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Eigenvector Proof

  • #1

Homework Statement


Let B = S^-1 * A * S and x be an eigenvector of B belonging to an eigenvalue [tex]\lambda[/tex]. Show S*x is an eigenvector of A belonging to [tex]\lambda[/tex].


Homework Equations





The Attempt at a Solution


The only place I can think of to start, is that B*x = [tex]\lambda[/tex]*x.
However, even starting with that, I can't figure out where to go next.
Could someone point me in the right direction?
 

Answers and Replies

  • #2
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5,199

Homework Statement


Let B = S^-1 * A * S and x be an eigenvector of B belonging to an eigenvalue [tex]\lambda[/tex]. Show S*x is an eigenvector of A belonging to [tex]\lambda[/tex].


Homework Equations





The Attempt at a Solution


The only place I can think of to start, is that B*x = [tex]\lambda[/tex]*x.
However, even starting with that, I can't figure out where to go next.
Could someone point me in the right direction?
That's a decent start. Next, show that A(Sx) = [itex]\lambda[/itex]x. That's what it means to say that Sx is an eigenvector of A corresponding to [itex]\lambda[/itex].
 
  • #3
What do I use to show that?

The only new information I've got that might be helpful is that A = S * B * S^-1

Multiplying on the left by S gives A*S = S*B

After doing that, I'm stuck again. I feel like this is the right track, but I don't know how to relate this back to what I'm trying to prove.
 
  • #4
vela
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That's a decent start. Next, show that A(Sx) = [itex]\lambda[/itex]x. That's what it means to say that Sx is an eigenvector of A corresponding to [itex]\lambda[/itex].
Slight correction: You want to show that A(Sx) = [itex]\lambda[/itex](Sx)
What do I use to show that?

The only new information I've got that might be helpful is that A = S * B * S^-1

Multiplying on the left by S gives A*S = S*B

After doing that, I'm stuck again. I feel like this is the right track, but I don't know how to relate this back to what I'm trying to prove.
Multiply by x now.
 
  • #5
I think I got it now.

After multiplying by x, I have ASx = SBx.

Bx has already been shown equal to [tex]\lambda[/tex]x, so I substitute that in, giving

ASx = S[tex]\lambda[/tex]x

[tex]\lambda[/tex] can be moved to the other side of S since it's a scalar, giving ASx = [tex]\lambda[/tex]Sx.
 
  • #6
33,521
5,199
Right.
 

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