# Introduction and Definitions

In this chapter, the linear transformations are from a given finite dimensional vector space 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 means for some positive integer

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

To solve this, consider the Lagrangian

Partially differentiating with respect to for we get

and so on, till

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

We therefore need to find a and such that for the extremal problem.

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

 (6.1.1)

where is a real matrix and is a column vector.
To get a solution, let us assume that

 (6.1.2)

is a solution of (6.1.1) and look into what and has to satisfy, i.e., we are investigating for a necessary condition on and so that (6.1.2) is a solution of (6.1.1). Note here that (6.1.1) has the zero solution, namely and so we are looking for a non-zero Differentiating (6.1.2) with respect to and substituting in (6.1.1), leads to

 (6.1.3)

So, (6.1.2) is a solution of the given system of differential equations if and only if and satisfy (6.1.3). That is, given an matrix we are this lead to find a pair such that and (6.1.3) is satisfied.

Let be a matrix of order In general, we ask the question:
For what values of there exist a non-zero vector such that

 (6.1.4)

Here, stands for either the vector space over or over Equation (6.1.4) is equivalent to the equation

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

So, to solve (6.1.4), we are forced to choose those values of for which Observe that is a polynomial in of degree We are therefore lead to the following definition.

DEFINITION 6.1.3 (Characteristic Polynomial)   Let be a matrix of order The polynomial is called the characteristic polynomial of and is denoted by The equation is called the characteristic equation of If is a solution of the characteristic equation then is called a characteristic value of

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

THEOREM 6.1.4   Let Suppose is a root of the characteristic equation. Then there exists a non-zero such that

Proof. Since is a root of the characteristic equation, This shows that the matrix is singular and therefore by Theorem 2.5.1 the linear system

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

Remark 6.1.5   Observe that the linear system has a solution for every So, we consider only those that are non-zero and are solutions of the linear system

DEFINITION 6.1.6 (Eigenvalue and Eigenvector)   If the linear system has a non-zero solution for some then
1. is called an eigenvalue of
2. is called an eigenvector corresponding to the eigenvalue of and
3. the tuple 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 Then the characteristic polynomial of is

Given the matrix recall the linear transformation defined by

1. If that is, if is considered a COMPLEX matrix, then the roots of in are So, has and as eigenpairs.
2. If that is, if is considered a REAL matrix, then has no solution in Therefore, if then has no eigenvalue but it has as characteristic values.

Remark 6.1.8   Note that if is an eigenpair for an matrix then for any non-zero is also an eigenpair for Similarly, if are eigenvectors of corresponding to the eigenvalue then for any non-zero it is easily seen that if , then is also an eigenvector of corresponding to the eigenvalue Hence, when we talk of eigenvectors corresponding to an eigenvalue we mean LINEARLY INDEPENDENT EIGENVECTORS.

Suppose is a root of the characteristic equation Then is singular and Suppose Then by Corollary 4.3.9, the linear system has linearly independent solutions. That is, has linearly independent eigenvectors corresponding to the eigenvalue whenever

EXAMPLE 6.1.9
1. Let with for Then is the characteristic equation. So, the eigenpairs are

2. Let Then Hence, the characteristic equation has roots That is is a repeated eigenvalue. Now check that the equation for is equivalent to the equation And this has the solution Hence, from the above remark, is a representative for the eigenvector. Therefore, HERE WE HAVE TWO EIGENVALUES MATHEND000# BUT ONLY ONE EIGENVECTOR.
3. Let Then The characteristic equation has roots Here, the matrix that we have is and we know that for every and we can CHOOSE ANY TWO LINEARLY INDEPENDENT VECTORS from to get and as the two eigenpairs.

In general, if are linearly independent vectors in then are eigenpairs for the identity matrix,

4. Let Then The characteristic equation has roots Now check that the eigenpairs are and 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 Then The characteristic equation has roots Hence, over the matrix has no eigenvalue. Over the reader is required to show that the eigenpairs are and

EXERCISE 6.1.10
1. Find the eigenvalues of a triangular matrix.
2. Find eigenpairs over for each of the following matrices:
and
3. Let and be similar matrices.
1. Then prove that and have the same set of eigenvalues.
2. Let be an eigenpair for and be an eigenpair for What is the relationship between the vectors and ?

[Hint: Recall that if the matrices and are similar, then there exists a non-singular matrix such that ]

4. Let be an matrix. Suppose that for all Then prove that is an eigenvalue of What is the corresponding eigenvector?
5. Prove that the matrices and have the same set of eigenvalues. Construct a matrix such that the eigenvectors of and are different.
6. Let be a matrix such that ( is called an idempotent matrix). Then prove that its eigenvalues are either 0 or or both.
7. Let be a matrix such that ( is called a nilpotent matrix) for some positive integer . Then prove that its eigenvalues are all 0 .

THEOREM 6.1.11   Let be an matrix with eigenvalues not necessarily distinct. Then and

Proof. Since are the eigenvalues of by definition,

 (6.1.5)

(6.1.5) is an identity in as polynomials. Therefore, by substituting in (6.1.5), we get

Also,
 (6.1.6) (6.1.7)

for some Note that the coefficient of comes from the product

So, by definition of trace.

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

 (6.1.8)

Therefore, comparing the coefficient of we have

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

EXERCISE 6.1.12
1. Let be a skew symmetric matrix of order Then prove that 0 is an eigenvalue of
2. Let be a orthogonal matrix .If , then prove that there exists a non-zero vector such that

Let be an matrix. Then in the proof of the above theorem, we observed that the characteristic equation is a polynomial equation of degree in Also, for some numbers it has the form

Note that, in the expression is an element of Thus, we can only substitute by elements of

It turns out that the expression

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 be a square matrix of order Then satisfies its characteristic equation. That is,

holds true as a matrix identity.

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

Remark 6.1.14
1. Let Then its characteristic polynomial is Also, for the function, and This shows that the condition for each eigenvalue of does not imply that
2. Suppose we are given a square matrix of order and we are interested in calculating where is large compared to Then we can use the division algorithm to find numbers and a polynomial such that

Hence, by the Cayley Hamilton Theorem,

That is, we just need to compute the powers of till

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

3. Let be a non-singular matrix of order Then note that and

This matrix identity can be used to calculate the inverse.
Note that the vector (as an element of the vector space of all matrices) is a linear combination of the vectors

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

THEOREM 6.1.16   If are distinct eigenvalues of a matrix with corresponding eigenvectors then the set is linearly independent.

Proof. The proof is by induction on the number of eigenvalues. The result is obviously true if 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 We prove the result for We consider the equation

 (6.1.9)

for the unknowns We have
 (6.1.10)

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

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

But the eigenvalues are distinct implies for We therefore get for Also, and therefore (6.1.9) gives

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 matrix are linearly independent.

EXERCISE 6.1.18
1. For an matrix prove the following.
1. and have the same set of eigenvalues.
2. If is an eigenvalue of an invertible matrix then is an eigenvalue of
3. If is an eigenvalue of then is an eigenvalue of for any positive integer
4. If and are matrices with nonsingular then and have the same set of eigenvalues.

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

2. Let and be matrices for which and
1. Do and have the same set of eigenvalues?
2. Give examples to show that the matrices and need not be similar.
3. Let be an eigenpair for a matrix and let be an eigenpair for another matrix
1. Then prove that is an eigenpair for the matrix
2. Give an example to show that if are respectively the eigenvalues of and then need not be an eigenvalue of
4. Let be distinct non-zero eigenvalues of an matrix Let be the corresponding eigenvectors. Then show that forms a basis of If then show that has the unique solution

A K Lal 2007-09-12