Eigenvectors & subspace spanning

[AngrySaki]

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

Homework Statement

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.]

Homework Equations

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

 

[vela]

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.

 

[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

 

[Office_Shredder]

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

 

[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?

 

[HallsofIvy]

Thanks, that helps a lot.

Does this makes sense as an answer?


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

More correctly, "if X is an eigenvector of A, with eigenvalue \lambda, then
(\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