# Eigenvalue and Eigenvector problem

1. May 8, 2006

### Mathman23

Hi

Given a 3x3 matrix

$$A = \[ \left[ \begin{array}{ccc} 0 & 0 & 1+2i \\ 0 & 5 & 0 \\ 1-2i & 0 & 4 \end{array} \right]$$

I need to a another 3x3 which satisfacies

D = U^-1 A U

Step 1.

Finding the eigenvalues

$$0 = det(A- \lambda I ) = (0- \lambda)(\lambda - 5) (\lambda -4 ), \lambda = 5,4,0$$

step 2.

Finding the eigenvectors.

A vector which satisfies (A-\lambda I) v = 0

For \lambda = 5

p(\lambda = 5) = $$\[ \left[ \begin{array}{ccc} -5 & 0 & 1+2i \\ 0 & 0 & 0 \\ 1-2i & 0 & -1 \end{array} \right]$$ ~ $$\[ \left[ \begin{array}{ccc} 1 & 0 & -1/5-2/5i \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right]$$

How do I read the eigenvector from the reduced matrix ???

Sincerely Fred

Last edited: May 8, 2006
2. May 8, 2006

### daveb

Your eigenvalues are wrong. Find the characteristic equation by working it out completely, and you'll find eigenvalues of -1, 5, and 5.

3. May 8, 2006

### Mathman23

Is 5 then a what is called a double root ?

/Fred

4. May 8, 2006

### Hammie

Kind of. The eigenvalue with the value of five is said to have a multiplicity of two.

5. May 8, 2006

### Mathman23

I found the first eigenvector for lambda = -1 to be [1+2i, 0, -1]^T

To find for lambda = 5, do I row reduce the matrix A-5I ?

If I put that matrix into reduced echelon form, I get [0,0,0], but that can't be right?

/Fred

Last edited: May 8, 2006