Eigenvectors and Eigenvalues

Question 1

Consider a linear transformation Tx = Ax from R^n to R^n. What does it mean for a matrix A to be diagonalizable?

Answer 1

A or T is said to be diagonalizable if the matrix B of T with respect to some basis is diagonalizable. The matrix is diagonalizable if (and only if) A is similar to some diagonal matrix B, meaning there exists an invertible matrix S such that S^-1AS = B is diagonal.

Question 2

How do you diagonalize a square matrix A?

Answer 2

In order to diagonalize a square matrix A, you must find an invertible matrix S and a diagonal matrix B such that S^-1AS = B

Question 3

Define an eigenvector and its eigenvalue

Answer 3

A nonzero vector v in R^n is called an eigenvector of A (or T) if Av = (lambda)v for some scalar lambda which is its associated eigenvalue

Question 4

Define an eigenbasis

Answer 4

A basis v1,…,vn of R^n is called an eigenbasis for A (or T) if the vectors v1,…,vn are eigenvectors of A, meaning that Av1 = (lambda)v1,….Avn = (lambda)vn for some scalars lambda

Question 5

Prove that A^2v = (lambda)^2v , A^3v = (lambda)^3v ,….. A^mv = (lambda)^mv

Answer 5

Lets consider the base case
A^1(v) = Av = (lambda)v = (lambda)^1(v)
Now
A^(m+1)(v) = A(A^m(v)) = A((lambda)^m(v)) = (lambda)^m(Av) = (lamda)^m(lambda)(v) = (lambda)^(m+1)(v)

Question 6

Define diagonalization for an invertible n x n matrix A using eigenbases

Answer 6

The matrix A is diagonalizable if (and only if) there exists an eigenbasis for A. If v1,….,vn is an eigenbasis for A, with Av=(lambda)v ….Avn=(lambda)n(vn), then the matrices
S = [v1,v2,….vn] and
B = [lambda1,lambda2,….lambdan] will diagonalize A meaning that S^-1AS = B

Question 7

Going back to defining diagonalization for a given matrix A using eigenbases, what can you say about the characteristics of the matrices S and B?

Answer 7

If matrices S and B diagonalize A, then the columns of S will form an eigenbasis for A, and the diagonal entries of B will be the associated eigenvalues.

Question 8

For an eigenbasis v1,….,vn for A, prove AS = SB

Answer 8

AS = A[v1,v2,….vn] = [Av1,Av2,….Avn]
=[lambda1v1,lambda2,v2,…..lambda_n(v_n)]
=[v1,v2,…,v_n][lambda1,lambda2,….lambda_n] = SB

Question 9

What makes lambda an eigenvalue of A?

Answer 9

Lambda is an eigenvalue of A if (and only if)

det(A-lambda(I)) = 0

Question 10

Given a triangular matrix, what are its eigenvalues?

Answer 10

The eigenvalues are simply the diagonal entries

Question 11

What is the trace of a square matrix?

Answer 11

The sum of the diagonal entries

Question 12

Define the characteristic equation of a 2 x 2 matrix A.

Answer 12

det(A-(lambda)(I)) = (lambda)^2-(trA)lambda + detA = 0

Question 13

Define the characteristic polynomial of any n x n matrix A

Answer 13

(-)^n(lambda^n)+(-1)^(n-1)(trA)(lambda)^(n-1)+….+detA

Question 14

Define the algebraic multiplicity of an eigenvalue

Answer 14

We say that an eigenvalue of a square matrix A has algebraic multiplicity k is lambda is a root of multiplicity k of the characteristic polynomial f(lambda) denoted by almu(lambda)
Ex.
f(lambda) = (5-lambda)^3 (4-lambda)^2
almu(5) = 3

Question 15

What can you say about the number of eigenvalues in an n x n matrix A?

Answer 15

An n x n matrix has at most n real eigenvalues, even if they are counted with their algebraic multiplicities. If n is ODD, then an n x n matrix has at least one real eigenvalue

Question 16

If an n x n matrix A has eigenvalues lambda1, lambda2,….lambda_n, define the relationship between the detA and A’s eigenvalues along with the trace of A and A’s eigenvalues

Answer 16

detA = the product of the eigenvalues

trA = the sum of the eigenvalues

Question 17

Define an eigenspace

Answer 17

Consider an eigenvalue lambda of an n x n matrix A. Then the kernel of the matrix A-lambda(I) is called the eigenspace associated with lambda, denoted by E_lambda
E_lambda = ker(A-(lambda)(I)) = {v in R^n: Av=lambdav}

Question 18

State the general steps in finding an eigenspace for a given square matrix A.

Answer 18

1. Find the eigenvalues of A
2. Determine the ker(A-(lambda)(I)) for each eigenvalue
3. These vectors form an eigenbasis for A. Check AS=SB to make sure

Question 19

Define the geometric multiplicity of a determined eigenvalue.

Answer 19

Consider an eigenvalue lambda of an n x n matrix A. The dimension of eigenspace
E_lambda = ker(A-lambda(I)) is called the geometric multiplicity of eigenvalue lambda, denoted by gemu(lambda)
Also helpful to remember
gemu(lambda)=nullity(A-lambda(I))=n-rank(A-lambda(I))
Sum of gemu = s

Question 20

Using geometric multiplicities, how do we know matrix A is diagonalizable?

Answer 20

Matrix A is diagonalizable if (and only if) the geometric multiplicities of the eigenvalues add up to n, that is s = n

Question 21

If an n x n matrix A has n eigenvalues, what can you say about matrix A?

Answer 21

The matrix is diagonalizable. We can construct an eigenbasis by finding an eigenvector for each eigenvalue.

Question 22

What are 4 important properties of the eigenvalues of similar matrices A and B?

Answer 22

1. Matrices A and B have the same characteristic polynomial
2. rankA=rankB and nullity A = nullity B
3. Matrices A and B have the same eigenvalues, with the same algebraic and geometric multiplicites (however eigenvectors need not be the same)
4. Matrices A and B have same det and trace

Question 23

Prove that if matrix A and B are similar, they have the same characteristic polynomial equation

Answer 23

B=S^-1AS then det(B-lambda(I))=det(S^-1AS-lambda(I))=det(S^-1(A-lambda(I)S)=det(S^-1)det(S)det(A-lambda(I)) = det(A-lambda(I))