Empty
  • Empty
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
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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UsukiDoll
  • UsukiDoll
both invertible matrices.. like non-singular...as in there's an inverse.
Miracrown
  • Miracrown
Essentially, we have to show that BAx = lambda x
UsukiDoll
  • UsukiDoll
O_O the Latex just exploded.

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Miracrown
  • Miracrown
\[(AB)^{-1} \space \space = B ^{-1} \space \space A ^{-1}\]
Miracrown
  • Miracrown
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
  • Empty
:)
Miracrown
  • Miracrown
Take y = Bx so y is just an intermediate vector variable So let's rewrite the ABx = lam x with y
Miracrown
  • Miracrown
|dw:1434863967669:dw|
Miracrown
  • Miracrown
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
  • nincompoop
no.
UsukiDoll
  • UsukiDoll
BAy =(lambda)xB ? like that?
Miracrown
  • Miracrown
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
  • Miracrown
\(\color{blue}{\text{Originally Posted by}}\) @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
  • UsukiDoll
BAy=(lambda)Bx
UsukiDoll
  • UsukiDoll
D": BAy=B(lambda)x
Miracrown
  • Miracrown
So you get lambda B x but now, you cannot switch B and x they will stay as Bx
Miracrown
  • Miracrown
BAy = (lambda) Bx
UsukiDoll
  • UsukiDoll
oh.. ok so it is BAy=(lambda)Bx
Miracrown
  • Miracrown
yes
Miracrown
  • Miracrown
the lambda is moved in front because it's a constant, so the matrix laws don't bother it
Miracrown
  • Miracrown
Now, remember how this started with saying "y = Bx"? So now on the right side, replace the Bx with y
UsukiDoll
  • UsukiDoll
BAy=(lambda)y
Miracrown
  • Miracrown
Yes :)
Miracrown
  • Miracrown
So by definition, lambda is an eigenvector of BA with eigenvector y
UsukiDoll
  • UsukiDoll
:O neat... it's like we started from the left side of the equation and then use theorems and definitions to try and get to the right side of the equation.
Miracrown
  • Miracrown
It kinda relies on making that correct substitution at the beginning. A lot of these proofs are fairly simple once you get them
UsukiDoll
  • UsukiDoll
somehow I understand doing proofs more in small groups rather than by myself T_____T!!!!!
Miracrown
  • Miracrown
Yep. There is actually a different way to prove it using the definition 0=(A - lambda I) I can quickly show you if you'd like
UsukiDoll
  • UsukiDoll
except Elementary Number Theory... I can do the proofs on my own because the theorems are easy to understand
Miracrown
  • Miracrown
You're not wrong
Empty
  • Empty
Yeah the substitution Bx=y was exactly what I was looking for, fun. =P I wouldn't mind watching or playing around with whatever you two want to continue showing or expanding upon.
UsukiDoll
  • UsukiDoll
<--- still a noob at proofs, but at least tries X_x
Miracrown
  • Miracrown
LOL usuki, you're more clever than me, I assure you! (;
Empty
  • Empty
Idk, Mira you're pretty clever, also nice to see you back, it's been a while. =P
Miracrown
  • Miracrown
It sure has, bud! Anyhoo another way using the formula for solving for eigenvalue rather than the definition of the eigenvector is:|dw:1434865293339:dw| That's the equation you solve to get eigenvalues. So now multiple by (AB) inverse on the right
Miracrown
  • Miracrown
|dw:1434865385356:dw| So now with the property of inverses from the very beginning, if we open the bracket, they will switch places
Miracrown
  • Miracrown
and I can be ditched because it's just the identity matrix |dw:1434865599306:dw| So what we need at the end is 0 = BA - lambda I
Miracrown
  • Miracrown
So here, because we have the B and the A inverses written without any other vectors what you want to do is multiply by B on the LEFT and by A on the RIGHT |dw:1434865665195:dw|
Miracrown
  • Miracrown
|dw:1434865717135:dw|
Miracrown
  • Miracrown
This looks better when it's done in several steps, but writing on the board is a little annoying..=/. So yes, the inverses go away and you are left with the BA But the trick is to get the order right :)

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