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  • one year ago

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

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  1. UsukiDoll
    • one year ago
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    both invertible matrices.. like non-singular...as in there's an inverse.

  2. Miracrown
    • one year ago
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    Essentially, we have to show that BAx = lambda x

  3. UsukiDoll
    • one year ago
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    O_O the Latex just exploded.

  4. Miracrown
    • one year ago
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    \[(AB)^{-1} \space \space = B ^{-1} \space \space A ^{-1}\]

  5. Miracrown
    • one year ago
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    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...

  6. Empty
    • one year ago
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    :)

  7. Miracrown
    • one year ago
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    Take y = Bx so y is just an intermediate vector variable So let's rewrite the ABx = lam x with y

  8. Miracrown
    • one year ago
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    |dw:1434863967669:dw|

  9. Miracrown
    • one year ago
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    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? :)

  10. nincompoop
    • one year ago
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    no.

  11. UsukiDoll
    • one year ago
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    BAy =(lambda)xB ? like that?

  12. Miracrown
    • one year ago
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    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.

  13. Miracrown
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @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.

  14. UsukiDoll
    • one year ago
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    BAy=(lambda)Bx

  15. UsukiDoll
    • one year ago
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    D": BAy=B(lambda)x

  16. Miracrown
    • one year ago
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    So you get lambda B x but now, you cannot switch B and x they will stay as Bx

  17. Miracrown
    • one year ago
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    BAy = (lambda) Bx

  18. UsukiDoll
    • one year ago
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    oh.. ok so it is BAy=(lambda)Bx

  19. Miracrown
    • one year ago
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    yes

  20. Miracrown
    • one year ago
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    the lambda is moved in front because it's a constant, so the matrix laws don't bother it

  21. Miracrown
    • one year ago
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    Now, remember how this started with saying "y = Bx"? So now on the right side, replace the Bx with y

  22. UsukiDoll
    • one year ago
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    BAy=(lambda)y

  23. Miracrown
    • one year ago
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    Yes :)

  24. Miracrown
    • one year ago
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    So by definition, lambda is an eigenvector of BA with eigenvector y

  25. UsukiDoll
    • one year ago
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    :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.

  26. Miracrown
    • one year ago
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    It kinda relies on making that correct substitution at the beginning. A lot of these proofs are fairly simple once you get them

  27. UsukiDoll
    • one year ago
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    somehow I understand doing proofs more in small groups rather than by myself T_____T!!!!!

  28. Miracrown
    • one year ago
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    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

  29. UsukiDoll
    • one year ago
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    except Elementary Number Theory... I can do the proofs on my own because the theorems are easy to understand

  30. Miracrown
    • one year ago
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    You're not wrong

  31. Empty
    • one year ago
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    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.

  32. UsukiDoll
    • one year ago
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    <--- still a noob at proofs, but at least tries X_x

  33. Miracrown
    • one year ago
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    LOL usuki, you're more clever than me, I assure you! (;

  34. Empty
    • one year ago
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    Idk, Mira you're pretty clever, also nice to see you back, it's been a while. =P

  35. Miracrown
    • one year ago
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    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

  36. Miracrown
    • one year ago
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    |dw:1434865385356:dw| So now with the property of inverses from the very beginning, if we open the bracket, they will switch places

  37. Miracrown
    • one year ago
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    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

  38. Miracrown
    • one year ago
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    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|

  39. Miracrown
    • one year ago
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    |dw:1434865717135:dw|

  40. Miracrown
    • one year ago
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    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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