Prove the following results for the eigenvalues of a n * n matrix A.

Question

Prove the following results for the eigenvalues of a n * n matrix A.

(a) 0 is an eigenvalue for A if and only if A is non invertible.

(b) A and A^T have the same eigenvalues.

Answer 1

Hi Sammy,

 

The eigenvalue problem is

 

Av = λv

 

where A is a matrix, v is an eigenvector and λ is a number called an eigenvalue. This is a very special relationship. For an nXn matrix A, there are be at most n eigenvalues, and there is a special eigenvector corresponding to each eigenvalue.

 

To find eigenvalues, you have to express what is called a characteristic equation. So, you can write

 

Av = λIv

 

where I is the n X n identity matrix. Then, you subtract λIv from each side to form:

 

(A - λI)v = θ

 

where θ is the zero vector.

 

If a matrix times a nonzero vector equals the zero vector, then that matrix must be singular. This is a theorem.

 

A itself it generally nonsingular. But, A - λI is the matrix that has to be singular. Now, if λ = 0 is an eigenvalue, then A - λI becomes simply A. So,

 

(A - λI)v = θ becomes Av = θ. This means that A itself must be singular. A singular matrix is also called an non-invertible matrix, to use your nomenclature.

 

(b) Let's go back to the following equation from part (a).

 

(A - λI)v = θ

 

To find eigenvalues, you form the characteristic equation by writing det(A - λI)  = 0, where det means determinant. This equation forms a polynomial in λ equal to zero that you need to solve for λ to find the eigenvalues. Lets take the transpose of the above eqpression.

 

det(A - λI)^T  = 0

 

det(A^T - λ^TI^T)  = 0

 

But, I^T = I. Also, since λ is a scalar, clearly λ^T = λ.

 

So, this becomes det(A^T - λI)  = 0.

 

This is the characteristic polynomial for the eigenvalue equation:

 

A^Tv = λv

 

Therefore, A^T and A have the same eigenvalues.

Answer 2

Hi Sammy!

 

To start, recall three things:

1) the determinant of a matrix equals the product of all its eigenvalues.  

2) in order for a matrix to be invertible, it must be a square matrix with a non-zero determinant

3) the Multiplication Property of Zero, which says that if you multiply any number by 0, the answer is 0.

 

So...

 

a) if 0 is an eigenvalue for A, then by #3, the product of all the eigenvalues must equal 0.  Therefore, by #1, the determinant must equal 0.  This violates the requirements set forth by #2 above, which therefore means that the matrix is non-invertible.  (The same argument can be made in reverse as well, which is why it's an if and only if statement.)

 

b) We can go into much more depth, but the nuts and bolts of this are: when creating a transpose matrix, the original columns become rows and the original rows become columns.  A and A^T have the same determinant and, it can be proved, the same eigenvalues.