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 $B$?