Linear Algebra: Similar matrices *Prove*

[tua96426]

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

Homework Equations

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 ... :(

 

[jbunniii]

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.

 

[tua96426]

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]

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.

 

[tua96426]

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]

I SEE IT!

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

cool!
Thanks a lot.

Isn't that "ah ha!" moment great?

Cheers!

 

[tua96426]

absolutely!