Linear Algebra - Characteristic polynomials and similar matrices question

[zeion]

Homework Statement

For each matrix A below, let T be the linear operator on R3 thathas matrix A relative to the basis A = {(1,0,0), (1,1,0), (1,1,1)}. Find the algebraic and geometric multiplicities of each eigenvalues, and a basis for each eigenspace.

a) A = <br /> \begin{bmatrix} 8&amp;5&amp;-5\\5&amp;8&amp;-5\\15&amp;15&amp;-12\end{bmatrix} <br /> <br />

Homework Equations

The Attempt at a Solution

So I tried to find the eigenvalues normally and turns out that was pretty hard.. So I know that similar matrices have the same eigenvalues, then can I just take the eigenvalues of the matrix <br /> \begin{bmatrix} 1&amp;1&amp;1\\0&amp;1&amp;1\\0&amp;0&amp;1\end{bmatrix} <br /> <br />

since it is similar to A? Or is it similar?

 

[VeeEight]

The eigenvalues of similar matrices are the same but the eigenvectors may be different.

 

[vela]

The second matrix isn't similar to A. Two matrices A and B are similar if you can write

B=P^{-1}AP

for some invertible matrix P. You could use your matrix with the basis-vector columns as P.

 

[zeion]

I calculated B but that doesn't seem to make finding eigenvalues any easier..?

 

[vela]

You just need to work it out. It's only a 3x3 matrix after all.

 

[zeion]

Okay nice it seems it's just because I made a mistake in calculating the inverse of P..
turns out it's much easier to find the eigenvalues of B.

..or not. I don't understand why there is suddenly such a computational question when everything else is hardly as hard..

 

[zeion]

Can I row-reduce a matrix before subtracting lambda and then find the determinant? Or do I have to subtract lambda first?

 

[vela]

You have to subtract \lambda first. Think about it. You can reduce any invertible matrix to the identity matrix. If you then subtracted \lambda, all the eigenvalues would be 1, which is obviously not the case for every invertible matrix.