Linear Algebra - Eigenvector question

1. Feb 15, 2010

zeion

1. The problem statement, all variables and given/known data

The linear operator T on R2 has the matrix
$$\begin{bmatrix}4&-5\\-4&-3 \end{bmatrix}$$ relative to the basis {(1,2), (0,1)}
Find the eigenvalues of T, and obtain an eigenvector corresponding to each eigenvalue.

2. Relevant equations

3. The attempt at a solution

So I solved the eigenvalues to be $$\lambda = 8, \lambda = -1$$

I know I normally just sub in the lambda to the matrix and then solve for the null space to get the eigenvector, but how do I do it with a different basis?

2. Feb 16, 2010

vela

Staff Emeritus
Is that the right matrix? I get different eigenvalues than you do.

Oh, and to answer your question, you just solve for them like you usually do. The eigenvectors you find will be represented in the given basis.

Last edited: Feb 16, 2010
3. Feb 16, 2010

zeion

Oh sorry it's supposed to be a 3 not -3..
So then I solve for it like this?

For lambda = 8
$$\begin{bmatrix}4&-5\\-4&3 \end{bmatrix} \rightarrow \begin{bmatrix}-4&-5\\-4& -5 \end{bmatrix} \rightarrow \begin{bmatrix}1&5/4\\0& 0 \end{bmatrix}$$

Then the vector is (-5/4, 1)?

But the answer in the book is (-5, -6) :/

Ok I think I see what they did.. just multiplied it by 4 to get rid of the fraction and then wrote it wrt to the standard basis?

Last edited: Feb 16, 2010
4. Feb 16, 2010

vela

Staff Emeritus
Yup, that's what they did.

5. Feb 21, 2010

TorcidaS

Hi, could you just help me with the part where they write it wrt to the basis? It sort of confuses me.

I understand we get (-5/4 1), and once clearing the fraction we get (-5, 4). But how does this jump to (-5, -6)?

6. Feb 22, 2010

zeion

-5(1,2) + 4(0,1)