1. ## Eigenvalue

I have this matrix:

2 1 -1 0
0 4 -2 0
0 3 -1 0
0 3 -2 1

I need to find the eigenvalues.

So I get the matrix

(x-2) 1 -2 1
0 (x-4) -4 2
0 3 (x+4) 3
0 3 -5 (x-4)

I take the determinant:
(x-2)*[(x-4)(x+4)(x-4)-36-30-6(x+4)+15(x-4)+12(x-4)]=
(x-2)*[(x-4)(x+4)(x-4)-66-6(x+4)+27(x-4)]

Clearly, 2 is an eigenvalue. I get messy other ones though so it doesn't seem right. How would you solve a complex cubic equation anyway?

Can anyone check this?

2. OK. Forget the top.

How could I show that the following two matrices are similar without mentioning eigenvalues?

2 1 -1 0
0 4 -2 0
0 3 -1 0
0 3 -2 1

and

2 1 -2 1
0 4 -4 2
0 3 -4 3
0 3 -5 4

They are similar, but what is the matrix P such that A=inv(P)*B*P?

3. Originally Posted by SwedishMan
I have this matrix:

2 1 -1 0
0 4 -2 0
0 3 -1 0
0 3 -2 1

I need to find the eigenvalues.

So I get the matrix

(x-2) 1 -2 1
0 (x-4) -4 2
0 3 (x+4) 3
0 3 -5 (x-4)
why not:

Code:
(2-x)    1     -1     0
0    (4-x)   -2     0
0      3   (-1-x)   0
0      3     -2    (1-x)
which has determinant (x-1)(x-2)(x^2-3x+2)

RonL

4. Yeah I figured that one out. I just want to save a lot of work and actually find out how to do this (show two matrices are similar). It would save time on an exam if he ever gave us another problem like this.

Basically I had to find the eigenvalues and eigenvectors of the two matrices shown in my second post. I'm almost 100% sure they are similar. Just how do I show it?