# Eigenvectors & subspace spanning

1. Jun 25, 2010

### AngrySaki

The question is at the end of a chapter on spanning vector spaces.

1. The problem statement, all variables and given/known data
Let P denote an invertible n x n matrix.
If $$\lambda$$ is a number, show that

$$E_{\lambda}(PAP^{-1}) = \left\{PX | X\;is\;in\;E_{\lambda}(A)\right\}$$

for each n x n matrix A. [Here $$E_{\lambda}(A)}$$ is the set of eigenvectors of A.]

2. Relevant equations

3. The attempt at a solution

I'm having trouble understanding what the equality means, or how to read it.
The left side looks to be the eigenvectors of a diagonal matrix, which I think are always the columns of an identity matrix.

On the right side, I assume the matrix P would be the eigenvectors of A, so I think it's the span of the products of P multiplied by each of it's columns.

I don't know what to make of those ideas, so I think I'm either missing something about eigenvectors or spanning (very possible), or reading the question wrong.

Thanks

2. Jun 25, 2010

### vela

Staff Emeritus
The matrix PAP-1 is diagonal if the columns of P are the eigenvectors of A, but you don't know anything about P other than it's invertible.

Let B=PAP-1. What the question is asking you to show is that the eigenvectors of B are of the form Px and that Px is an eigenvector of B, where x is an eigenvector of A.

3. Jun 26, 2010

### AngrySaki

Thanks, that helps a lot.

Does this makes sense as an answer?

Eigenvalues of A can be written:
$$(\lambda I -A)X=0$$

So the right side of the original equation is:
$$P(\lambda I -A)X=0$$

Move the P inside:
$$(\lambda P -PA)X=0$$

Multiply by I:
$$(\lambda P -PA)P^{-1}PX=0$$

Move P inverse inside:
$$(\lambda PP^{-1} -PAP^{-1})PX=0$$

Eigenvalues of PAP-1 are of the form PX:
$$(\lambda I -PAP^{-1})PX=0$$

4. Jun 26, 2010

### Office_Shredder

Staff Emeritus
Except they're not eigenvalues, they're eigenvectors.

5. Jun 26, 2010

### AngrySaki

Oh, oops. I mean to write eigenvectors.

Just to be clear, does the method makes sense if I had written the word eigenvectors instead of eigenvalues?

6. Jun 26, 2010

### HallsofIvy

More correctly, "if X is an eigenvector of A, with eigenvalue $\lambda$, then
$$(\lambda I- A)X= 0$$"