How can I show that y is an eigenvalue of B?

[gimpy]

Ok well i have two questions.

1) If B = P^-1AP and let X be an eigenvector of A corresponding to the eigenvalue y. Show that y is an eigenvalue of B and find a corresponding eigenvector.

This is what i did.
AX=yI and since B = P^-1Ap -> A = PBP^-1
so (PBP^-1)X=yI
Now this is the part where i get lost. Am i on the right track?

2) If A and B are nxn matrices, A is invertable, show that BA is similar to AB.

So BA = P^-1ABP because BA is similar to AB. But I am kinda lost now. I'm sure i have to do something with the fact that A is invertable. Umm... A^-1...

 

[Hurkyl]

1)

AX=yI

Here's your mistake; what you should have is Ax = yx.

Recall that the definition is:

\vec{v} is an eigenvector of A corresponding to eigenvalue \lambda iff A\vec{v} = \lambda \vec{v}.


2)

Ok, you know to be similar, you need

BA = P^{-1}ABP

What's the simplest guess as to what P should be to make this equation hold?

 

[gimpy]

Originally posted by Hurkyl
1)



Here's your mistake; what you should have is Ax = yx.

Recall that the definition is:

\vec{v} is an eigenvector of A corresponding to eigenvalue \lambda iff A\vec{v} = \lambda \vec{v}.


2)

Ok, you know to be similar, you need

BA = P^{-1}ABP

What's the simplest guess as to what P should be to make this equation hold?

lol

i meant A\vec{v} = \lambda \vec{v}. Stupid typo by me. But i still don't get it [b(]

And for the other one, I am lost as to what P should be to make this equation hold. It says that A and B are nxn matrices. A is invertable. show that BA is similar to AB.. umm...

 

[Hurkyl]

1)

You know that you need to find something of the form B\vec{w}=\lambda\vec{w}...

but you have something of the form P^{-1}BP\vec{v} = y \vec{v}...

The first thing I notice is that the LHS must be the matrix B times some vector... so can you rewrite what you do have in such a way that the LHS is B times some vector?


2)

You need to find a P such that:

BA = P^{-1}ABP


The first thing I notice is that I need to have an A on the right of B... you have a B on the right hand side, can you select a value for P so that we have an A on the right of the B?