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 = [tex]
\begin{bmatrix} 8&5&-5\\5&8&-5\\15&15&-12\end{bmatrix}

[/tex]

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 [tex]
\begin{bmatrix} 1&1&1\\0&1&1\\0&0&1\end{bmatrix}

[/tex]

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

[tex]B=P^{-1}AP[/tex]

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