Proof: Eigenvector of B Belonging to \lambda for A*S*x

[WTFsandwich]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

 

[Mark44]

Homework Statement

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

Homework Equations

The Attempt at a Solution

The only place I can think of to start, is that B*x = $$\lambda$$*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) = $\lambda$x. That's what it means to say that Sx is an eigenvector of A corresponding to $\lambda$.

 

[WTFsandwich]

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.

 

[vela]

That's a decent start. Next, show that A(Sx) = $\lambda$x. That's what it means to say that Sx is an eigenvector of A corresponding to $\lambda$.

Slight correction: You want to show that A(Sx) = $\lambda$(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.

 

[WTFsandwich]

I think I got it now.

After multiplying by x, I have ASx = SBx.

Bx has already been shown equal to $$\lambda$$x, so I substitute that in, giving

ASx = S$$\lambda$$x

$$\lambda$$ can be moved to the other side of S since it's a scalar, giving ASx = $$\lambda$$Sx.

 

[Mark44]

Right.