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

Essentially, we have to show that BAx = lambda x

O_O the Latex just exploded.

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

:)

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

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no.

BAy =(lambda)xB ? like that?

BAy=(lambda)Bx

D": BAy=B(lambda)x

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

BAy = (lambda) Bx

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

yes

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

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

BAy=(lambda)y

Yes :)

So by definition, lambda is an eigenvector of BA with eigenvector y

somehow I understand doing proofs more in small groups rather than by myself T_____T!!!!!

You're not wrong

<--- still a noob at proofs, but at least tries X_x

LOL usuki, you're more clever than me, I assure you! (;

Idk, Mira you're pretty clever, also nice to see you back, it's been a while. =P

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