Here's a fun question I found that has a simple fun answer, try it out.

$\bar x$ is an eigenvector with corresponding eigenvalue$\lambda$ of the matrix product AB where A and B are both invertible matrices:

$$AB \bar x = \lambda \bar x$$

Show that $\lambda$ is an eigenvalue of the matrix BA

Question

Here's a fun question I found that has a simple fun answer, try it out.

$\bar x$ is an eigenvector with corresponding eigenvalue$\lambda$ of the matrix product AB where A and B are both invertible matrices:

$$AB \bar x = \lambda \bar x$$

Show that $\lambda$ is an eigenvalue of the matrix BA

usukidoll · Answer

both invertible matrices.. like non-singular...as in there's an inverse.

miracrown · Answer

Essentially, we have to show that BAx = lambda x

usukidoll · Answer

O_O the Latex just exploded.

miracrown · Answer

$$(AB)^{-1} \space \space = B ^{-1} \space \space A ^{-1}$$

miracrown · Answer

I know it will be a trick with the order of multiplication. And I just realized that x doesn't need to be an eigenvector of BA, just the eigenvalue has to be the same...

empty · Answer

:)

miracrown · Answer

Take y = Bx so y is just an intermediate vector variable
So let's rewrite the ABx = lam x with y

miracrown · Answer

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miracrown · Answer

Now multiply both side by B. And this is the part that you have to know the commutative property well. Can you multiply these for me? :)

nincompoop · Answer

no.

usukidoll · Answer

BAy =(lambda)xB ? like that?

miracrown · Answer

When you multiply matrices, you know that AB and BA will not always be equal.So that means it matters which side you multiply by. So if you multiply from the LEFT side...
you have to keep B on the left.

miracrown · Answer

$\color{blue}{\text{Originally Posted by}}$ @UsukiDoll
BAy =(lambda)xB ? like that?
$\color{blue}{\text{End of Quote}}$
When you first multiply by B, yes. But it has to stay left of 'x'
So B lambda x And then because lambda is a constant, that how you can justify it moving it in front of B.

usukidoll · Answer

BAy=(lambda)Bx

usukidoll · Answer

D": BAy=B(lambda)x

miracrown · Answer

So you get lambda B x but now, you cannot switch B and x they will stay as Bx

miracrown · Answer

BAy = (lambda) Bx

usukidoll · Answer

oh.. ok so it is BAy=(lambda)Bx

miracrown · Answer

yes

miracrown · Answer

the lambda is moved in front because it's a constant, so the matrix laws don't bother it

miracrown · Answer

Now, remember how this started with saying "y = Bx"?
So now on the right side, replace the Bx with y

usukidoll · Answer

BAy=(lambda)y

miracrown · Answer

Yes :)