# Linear Algebra: Similar matrices *Prove*

## Homework Statement

if A is similar to B with P^-1 A P = B and x is an eigenvector of A, prove that P^-1x is an eigenvector of B

## The Attempt at a Solution

so far I did this:

A = PBP^-1
Ax = PBP^-1x
P^-1Ax = BP^-1x
B^-1P^-1Ax = P^-1x

is that sufficient to prove that P^-1x is the eigenvector of B? ??

I don't think so .. if yes please explain how.. or else please give a proper "proof" .

I guess the reason I don't think this is right is becaues I am not able to bring it to the format of Ax = tx where t is an eigenvalue and x is the eigenvector. Any help would be appreciated.

Final exam on tuesday ... :(

## The Attempt at a Solution

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jbunniii
Homework Helper
Gold Member

## Homework Statement

if A is similar to B with P^-1 A P = B and x is an eigenvector of A, prove that P^-1x is an eigenvector of B

so far I did this:

A = PBP^-1
Ax = PBP^-1x
P^-1Ax = BP^-1x
B^-1P^-1Ax = P^-1x

is that sufficient to prove that P^-1x is the eigenvector of B? ??

I don't think so .. if yes please explain how.. or else please give a proper "proof" .
No, but you're on the right track. You got as far as

$$P^{-1}Ax = BP^{-1}x$$

Now you need to use the fact that x is an eigenvector of A, so you can replace Ax in this equation with something else, and then you're essentially done.

No, but you're on the right track. You got as far as

$$P^{-1}Ax = BP^{-1}x$$

Now you need to use the fact that x is an eigenvector of A, so you can replace Ax in this equation with something else, and then you're essentially done.
so that would mean:
Ax = tx Where t is the eigenvalue:

so $$P^{-1}tx = BP^{-1}x$$
$$tx = PBP^{-1}x$$
and we know $$PBP^{-1}$$ is A
so we essentially have $$tx = Ax$$

but how does this show that $$P^{-1}x$$ is the eigenvector of B?

jbunniii
Homework Helper
Gold Member
so that would mean:
Ax = tx Where t is the eigenvalue:

so $$P^{-1}tx = BP^{-1}x$$
$$tx = PBP^{-1}x$$
and we know $$PBP^{-1}$$ is A
so we essentially have $$tx = Ax$$

but how does this show that $$P^{-1}x$$ is the eigenvector of B?
You can see it from the first equation that you wrote:

$$P^{-1}tx = BP^{-1}x$$

Rearrange the left hand slightly, then add some parentheses in appropriate places on both sides, until you see something that looks like an eigenvalue/eigenvector relationship. Notice that this shows something interesting beyond what was asked for: in particular, compare the eigenVALUE of B corresponding to $$P^{-1}x$$ to the eigenvalue of A corresponding to x.

You can see it from the first equation that you wrote:

$$P^{-1}tx = BP^{-1}x$$

Rearrange the left hand slightly, then add some parentheses in appropriate places on both sides, until you see something that looks like an eigenvalue/eigenvector relationship.
I SEE IT!!!

$$t(P^{-1}x) = B(P^{-1}x)$$

cool!
Thanks a lot.

jbunniii
Homework Helper
Gold Member
I SEE IT!!!

$$t(P^{-1}x) = B(P^{-1}x)$$

cool!
Thanks a lot.
Isn't that "ah ha!" moment great?

Cheers!

absolutely!