Introduction and Definitions

In this chapter, the linear transformations are from a given finite dimensional vector space $V$ to itself. Observe that in this case, the matrix of the linear transformation is a square matrix. So, in this chapter, all the matrices are square matrices and a vector ${\mathbf x}$ means ${\mathbf x}=(x_1,x_2,\ldots,x_n)^t$ for some positive integer $n.$

EXAMPLE 6.1.1   Let $A$ be a real symmetric matrix. Consider the following problem:

$${\mbox{ Maximize (Minimize)}} \;\;{\mathbf x}^t A {\mathbf x}{\mb... ... }} {\mathbf x} \in {\mathbb{R}}^n {\mbox{ and }} {\mathbf x}^t {\mathbf x}= 1.$$

To solve this, consider the Lagrangian

$$L({\mathbf x}, \lambda) = {\mathbf x}^t A {\mathbf x}- \lambda ( ... ...\sum_{i=1}^n\sum_{j=1}^n a_{ij} x_i x_j - \lambda (\sum_{i=1}^n x_i^2 \;\; -1).$$

Partially differentiating $L({\mathbf x}, \lambda)$ with respect to $x_i$ for $1 \leq i \leq n,$ we get

$$\frac{\partial L}{\partial x_1} = 2 a_{11} x_1 + 2 a_{12} x_2 + \cdots + 2 a_{1n} x_n - 2 \lambda x_1,$$

$$\frac{\partial L}{\partial x_2} = 2 a_{21} x_1 + 2 a_{22} x_2 + \cdots + 2 a_{2n} x_n - 2 \lambda x_2,$$

and so on, till

$$\frac{\partial L}{\partial x_n} = 2 a_{n1} x_1 + 2 a_{n2} x_2 + \cdots + 2 a_{nn} x_n - 2 \lambda x_n.$$

Therefore, to get the points of extrema, we solve for

$$(0,0,\ldots,0)^t = (\frac{\partial L}{\partial x_1}, \frac{\parti... ...rac{\partial L}{\partial {\mathbf x}} = 2 (A {\mathbf x}- \lambda {\mathbf x}).$$

We therefore need to find a $\lambda \in {\mathbb{R}}$ and ${\mathbf 0}\neq {\mathbf x}\in {\mathbb{R}}^n$ such that $A {\mathbf x}= \lambda {\mathbf x}$ for the extremal problem.

EXAMPLE 6.1.2   Consider a system of $n$ ordinary differential equations of the form

$$\frac{d \; {\mathbf y}(t)}{d t} = A {\mathbf y}, \; t \geq 0;$$ (6.1.1)

where $A$ is a real $n \times n$ matrix and ${\mathbf y}$ is a column vector.
To get a solution, let us assume that

$${\mathbf y}(t) = {\mathbf c}e^{ {\lambda}t}$$ (6.1.2)

is a solution of (6.1.1) and look into what ${\lambda}$ and ${\mathbf c}$ has to satisfy, i.e., we are investigating for a necessary condition on ${\lambda}$ and ${\mathbf c}$ so that (6.1.2) is a solution of (6.1.1). Note here that (6.1.1) has the zero solution, namely $y(t) \equiv 0$ and so we are looking for a non-zero ${\mathbf c}.$ Differentiating (6.1.2) with respect to $t$ and substituting in (6.1.1), leads to

$${\lambda}e^{{\lambda}t} {\mathbf c}= A e^{{\lambda}t} {\mathbf c}\;\; {\mbox{or equivalently}} \;\; (A - {\lambda}I) {\mathbf c}= {\mathbf 0}.$$ (6.1.3)

So, (6.1.2) is a solution of the given system of differential equations if and only if ${\lambda}$ and ${\mathbf c}$ satisfy (6.1.3). That is, given an $n \times n$ matrix $A,$ we are this lead to find a pair $({\lambda}, {\mathbf c})$ such that ${\mathbf c}\neq {\mathbf 0}$ and (6.1.3) is satisfied.

Let $A$ be a matrix of order $n.$ In general, we ask the question:
For what values of ${\lambda}\in {\mathbb{F}},$ there exist a non-zero vector ${\mathbf x}\in {\mathbb{F}}^n$ such that

$$A {\mathbf x}= \lambda {\mathbf x}?$$ (6.1.4)

Here, ${\mathbb{F}}^n$ stands for either the vector space ${\mathbb{R}}^n$ over ${\mathbb{R}}$ or ${\mathbb{C}}^n$ over ${\mathbb{C}}.$ Equation (6.1.4) is equivalent to the equation

$$(A - \lambda I) {\mathbf x}= {\mathbf 0}.$$

By Theorem 2.5.1, this system of linear equations has a non-zero solution, if

$${\mbox{rank }} (A - \lambda I) < n, \;\; {\mbox{ or equivalently }} \;\; \det(A - {\lambda}I) = 0.$$

So, to solve (6.1.4), we are forced to choose those values of $\lambda \in {\mathbb{F}}$ for which $\det (A - \lambda I) = 0.$ Observe that $\det(A - {\lambda}I)$ is a polynomial in ${\lambda}$ of degree $n.$ We are therefore lead to the following definition.

DEFINITION 6.1.3 (Characteristic Polynomial)   Let $A$ be a matrix of order $n.$ The polynomial $\det(A - {\lambda}I)$ is called the characteristic polynomial of $A$ and is denoted by $p({\lambda}).$ The equation $p({\lambda}) = 0$ is called the characteristic equation of $A.$ If ${\lambda}\in {\mathbb{F}}$ is a solution of the characteristic equation $p({\lambda}) = 0,$ then ${\lambda}$ is called a characteristic value of $A.$

Some books use the term EIGENVALUE in place of characteristic value.

THEOREM 6.1.4   Let $A= [a_{ij}]; \; a_{ij} \in {\mathbb{F}}, \; {\mbox{ for }} 1 \leq i, j \leq n.$ Suppose $\lambda = \lambda_0 \in {\mathbb{F}}$ is a root of the characteristic equation. Then there exists a non-zero ${\mathbf v}\in {{\mathbb{F}}}^n$ such that $A {\mathbf v}= \lambda_0 {\mathbf v}.$

Proof. Since $\lambda_0$ is a root of the characteristic equation, $\det (A - \lambda_0 I) = 0.$ This shows that the matrix $A - \lambda_0 I$ is singular and therefore by Theorem 2.5.1 the linear system

$$(A - \lambda_0 I_n) {\mathbf x}= {\mathbf 0}$$

has a non-zero solution. height6pt width 6pt depth 0pt

Remark 6.1.5   Observe that the linear system $A {\mathbf x}= {\lambda}{\mathbf x}$ has a solution ${\mathbf x}={\mathbf 0}$ for every ${\lambda}\in {\mathbb{F}}.$ So, we consider only those ${\mathbf x}\in {\mathbb{F}}^n$ that are non-zero and are solutions of the linear system $A {\mathbf x}= {\lambda}{\mathbf x}.$

DEFINITION 6.1.6 (Eigenvalue and Eigenvector)   If the linear system $A {\mathbf x}= {\lambda}{\mathbf x}$ has a non-zero solution ${\mathbf x}\in {\mathbb{F}}^n$ for some ${\lambda}\in {\mathbb{F}},$ then

1. $\lambda \in {\mathbb{F}}$ is called an eigenvalue of $A,$
2. ${\mathbf 0}\neq {\mathbf x}\in {\mathbb{F}}^n$ is called an eigenvector corresponding to the eigenvalue ${\lambda}$ of $A,$ and
3. the tuple $(\lambda, {\mathbf x})$ is called an eigenpair.

Remark 6.1.7   To understand the difference between a characteristic value and an eigenvalue, we give the following example.

Consider the matrix $A = \begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}.$ Then the characteristic polynomial of $A$ is

$$p({\lambda}) = {\lambda}^2 + 1.$$

Given the matrix $A,$ recall the linear transformation $T_A: {\mathbb{F}}^2 {\longrightarrow}{\mathbb{F}}^2$ defined by

$$T_A({\mathbf x}) = A {\mathbf x}\;\; {\mbox{ for every }} \;\; {\mathbf x}\in {\mathbb{F}}^2.$$

1. If $\; {\mathbb{F}}= {\mathbb{C}},$ that is, if $A$ is considered a COMPLEX matrix, then the roots of $p({\lambda}) = 0$ in ${\mathbb{C}}$ are $\pm i.$ So, $A$ has $(i, (1,i)^t)$ and $(-i, (i, 1)^t)$ as eigenpairs.
2. If $\; {\mathbb{F}}= {\mathbb{R}},$ that is, if $A$ is considered a REAL matrix, then $p({\lambda}) = 0$ has no solution in ${\mathbb{R}}.$ Therefore, if ${\mathbb{F}}= {\mathbb{R}},$ then $A$ has no eigenvalue but it has $\pm i$ as characteristic values.

Remark 6.1.8   Note that if $(\lambda, {\mathbf x})$ is an eigenpair for an $n \times n$ matrix $A$ then for any non-zero $c \in {\mathbb{F}},\;c \neq 0, \; (\lambda, c {\mathbf x})$ is also an eigenpair for $A.$ Similarly, if ${\mathbf x}_1, {\mathbf x}_2, \ldots, {\mathbf x}_r$ are eigenvectors of $A$ corresponding to the eigenvalue ${\lambda},$ then for any non-zero $(c_1, c_2, \ldots, c_r) \in {\mathbb{F}}^r,$ it is easily seen that if $\sum\limits_{i=1}^r c_i {\mathbf x}_i \neq {\mathbf 0}$ , then $\sum\limits_{i=1}^r c_i {\mathbf x}_i$ is also an eigenvector of $A$ corresponding to the eigenvalue ${\lambda}.$ Hence, when we talk of eigenvectors corresponding to an eigenvalue $\lambda,$ we mean LINEARLY INDEPENDENT EIGENVECTORS.

Suppose ${\lambda}_0 \in {\mathbb{F}}$ is a root of the characteristic equation $\det(A - {\lambda}_0 I) = 0.$ Then $A - {\lambda}_0 I$ is singular and ${\mbox{rank }}(A - {\lambda}_0 I) < n.$ Suppose ${\mbox{rank }}(A - {\lambda}_0 I)=r < n.$ Then by Corollary 4.3.9, the linear system $(A - {\lambda}_0 I) {\mathbf x}= {\mathbf 0}$ has $n-r$ linearly independent solutions. That is, $A$ has $n-r$ linearly independent eigenvectors corresponding to the eigenvalue ${\lambda}_0$ whenever ${\mbox{rank }}(A - {\lambda}_0 I)=r < n.$

EXAMPLE 6.1.9  

1. Let $A = {\mbox{diag}}(d_1, d_2, \ldots, d_n)$ with $d_i \in {\mathbb{R}}$ for $1 \leq i \leq n.$ Then $p({\lambda}) = \prod_{i=1}^n ({\lambda}- d_i)$ is the characteristic equation. So, the eigenpairs are
   $$(d_1, (1,0,\ldots, 0)^t), (d_2, (0,1,0,\ldots,0)^t), \ldots, (d_n, (0,\ldots, 0,1)^t).$$

2. Let $A = \begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}.$ Then $\det (A - \lambda I_2) = (1-\lambda)^2.$ Hence, the characteristic equation has roots $1, 1.$ That is $1$ is a repeated eigenvalue. Now check that the equation $(A - I_2) {\mathbf x}= {\mathbf 0}$ for ${\mathbf x}= (x_1, x_2)^{t}$ is equivalent to the equation $x_2 = 0.$ And this has the solution ${\mathbf x}= (x_1, 0)^{t}.$ Hence, from the above remark, $(1, 0)^{t}$ is a representative for the eigenvector. Therefore, HERE WE HAVE TWO EIGENVALUES MATHEND000# BUT ONLY ONE EIGENVECTOR.
3. Let $A = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}.$ Then $\det (A - \lambda I_2) = (1-\lambda)^2.$ The characteristic equation has roots $1, 1.$ Here, the matrix that we have is $I_2$ and we know that $I_2 {\mathbf x}= {\mathbf x}$ for every ${\mathbf x}^{t} \in {\mathbb{R}}^2$ and we can CHOOSE ANY TWO LINEARLY INDEPENDENT VECTORS ${\mathbf x}^{t}, {\mathbf y}^{t}$ from ${\mathbb{R}}^2$ to get $(1, {\mathbf x})$ and $(1, {\mathbf y})$ as the two eigenpairs.

   In general, if ${\mathbf x}_1, {\mathbf x}_2, \ldots, {\mathbf x}_n$ are linearly independent vectors in ${\mathbb{R}}^n,$ then $(1, {\mathbf x}_1), \; (1, {\mathbf x}_2), \; \ldots, (1, {\mathbf x}_n)$ are eigenpairs for the identity matrix, $I_n.$

4. Let $A = \begin{bmatrix}1 & 2 \\ 2 & 1 \end{bmatrix}.$ Then $\det (A - \lambda I_2) = (\lambda- 3)(\lambda + 1).$ The characteristic equation has roots $3, -1.$ Now check that the eigenpairs are $(3, (1,1)^{t}),$ and $(-1, (1, -1)^{t}).$ In this case, we have TWO DISTINCT EIGENVALUES AND THE CORRESPONDING EIGENVECTORS ARE ALSO LINEARLY INDEPENDENT. The reader is required to prove the linear independence of the two eigenvectors.
5. Let $A = \begin{bmatrix}1 & -1 \\ 1 & 1 \end{bmatrix}.$ Then $\det (A - \lambda I_2) = \lambda^2 - 2 \lambda + 2.$ The characteristic equation has roots $1 + i, 1 - i.$ Hence, over ${\mathbb{R}},$ the matrix $A$ has no eigenvalue. Over ${\mathbb{C}},$ the reader is required to show that the eigenpairs are $(1+i, (i,1)^t)$ and $(1-i, (1,i)^t).$

EXERCISE 6.1.10  

1. Find the eigenvalues of a triangular matrix.
2. Find eigenpairs over ${\mathbb{C}},$ for each of the following matrices:
   $\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}, \hspace{0.1in} \begin{bmatrix}1 &... ...bmatrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{bmatrix},$ and $\;\;\begin{bmatrix}\cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{bmatrix}.$
3. Let $A$ and $B$ be similar matrices.
   1. Then prove that $A$ and $B$ have the same set of eigenvalues.
   2. Let $({\lambda}, {\mathbf x})$ be an eigenpair for $A$ and $({\lambda}, {\mathbf y})$ be an eigenpair for $B.$ What is the relationship between the vectors ${\mathbf x}$ and ${\mathbf y}$ ?

      [Hint: Recall that if the matrices $A$ and $B$ are similar, then there exists a non-singular matrix $P$ such that $B = P A P^{-1}.$ ]

4. Let $A=(a_{ij})$ be an $n \times n$ matrix. Suppose that for all $i, \; 1 \leq i \leq n, \; \sum\limits_{j=1}^n a_{ij} = a.$ Then prove that $a$ is an eigenvalue of $A.$ What is the corresponding eigenvector?
5. Prove that the matrices $A$ and $A^t$ have the same set of eigenvalues. Construct a $2 \times 2$ matrix $A$ such that the eigenvectors of $A$ and $A^t$ are different.
6. Let $A$ be a matrix such that $A^2 = A$ ($A$ is called an idempotent matrix). Then prove that its eigenvalues are either 0 or $1$ or both.
7. Let $A$ be a matrix such that $A^k = {\mathbf 0}$ ($A$ is called a nilpotent matrix) for some positive integer $k \ge 1$ . Then prove that its eigenvalues are all 0 .

THEOREM 6.1.11   Let $A=[a_{ij}]$ be an $n \times n$ matrix with eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n,$ not necessarily distinct. Then $\det (A) = \prod\limits_{i=1}^n \lambda_i$ and ${\mbox{ tr}}(A) = \sum\limits_{i=1}^n a_{ii} = \sum\limits_{i=1}^n \lambda_i.$

Proof. Since $\lambda_1, \lambda_2, \ldots, \lambda_n$ are the $n$ eigenvalues of $A,$ by definition,

$$\det (A - \lambda I_n) = p({\lambda}) = (-1)^n (\lambda - \lambda_1) (\lambda - \lambda_2) \cdots (\lambda - \lambda_n).$$ (6.1.5)

(6.1.5) is an identity in ${\lambda}$ as polynomials. Therefore, by substituting $\lambda = 0$ in (6.1.5), we get

$$\det (A) = (-1)^n (-1)^n \prod_{i=1}^n \lambda_i = \prod_{i=1}^n \lambda_i.$$

Also,

$$\det (A - \lambda I_n) = \begin{bmatrix}a_{11} - \lambda & a_{12} & \cdots & a_{1n} \\ a_{... ... & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} - \lambda \end{bmatrix}$$ (6.1.6)

$$= a_0 - \lambda a_1 + \lambda^2 a_2 + \cdots$$

$$+(-1)^{n-1} {\lambda}^{n-1} a_{n-1} + (-1)^n \lambda^n$$ (6.1.7)

for some $a_0, a_1, \ldots, a_{n-1} \in {\mathbb{F}}.$ Note that $a_{n-1},$ the coefficient of $(-1)^{n-1} {\lambda}^{n-1},$ comes from the product

$$(a_{11} - {\lambda})(a_{22} - {\lambda}) \cdots (a_{nn} - {\lambda}).$$

So, $a_{n-1} = \sum\limits_{i=1}^n a_{ii} = {\mbox{tr}}(A)$ by definition of trace.

But , from (6.1.5) and (6.1.7), we get

$$a_0 - \lambda a_1 + \lambda^2 a_2 + \cdots + (-1)^{n-1} {\lambda}^{n-1} a_{n-1} + (-1)^n \lambda^n$$

$$= (-1)^n (\lambda - \lambda_1) (\lambda - \lambda_2) \cdots (\lambda - \lambda_n).$$ (6.1.8)

Therefore, comparing the coefficient of $(-1)^{n-1} \lambda^{n-1},$ we have

$${\mbox{tr}}(A)= a_{n-1} = (-1) \{ (-1)\sum\limits_{i=1}^n \lambda_i \} = \sum\limits_{i=1}^n \lambda_i.$$

Hence, we get the required result. height6pt width 6pt depth 0pt

EXERCISE 6.1.12  

1. Let $A$ be a skew symmetric matrix of order $2n + 1.$ Then prove that 0 is an eigenvalue of $A.$
2. Let $A$ be a $3 \times 3$ orthogonal matrix $(A A^t = I)$ .If $\det(A) = 1$ , then prove that there exists a non-zero vector ${\mathbf v}\in {\mathbb{R}}^3$ such that $A {\mathbf v}= {\mathbf v}.$

Let $A$ be an $n \times n$ matrix. Then in the proof of the above theorem, we observed that the characteristic equation $\det(A - {\lambda}I) = 0$ is a polynomial equation of degree $n$ in ${\lambda}.$ Also, for some numbers $a_0, a_1, \ldots, a_{n-1} \in {\mathbb{F}},$ it has the form

$${\lambda}^n + a_{n-1} {\lambda}^{n-1} + a_{n-2} {\lambda}^2 + \cdots a_1 {\lambda}+ a_0 = 0.$$

Note that, in the expression $\det(A - {\lambda}I)= 0, \;\; {\lambda}$ is an element of ${\mathbb{F}}.$ Thus, we can only substitute ${\lambda}$ by elements of ${\mathbb{F}}.$

It turns out that the expression

$$A^n + a_{n-1} A^{n-1} + a_{n-2} A^2 + \cdots a_1 A + a_0 I = {\mathbf 0}$$

holds true as a matrix identity. This is a celebrated theorem called the Cayley Hamilton Theorem. We state this theorem without proof and give some implications.

THEOREM 6.1.13 (Cayley Hamilton Theorem)   Let $A$ be a square matrix of order $n.$ Then $A$ satisfies its characteristic equation. That is,

$$A^n + a_{n-1} A^{n-1} + a_{n-2} A^2 + \cdots a_1 A + a_0 I = {\mathbf 0}$$

holds true as a matrix identity.

Some of the implications of Cayley Hamilton Theorem are as follows.

Remark 6.1.14  

1. Let $A = \left[\begin{array}{cc}0&1 \\ 0 & 0 \end{array}\right].$ Then its characteristic polynomial is $p({\lambda}) = {\lambda}^2.$ Also, for the function, $f(x) = x,$ $\;f(0) = 0,$ and $\; f(A) = A \neq {\mathbf 0}.$ This shows that the condition $f({\lambda}) = 0$ for each eigenvalue ${\lambda}$ of $A$ does not imply that $f(A) = {\mathbf 0}.$
2. Suppose we are given a square matrix $A$ of order $n$ and we are interested in calculating $A^{\ell}$ where $\ell$ is large compared to $n.$ Then we can use the division algorithm to find numbers ${\alpha}_0, {\alpha}_1, \ldots, {\alpha}_{n-1}$ and a polynomial $f({\lambda})$ such that
   

   $${\lambda}^{\ell} = f({\lambda}) \bigl( {\lambda}^n + a_{n-1} {\lambda}^{n-1} + a_{n-2} {\lambda}^2 + \cdots a_1 {\lambda}+ a_0 \bigr)$$

   $$+ {\alpha}_0 + {\lambda}{\alpha}_1 + \cdots + {\lambda}^{n-1} {\alpha}_{n-1}.$$

   
   Hence, by the Cayley Hamilton Theorem,
   $$A^{\ell} = {\alpha}_0 I + {\alpha}_1 A + \cdots + {\alpha}_{n-1} A^{n-1}.$$
   That is, we just need to compute the powers of $A$ till $n-1.$

   In the language of graph theory, it says the following:
   ``Let $G$ be a graph on $n$ vertices. Suppose there is no path of length $n-1$ or less from a vertex $v$ to a vertex $u$ of $G.$ Then there is no path from $v$ to $u$ of any length. That is, the graph $G$ is disconnected and $v$ and $u$ are in different components."

3. Let $A$ be a non-singular matrix of order $n.$ Then note that $a_n = \det (A) \neq 0$ and
   $$A^{-1} = \frac{-1}{a_n} [ A^{n-1} + a_{n-1} A^{n-2} + \cdots + a_1 I ].$$
   This matrix identity can be used to calculate the inverse.
   Note that the vector $A^{-1}$ (as an element of the vector space of all $n \times n$ matrices) is a linear combination of the vectors $I, A, \ldots, A^{n-1}.$

EXERCISE 6.1.15   Find inverse of the following matrices by using the Cayley Hamilton Theorem

$$i)\;\; \begin{bmatrix}2&3&4 \\ 5&6&7\\ 1&1&2 \end{bmatrix}\;\;\;\... ...\; iii) \begin{bmatrix} 1 & -2 & -1 \\ -2 & 1 & -1 \\ 0 & -1 & 2 \end{bmatrix}.$$

THEOREM 6.1.16   If ${\lambda}_1, {\lambda}_2, \ldots, {\lambda}_k$ are distinct eigenvalues of a matrix $A$ with corresponding eigenvectors ${\mathbf x}_1, {\mathbf x}_2, \ldots, {\mathbf x}_k,$ then the set $\{{\mathbf x}_1, {\mathbf x}_2, \ldots, {\mathbf x}_k\}$ is linearly independent.

Proof. The proof is by induction on the number $m$ of eigenvalues. The result is obviously true if $m = 1$ as the corresponding eigenvector is non-zero and we know that any set containing exactly one non-zero vector is linearly independent.

Let the result be true for $m, \; 1 \leq m < k.$ We prove the result for $m+1.$ We consider the equation

$$c_1 x_1 + c_2 x_2 + \cdots + c_{m+1} x_{m+1} = {\mathbf 0}$$ (6.1.9)

for the unknowns $c_1, c_2, \ldots, c_{m+1}.$ We have

$${\mathbf 0}= A {\mathbf 0} = A ( c_1 x_1 + c_2 x_2 + \cdots + c_{m+1} x_{m+1} )$$

$$= c_1 A x_1 + c_2 A x_2 + \cdots + c_{m+1} A x_{m+1}$$

$$= c_1 \lambda_1 x_1 + c_2 \lambda_2 x_2 + \cdots + c_{m+1} \lambda_{m+1} x_{m+1}.$$ (6.1.10)

From Equations (6.1.9) and (6.1.10), we get

$$c_2 (\lambda_2 - \lambda_1 ) {\mathbf x}_2 + c_3 (\lambda_3 - \la... ...+ \cdots + c_{m+1} (\lambda_{m+1} - \lambda_1) {\mathbf x}_{m+1} = {\mathbf 0}.$$

This is an equation in $m$ eigenvectors. So, by the induction hypothesis, we have

$$c_i (\lambda_i - \lambda_1) = 0 \;\; {\mbox{ for }} \;\; 2 \leq i \leq m+1.$$

But the eigenvalues are distinct implies $\lambda_i - {\lambda}_1 \neq 0$ for $2 \leq i \leq m+1.$ We therefore get $c_i = 0$ for $2 \leq i \leq m+1.$ Also, ${\mathbf x}_1 \neq {\mathbf 0}$ and therefore (6.1.9) gives $c_1 = 0.$

Thus, we have the required result. height6pt width 6pt depth 0pt

We are thus lead to the following important corollary.

COROLLARY 6.1.17   The eigenvectors corresponding to distinct eigenvalues of an $n \times n$ matrix $A$ are linearly independent.

EXERCISE 6.1.18  

1. For an $n \times n$ matrix $A,$ prove the following.
   1. $A$ and $A^{t}$ have the same set of eigenvalues.
   2. If ${\lambda}$ is an eigenvalue of an invertible matrix $A$ then $$\frac{1}{{\lambda}}$$ is an eigenvalue of $A^{-1}.$
   3. If ${\lambda}$ is an eigenvalue of $A$ then ${\lambda}^{k}$ is an eigenvalue of $A^k$ for any positive integer $k.$
   4. If $A$ and $B$ are $n \times n$ matrices with $A$ nonsingular then $B A^{-1}$ and $A^{-1} B$ have the same set of eigenvalues.

      In each case, what can you say about the eigenvectors?

2. Let $A$ and $B$ be $2 \times 2$ matrices for which $\det(A) = \det(B)$ and ${\mbox{tr}}(A) = {\mbox{tr}}(B).$
   1. Do $A$ and $B$ have the same set of eigenvalues?
   2. Give examples to show that the matrices $A$ and $B$ need not be similar.
3. Let $({\lambda}_1, {\mathbf u})$ be an eigenpair for a matrix $A$ and let $({\lambda}_2, {\mathbf u})$ be an eigenpair for another matrix $B.$
   1. Then prove that $({\lambda}_1+{\lambda}_2, {\mathbf u})$ is an eigenpair for the matrix $A+B.$
   2. Give an example to show that if ${\lambda}_1, {\lambda}_2$ are respectively the eigenvalues of $A$ and $B,$ then ${\lambda}_1+{\lambda}_2$ need not be an eigenvalue of $A+B.$
4. Let ${\lambda}_i, 1 \leq i \leq n$ be distinct non-zero eigenvalues of an $n \times n$ matrix $A.$ Let ${\mathbf u}_i, 1 \leq i \leq n$ be the corresponding eigenvectors. Then show that ${\cal {B}} = \{{\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n \}$ forms a basis of ${\mathbb{F}}^n ({\mathbb{F}}).$ If $[{\mathbf b}]_{{\cal B}} = (c_1, c_2, \ldots, c_n)^t$ then show that $A {\mathbf x}= {\mathbf b}$ has the unique solution
   $${\mathbf x}= \frac{c_1}{{\lambda}_1} {\mathbf u}_1 + \frac{c_2}{{\lambda}_2} {\mathbf u}_2 + \cdots + \frac{c_n}{{\lambda}_n} {\mathbf u}_n.$$