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Linear Algebra: Similar matrices *Prove*

  • Thread starter tua96426
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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 ... :(

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
jbunniii
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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

[tex]P^{-1}Ax = BP^{-1}x[/tex]

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.
 
  • #3
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No, but you're on the right track. You got as far as

[tex]P^{-1}Ax = BP^{-1}x[/tex]

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

but how does this show that [tex] P^{-1}x [/tex] is the eigenvector of B?
 
  • #4
jbunniii
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so that would mean:
Ax = tx Where t is the eigenvalue:

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

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

[tex]P^{-1}tx = BP^{-1}x[/tex]

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 [tex]P^{-1}x[/tex] to the eigenvalue of A corresponding to x.
 
  • #5
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You can see it from the first equation that you wrote:

[tex]P^{-1}tx = BP^{-1}x[/tex]

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!!!

[tex] t(P^{-1}x) = B(P^{-1}x) [/tex]

cool!
Thanks a lot.
 
  • #6
jbunniii
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I SEE IT!!!

[tex] t(P^{-1}x) = B(P^{-1}x) [/tex]

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

Cheers!
 
  • #7
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absolutely!
 

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