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KonradZuse Group Title

diagonalizing a matrix

  • 2 years ago
  • 2 years ago

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  1. KonradZuse Group Title
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    Trying to figure out how this works, wolfram is giving me a different setup than my book... http://www.wolframalpha.com/input/?i=diagonalize+%7B%7B1%2C0%7D%2C%7B6%2C-1%7D%7D The book says p = [1/3][0] [1 ][1] and P^-1AP = [1][0] [0][-1] Which is the inverse of J...

    • 2 years ago
  2. KonradZuse Group Title
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    @UnkleRhaukus any idea sir?

    • 2 years ago
  3. UnkleRhaukus Group Title
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    \[S=P^{-1}\]?

    • 2 years ago
  4. KonradZuse Group Title
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    crap I forgot to post what the book said h/o.

    • 2 years ago
  5. KonradZuse Group Title
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    • 2 years ago
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  6. KonradZuse Group Title
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    @UnkleRhaukus

    • 2 years ago
  7. UnkleRhaukus Group Title
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    whats your question?

    • 2 years ago
  8. KonradZuse Group Title
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    I'm trying to do #14.

    • 2 years ago
  9. KonradZuse Group Title
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    I'm confused on how to do this... Wolfram gives a different definition than my book, and I'm not really sure what my book is trying to show...

    • 2 years ago
  10. UnkleRhaukus Group Title
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    http://www.wolframalpha.com/input/?i=diagonalise+matrix+%7B%7B1%2C0%2C0%7D%2C%7B+0%2C1%2C1%7D%2C%7B0%2C1%2C1%7D%7D

    • 2 years ago
  11. KonradZuse Group Title
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    mhm yeah I entered that in wolfram, but I'm not sure what SJ and M are... They are different from P and A....

    • 2 years ago
  12. KonradZuse Group Title
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    @lgbasallote

    • 2 years ago
  13. lgbasallote Group Title
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    @KonradZuse not my habit to reply to questions already being answered

    • 2 years ago
  14. UnkleRhaukus Group Title
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    i think that S and P are inverses of each other

    • 2 years ago
  15. KonradZuse Group Title
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    that's what I was looking at. @iggy we are in the process yo...

    • 2 years ago
  16. KonradZuse Group Title
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    but according to wolfram and the example 7 I showed above, there is no 1/3 in wolfram.... Is it just a different way to do it....?

    • 2 years ago
  17. KonradZuse Group Title
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    err # 13.

    • 2 years ago
  18. KonradZuse Group Title
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    yeah Idk about S and P...

    • 2 years ago
  19. KonradZuse Group Title
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    it look slike J = P^-1AP just flipped around... Same thing happened when I did #15.

    • 2 years ago
  20. KonradZuse Group Title
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    :(

    • 2 years ago
  21. phi Group Title
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    diagonalizing a matrix M uses eigenvectors x and eigenvalues \(\lambda\) the "big equation" is \( M x = \lambda x\) if we put all the eigenvectors x into the columns of a matrix P and the eigenvalues on the diagonal of matrix \( \Lambda \) we can say \( MP = P\Lambda \) note the order we must multiply P and \( \Lambda \) to get the right result multiplying by the inverse of P we have \(\Lambda = P^{-1}MP \) This says we can diagonalize matrix M by multiplying by the eigenvector matrix P It returns a diagonal matrix with entries being the eigenvalues of M

    • 2 years ago
  22. phi Group Title
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    When we do this diagonalizing, note that there is flexibility in defining an eigenvector. Eigenvectors represent a "direction", and if we scale them, they are still an eigenvector. Example: if eigenvector x= [ 1/3 1] corresponds to eigenvalue λ and we scale x by 3 to get x' = [1 3] , it is still an eigenvector associated with λ M(3x)=λ(3x) still works. We would scale x to get rid of fractions. Also, when diagonalizing, there is no forced order on the eigenvalues in matrix \(\Lambda\) The only requirement is that the eigenvalue in position n,n of matrix \(\Lambda\) corresponds to the eigenvector in column n of matrix P Often, though, people put the eigenvalues in ascending (or descending ) order. The point is we can get different variations of P and Λ, and all are correct, unless the order of the eigenvalues in Λ are stipulated to be in some order.

    • 2 years ago
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