}}_2$ are submatrices of orders $m\times p$ and $p\times p$ , respectively, where $p=n-m$ , and $O$ is the zero $p\times m$ matrix.

If $V=U\oplus W$ is a decomposition with both $U$ and $W$ invariant under $S$, then choose a basis $\{e_1,\dots ,e_m,e_{m+1},\dots ,e_n\}$ of $V$ such that the first $m$ vectors span $U$ while the last $p=n-m$ vectors span $W$. Then $S_i^k=0$ whenever $i>m$ and $k\le m$ , and the matrix of the operator $S$ has block diagonal form

$$\mathsf{S}=\left(\begin{array}{cc}{\mathsf{S}}_1& \mathsf{O}\\ \mathsf{O}& {\mathsf{S}}_2\end{array}\right).$$

Example 4.2

In Example 4.1 set $f_3=e_3$ . The vectors $f_1,f_2,f_3$ form a basis adapted to the invariant subspace spanned by $f_1$ and $f_2$ ,

$$S{f}_1=-f_1,S{f}_2=2f_2,S{f}_3=f_2-f_3,$$

and the matrix of $S$ has the upper block diagonal form

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

On the other hand, the one-dimensional subspace $W$ spanned by $\begin{array}{r}{f}_3^{\prime }=e_3-\frac{1}{2}{e}_1-\frac{1}{2}{e}_2\end{array}$ is invariant since $S{f}_3^{\prime }=-f_3^{\prime }$ , and in the basis $\{f_1,f_2,f_3^{\prime }\}$ adapted to the invariant decomposition $V=U\oplus W$ the matrix $S$ takes on block diagonal form

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

Eigenvectors and eigenvalues

Given an operator $S:V\to V,$ , a scalar $\lambda \in \mathbb{K}$ is said to be an eigenvalue of $S$ if there exists a non-zero vector $v$ such that

$$Sv=\lambda v(v\ne 0)\text{\hspace{1em}(4.6)}$$

and $v$ is called an eigenvector of $S$ corresponding to the eigenvalue $\lambda$. Eigenvectors are those non-zero vectors that are ‘stretched’ by an amount $\lambda$ on application of the operator $S$. It is important to stipulate $v\ne 0$ since the equation Eq. (4.6) always holds for the zero vector, $\mathcal{S}0=0=\lambda 0.$

For any scalar $\lambda \in \mathbb{K}$ , let

$$V_{\lambda }=\{u\mid Su=\lambda u\}.\text{\hspace{1em}(4.7)}$$

The set $V_{\lambda }$ is a vector subspace, for

$$\begin{array}{rl}Su=\lambda u\text{ and }Sv=\lambda v& ⟹S(u+av)=Su+aSv\\ & ⟹S(u+av)=\lambda u+a\lambda v=\lambda (u+av)\text{ for all }a\in \mathbb{K}.\end{array}$$

For every $\lambda ,$ the subspace $V_{\lambda }$ is invariant under $S$,

$$u\in V_{\lambda }⟹Su=\lambda u⟹S(Su)=\lambda Su⟹Su\in V_{\lambda }.$$

$V_{\lambda }$ consists of the set of all eigenvectors having eigenvalue $\lambda$, supplemented with the zero vector 0 . If $\lambda$ is not an eigenvalue of $S,$ , then $V_{\lambda }=\{0\}$

If $\{e_1,e_2,\dots ,e_n\}$ is any basis of the vector space $V,$ , and $v=v^i{e}_i$ any vector of $V,$ , let $v$ be the column vector of components

$$\mathbf{v}=\left(\begin{array}{c}{v}^1\\ v^2\\ ⋮\\ v^n\end{array}\right).$$

By Eq. (3.8) the matrix equivalent of Eq. (4.6) is

$$\mathbf{Sv}=\lambda \mathbf{v},\text{\hspace{1em}(4.8)}$$

where $S$ is the matrix of components of $S$. Under a change of basis Eq. (4.2) we have from Eqs. (3.18) and Eq. (4.5),

$${\mathbf{v}}^{\prime }=\mathsf{Av},{\mathsf{S}}^{\prime }={\mathsf{ASA}}^{-1}.$$

Hence, if $v$ satisfies Eq. (4.8) then ${\mathbf{v}}^{\prime }$ is an eigenvector of ${\mathsf{S}}^{\prime }$ with the same eigenvalue $\lambda$,

$${\mathsf{S}}^{\prime }{\mathbf{v}}^{\prime }={\mathsf{ASA}}^{-1}\mathsf{Av}=\mathsf{ASv}=\lambda \mathsf{Av}=\lambda {\mathbf{v}}^{\prime }.$$

This result is not unexpected, since Eq. (4.8) and its primed version are simply representations with respect to different bases of the same basis-independent equation Eq. (4.6).

Define the $n$th power $S^n$ of an operator inductively, by setting $S^0={\mathrm{id}}_V$ and

$$S^n=S\circ S^{n-1}=S{S}^{n-1}.$$

Thus $S^1=S$ and $S^2=SS$ , etc. If $p(x)=a_0+a_{1}x+a_2{x}^2+\cdot \cdot \cdot +a_n{x}^n$ is any polynomial with coefficients $a_i\in \mathbb{K}$ , the operator polynomial $p(S)$ is defined in the obvious way,

$$p(S)=a_0+a_{1}S+a_2{S}^2+\cdots +a_n{S}^n.$$

If $\lambda$ is an eigenvalue of $S$ and $v$ a corresponding eigenvector, then $v$ is an eigenvector of any power $S^n$ corresponding to eigenvalue ${\lambda }^n$ . For $n=0$

$$S^{0}v={\mathrm{id}}_{V}v={\lambda }^{0}v\text{ since }{\lambda }^0=1,$$

and the proof follows by induction: assume $S^{n-1}v={\lambda }^{n-1}$ $v$ then by linearity

$$S^{n}v=S{S}^{n-1}v=S{\lambda }^{n-1}v={\lambda }^{n-1}Sv={\lambda }^{n-1}\lambda v={\lambda }^{n}v.$$

For a polynomial $p(x)$ , it follows immediately that $v$ is an eigenvector of the operator $p(S)$ with eigenvalue $p(\lambda )$ ,

$$p(S)v=p(\lambda )v.\text{\hspace{1em}(4.9)}$$

Characteristic equation

The matrix equation Eq. (4.8) can be written in the form

$$(S-\lambda l)\mathbf{v}=0.\text{\hspace{1em}(4.10)}$$

A necessary and sufficient condition for this equation to have a non-trivial solution $\mathbf{v}\ne 0$ is

$$f(\lambda )=\det (\mathsf{S}-\lambda \mathrm{l})=\left|\begin{array}{cccc}{S}_1^1-\lambda & S_2^1& \dots & S_n^1\\ S_1^2& S_2^2-\lambda & \dots & S_n^2\\ ⋮& ⋮& & ⋮\\ S_1^n& S_2^n& \dots & S_n^n-\lambda \end{array}\right|=0,\text{\hspace{1em}(4.11)}$$

called the characteristic equation of $S$. The function $f(\lambda )$ is a polynomial of degree $n$ in $\lambda ,$

$$f(\lambda )=(-1{)}^n\left({\lambda }^n-S_k^k{\lambda }^{n-1}+\cdots +(-1{)}^n\det \mathsf{S}\right),\text{\hspace{1em}(4.12)}$$

known as the characteristic polynomial of $S$.

If the field of scalars is the complex numbers, $\mathbb{K}=\mathbb{C}.$ , then the fundamental theorem of algebra implies that there exist complex numbers ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$ such that

$$f(\lambda )=(-1{)}^n(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)\dots (\lambda -{\lambda }_n).$$

As some of these roots of the characteristic equation may appear repeatedly, we can write the characteristic polynomial in the form

$$\begin{array}{c}f(z)=(-1{)}^n(z-{\lambda }_1{)}^{p_1}(z-{\lambda }_2{)}^{p_2}\dots (z-{\lambda }_m{)}^{p_m}\\ \text{ where }{p}_1+p_2+\cdots +p_m=n.\end{array}\text{\hspace{1em}(4.13)}$$

Since for each $\lambda ={\lambda }_i$ there exists a non-zero complex vector solution $v$ to the linear set of equations given by (4.10), the eigenvalues of $S$ must all come from the set of roots $\{{\lambda }_1,\dots ,{\lambda }_m\}$ . The positive integer $p_i$ is known as the multiplicity of the eigenvalue ${\lambda }_i$ .

Example 4.3

When the field of scalars is the real numbers $R$ there will not in general be real eigenvectors corresponding to complex roots of the characteristic equation. For example, let $A$ be the operator on ${\mathbb{R}}^2$ defined by the following action on the standard basis vectors $e_1=(1,0)$ and $e_2=(0,1)$ ,

$$A{e}_1=e_2,A{e}_2=-e_1⟹\mathsf{A}=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right).$$

The characteristic polynomial is

$$f(z)=\left|\begin{array}{cc}-z& -1\\ 1& -z\end{array}\right|=z^2+1,$$

whose roots are $z=\pm i$ . The operator $A$ thus has no real eigenvalues and eigenvectors. However, if we regard the field of scalars as being $\mathbb{C}$ and treat $A$ as operating on ${\mathbb{C}}^2$ , then it has complex eigenvectors

$$\begin{array}{ll}u=e_1-i{e}_2,& Au=iu,\\ v=e_1+i{e}_2,& Av=-iv.\end{array}$$

It is worth noting that, since $A^2{e}_1,=A{e}_2=-e_1$ and $A^2{e}_2=-A{e}_1=-e_2$ , the operator $A$ satisfies its own characteristic equation

$$A^2+{\mathrm{id}}_{{\mathbb{R}}^2}=0⟹{\mathsf{A}}^2+\mathsf{I}=0.$$

This is a simple example of the important Cayley–Hamilton theorem – see Theorem 4.3 below.

Example 4.4

Let $V$ be a three-dimensional complex vector space with basis $e_1,e_2,e_3$ , and $S:V\to V$ the operator whose matrix with respect to this basis is

$$\mathsf{S}=\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right).$$

The characteristic polynomial is

$$f(z)=\left|\begin{array}{ccc}1-z& 1& 0\\ 0& 1-z& 0\\ 0& 0& 2-z\end{array}\right|=-(z-1{)}^2(z-2).$$

Hence the eigenvalues are 1 and 2, and it is trivial to check that the eigenvector corresponding to 2 is $e_3$ . Let $\mathbf{u}=x{\mathbf{e}}_1+y{\mathbf{e}}_2+z{\mathbf{e}}_3$ be an eigenvector with eigenvalue 1,

$$\mathbf{Su}=\mathbf{u}\text{ where }\mathbf{u}=\left(\begin{array}{c}x\\ y\\ z\end{array}\right),$$

then

$$x+y=x,y=y,2z=z.$$

Hence $y=z=0$ and $u=x{e}_1$ . Thus, even though the eigenvalue $\lambda =1$ has multiplicity 2, all corresponding eigenvectors are multiples of $e_1$ .

Note that while $e_2$ is not an eigenvector, it is annihilated by $(S-{\mathrm{id}}_V{)}^2$ , for

$$\begin{array}{c}S{e}_2=e_1+e_2⟹(S-{\mathrm{id}}_V)e_2=e_1\\ ⟹(S-{\mathrm{id}}_V{)}^2{e}_2=(S-{\mathrm{id}}_V)e_1=0.\end{array}$$

Operators of the form $S-{\lambda }_i{\mathrm{id}}_V$ and their powers $(S-\lambda {\mathrm{id}}_V{)}^m$ , where ${\lambda }_i$ are eigenvalues of $S$, will make regular appearances in what follows. These operators evidently commute with each other and there is no ambiguity in writing them as $(S-{\lambda }_i{)}^m$

Theorem 4.1 · Linear independence of eigenvectors for distinct eigenvalues

Any set of eigenvectors corresponding to distinct eigenvalues of an operator $S$ is linearly independent.

Proof

Let $\{f_1,f_2,\dots ,f_k\}$ be a set of eigenvectors of $S$ corresponding to eigenvalues ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_k$ , no pair of which are equal,

$$S{f}_i={\lambda }_i{f}_i(i=1,\dots ,k),$$

and let $c_1,c_2,\dots ,c_k$ be scalars such that

$$c_1{f}_1+c_2{f}_2+\cdots +c_k{f}_k=0.$$

If we apply the polynomial $P_1(S)=(S-{\lambda }_2)(S-{\lambda }_3)\dots (S-{\lambda }_k)$ ) to this equation, then al terms except the first are annihilated, leaving

$$c_1{P}_1({\lambda }_1)f_1=0.$$

Hence

$$c_1\left({\lambda }_1-{\lambda }_2\right)\left({\lambda }_1-{\lambda }_3\right)\dots \left({\lambda }_1-{\lambda }_k\right)f_1=0,$$

and since $f_1\ne 0$ and all the factors $({\lambda }_1-{\lambda }_i)\ne 0$ for $i=2,\dots ,k.$ , it follows that $c_1=0.$ Similarly, $c_2=\cdots =c_k=0$ , proving linear independence of $f_1,\dots ,f_k$ .

If the operator $S:V\to V$ has $n$ distinct eigenvalues ${\lambda }_1,\dots ,{\lambda }_n$ where $n=\dim V$, then Theorem 4.1 shows the eigenvectors $f_1,f_2,\dots ,f_n$ are l.i. and form a basis of $V$. With respect to this basis the matrix of $S$ is diagonal and its eigenvalues lie along the diagonal,

$$\mathsf{S}=\left(\begin{array}{cccc}{\lambda }_1& 0& \dots & 0\\ 0& {\lambda }_2& \dots & 0\\ ⋮& & \ddots & ⋮\\ 0& 0& \dots & {\lambda }_n\end{array}\right).$$

Conversely, any operator whose matrix is diagonalizable has a basis of eigenvectors (the eigenvalues need not be distinct for the converse). The more difficult task lies in the classification of those cases such as Example 4.4, where an eigenvalue $\lambda$ has multiplicity $p>1$ but there are less than $p$ independent eigenvectors corresponding to it.

Minimal annihilating polynomial

The space of linear operators $L(V,V)$ is a vector space of dimension $n^2$ since it can be put into one-to-one correspondence with the space of $n\times n$ matrices. Hence the first $n^2$ powers $\mathcal{I}\equiv {\mathrm{id}}_V=S^0,S=S^1,S^2,...,S^{n^2}$ ofany linear operator $S$ on $V$ cannot be linearly independent since there are $n^2+1$ operators in all. Thus $S$ must satisfy a polynomial equation,

$$P(S)=c_{0}I+c_{1}S+c_2{S}^2+\cdots +c_{n^2}{S}^{n^2}=0,$$

not all of whose coefficients $c_0,c_1,\dots ,c_n$ vanish.

Exercise

Show that the matrix equivalent of any such polynomial equation is basis-independent by showing that any similarity transform ${\mathsf{S}}^{\prime }=\mathsf{AS}{\mathsf{A}}^{-1}$ of $S$ satisfies the same polynomial equation, $P({\mathsf{S}}^{\prime })=0$

Let

$$\Delta (S)=S^k+c_1{S}^{k-1}+\cdots +c_{k}I=0$$

be the polynomial equation with leading coefficient 1 of lowest degree $k\le n^2$ , satisfied by $S$. The polynomial $\Delta (S)$ is unique, for if

$${\Delta }^{\prime }(S)=S^k+c_1^{\prime }{S}^{k-1}+\cdots +c_k^{\prime }I=0$$

is another such polynomial equation, then on subtracting these two equations we have

$$(\Delta -{\Delta }^{\prime })(S)=(c_1-c_1^{\prime })S^{k-1}+(c_2-c_2^{\prime })S^{k-2}+\cdots +(c_k-c_k^{\prime })=0,$$

which is a polynomial equation of degree $<k$ satisfied by $S$. Hence $c_1=c_1^{\prime },c_2=$ $c_2^{\prime },\dots ,c_k=c_k^{\prime }$ . The unique polynomial $\Delta (z)=z^k+c_1{z}^{k-1}+\cdot \cdot \cdot +c_k$ is called the minimal annihilating polynomial of $S$.

Theorem 4.2 · Eigenvalues and the minimal annihilating polynomial

A scalar $\lambda$ is an eigenvalue of an operator $S$ over a vector space $V$ if and only if it is a root of the minimal annihilatingpolynomial $\Delta (z)$

Proof

If $\lambda$ is an eigenvalue of $S$, let $u\ne 0$ be any corresponding eigenvector, $Su=\lambda u$ Since $0=\Delta (S)u=\Delta (\lambda )u$ it follows that $\Delta (\lambda )=0$

Conversely, if $\lambda$ is a root of $\Delta (z)=0$ then there exists a polynomial ${\Delta }^{\prime }(z)$ such that

$$\Delta (z)=(z-\lambda ){\Delta }^{\prime }(z),$$

and since ${\Delta }^{\prime }(z)$ has lower degree than $\Delta (z)$ it cannot annihilate $S,$

$${\Delta }^{\prime }(S)\ne 0.$$

Therefore, there exists a vector $u\in V$ such that ${\Delta }^{\prime }(S)u=v\ne 0$ , and

$$0=\Delta (S)u=(S-\lambda ){\Delta }^{\prime }(S)u=(S-\lambda )v.$$

Hence $Sv=\lambda v$ , and $\lambda$ is an eigenvalue of $S$ with eigenvector $v$.

It follows from this theorem that the minimal annihilating polynomial of an operator $S$ on a complex vector space can be written in the form

$$\Delta (z)={\left(z-{\lambda }_1\right)}^{k_1}{\left(z-{\lambda }_2\right)}^{k_2}\dots {\left(z-{\lambda }_m\right)}^{k_m}\left(k_1+k_2+\cdots +k_m=k\right)\text{\hspace{1em}(4.14)}$$

where ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m$ run through all the distinct eigenvalues of $S$. The various factors ${\left(z-{\lambda }_i\right)}^{k_i}$ are called the elementary divisors of $S$. The following theorem shows that the characteristic polynomial is always divisible by the minimal annihilating polynomial; that is, for each $i=1,2,\dots ,m$ the coefficient $k_i\le p_i$ where $p_i$ is the multiplicity of the $i$th eigenvalue.

Theorem 4.3 · Cayley–Hamilton theorem

(Cayley–Hamilton) Every linear operator $S$ over a finite dimensional vec tor space $V$ satisfies its own characteristic equation

$$f(S)={\left(S-{\lambda }_1\right)}^{p_1}{\left(S-{\lambda }_2\right)}^{p_2}\dots {\left(S-{\lambda }_m\right)}^{p_m}=0.$$

Equivalently, every $n\times n$ matrix $S$ satisfies its own characteristic equation

$$f(\mathsf{S})={\left(\mathsf{S}-{\lambda }_1\mathsf{I}\right)}^{p_1}{\left(\mathsf{S}-{\lambda }_2\mathsf{I}\right)}^{p_2}\dots {\left(\mathsf{S}-{\lambda }_m\mathsf{I}\right)}^{p_m}=0.$$

Proof

Let $e_1,e_2,\dots ,e_n$ be any basis of $V$, and le $\mathsf{S}=[S_j^k]$ be the matrix ofcomponents of $S$ with respect to this basis,

$$S{e}_j=S_j^k{e}_k.$$

This equation can be written as

$$(S_j^k{\mathrm{id}}_V-{\delta }_j^{k}S)e_k=0,$$

or alternatively as

$$T_j^k(S)e_k=0,\text{\hspace{1em}(4.15)}$$

where

$$T_j^k(z)=S_j^k-{\delta }_j^{k}z.$$

Set $\mathsf{R}(z)=[R_j^k(z)]$ to be the matrix of cofactors of $\mathsf{T}(z)=[T_j^k(z)]$ , such that

$$R_i^j(z)T_j^k(z)={\delta }_i^k\det \mathsf{T}(z)={\delta }_i^{k}f(z).$$

The components $R_j^k(z)$ are polynomials of degree $\le (n-1)$ in $z$, and multiplying both sides of Eq. (4.15) by $R_i^j(S)$ gives

$$R_i^j(S)T_j^k(S)e_k={\delta }_i^{k}f(S)e_k=f(S)e_i=0.$$

Since the $e_i$ span $V$ we have the desired result, $f(S)=0$ . The matrix version is simply the component version of this equation.

Example 4.5

Let $A$ be the matrix operator on the space of complex $4\times 1$ column vectors given by

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

Successive powers of $A$ are

$${\mathsf{A}}^2=\left(\begin{array}{cccc}-1& 2i\alpha & 0& 0\\ 0& -1& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& 1\end{array}\right),{\mathsf{A}}^3=\left(\begin{array}{cccc}-i& -3\alpha & 0& 0\\ 0& -i& 0& 0\\ 0& 0& -i& 0\\ 0& 0& 0& -1\end{array}\right)$$

and it is straightforward to verify that the matrices $I$, $A$, ${\mathsf{A}}^2$ are linearly independent, while

$${\mathsf{A}}^3=(-1+2i){\mathsf{A}}^2+(1+2i)\mathsf{A}+\mathsf{I}.$$

Hence the minimal annihilating polynomial of $A$ is

$$\Delta (z)=z^3+(1-2i)z^2-(1+2i)z-1=(z+1)(z-i{)}^2.$$

The elementary divisors of $A$ are thus $z+1$ and $(z-i{)}^2$ and the eigenvalues are $-1$ and $i$. Computation of the characteristic polynomial reveals that

$$f(z)=\det (\mathsf{A}-z\mathsf{I})=(z+1)(z-i{)}^3,$$

which is divisible by $\Delta (z)$ in agreement with Theorem 4.3.

Problem

Problem 4.1

The trace of an $n\times n$ matrix $\mathsf{T}=[T_j^i]$ is defined as the sum of its diagonal elements,

$$\mathrm{tr}\mathsf{T}=T_i^i=T_1^1+T_2^2+\cdots +T_n^n.$$

Show that

(a) $\mathrm{tr}(\mathsf{ST})=\mathrm{tr}(\mathsf{TS})$

(b) $\mathrm{tr}({\mathsf{ATA}}^{-1})=\mathrm{tr}T.$

(c) If $T:V\to V$ is any operator define its trace to be the trace of its matrix with respect to a basis $\{e_i\}$ . Show that this definition is independent of the choice of basis, so that there is no ambiguity in writing $\mathrm{tr}$ $T$.

(d) If $f(z)=a_0+a_{1}z+a_2{z}^2+\cdot \cdot \cdot +(-1{)}^n{z}^n$ is the characteristic polynomial of the operator $T$, show that $\mathrm{tr}$ $T=(-1{)}^{n-1}{a}_{n-1}.$

(e) If $T$ has eigenvalues ${\lambda }_1,\dots ,{\lambda }_m$ with multiplicities $p_1,\dots ,p_m$ , show that

$$\mathrm{tr}T=\sum _{i=1}^m{p}_i{\lambda }_i.$$