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KonradZuse Group Title

In Exercises 2–9, find a matrix P that orthogonally diagonalizes A, and determine P^-1 AP.

  • one year ago
  • one year ago

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  1. KonradZuse Group Title
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    \[A = \left[\begin{matrix}6 & -2 \\ -2 & 3\end{matrix}\right]\]

    • one year ago
  2. KonradZuse Group Title
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    @phi @kropot72

    • one year ago
  3. phi Group Title
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    Are you learning about SVD (singular value decomposition)?

    • one year ago
  4. KonradZuse Group Title
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    nope....

    • one year ago
  5. phi Group Title
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    but A is symmetric, so I think its eig vectors are orthogonal

    • one year ago
  6. KonradZuse Group Title
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    basically we learned about diagonalization in chaptrer 5 and now it';s back in 7. I did the first part fidning the eigenvalues.

    • one year ago
  7. KonradZuse Group Title
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    but I'm not sure about this finding P business.....

    • one year ago
  8. phi Group Title
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    so find its e-vecs and lambdas AS= S L S is the matrix of e-vecs (as cols) L is the diag matrix of lambdas and S^-1 A S = L A is "diagonalized" to its eigen values

    • one year ago
  9. KonradZuse Group Title
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    The way trhe book started out was finding the eigen values which I got to be 3 and 6.

    • one year ago
  10. phi Group Title
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    are you sure about the eigenvalues??

    • one year ago
  11. KonradZuse Group Title
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    yeah

    • one year ago
  12. phi Group Title
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    because I do not get 3 and 6

    • one year ago
  13. KonradZuse Group Title
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    actually h/o.

    • one year ago
  14. KonradZuse Group Title
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    I guess I did that wrong lol now i'm konfused with that :P

    • one year ago
  15. phi Group Title
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    Let me fix some typo's A x = λ x (Ax - λ x)= 0 (A - λ I) x =0 x (e-vec) is in the null space of (A - λ I) reduce to rref to find the null space vector (x) for x to exist (not be just the 0 vector) the matrix (A - λ I) cannot be full rank that means its determinant is 0 to find the lambdas solve det ( (A - λ I) ) =0 | 6-λ -2 | | -2 3-λ | = 0

    • one year ago
  16. KonradZuse Group Title
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    yeah that's what I thought... (6-L) (3-L) - (-2)(-2) = 18 - 9L +L^2 -4

    • one year ago
  17. phi Group Title
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    18 - 9L +L^2 -4=0

    • one year ago
  18. KonradZuse Group Title
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    yeah :)

    • one year ago
  19. phi Group Title
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    L^2 -9L +14=0

    • one year ago
  20. phi Group Title
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    this factors

    • one year ago
  21. KonradZuse Group Title
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    mhm

    • one year ago
  22. phi Group Title
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    you don't know how to factor?

    • one year ago
  23. phi Group Title
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    the + in +14 means the values are the same the - in -9 (coeff of L) means the largest is -. so both are negative list the pairs of factors of 14: 1,14 2,7 which pair added together with the same sign (- in this case) give -9 yes -2 and -7 so (L-2)(L-7)=0 this means L= 2 or L=7

    • one year ago
  24. KonradZuse Group Title
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    Sorry I had to do some holiday stuff.....

    • one year ago
  25. KonradZuse Group Title
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    and that makes sense yeah.

    • one year ago
  26. KonradZuse Group Title
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    @ganeshie8 @Hero

    • one year ago
  27. KonradZuse Group Title
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    not completely, I kind of left mid way for something.

    • one year ago
  28. KonradZuse Group Title
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    @tkhunny

    • one year ago
  29. KonradZuse Group Title
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    Save me please :). 2 Questions 1 hour... :(.

    • one year ago
  30. tkhunny Group Title
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    Eigenvalues are 7 and 2. Normal Eigenvectors are \(\left[-\dfrac{2}{\sqrt{5}},\dfrac{1}{\sqrt{5}}\right]^{T}\) and \(\left[-\dfrac{1}{\sqrt{5}},-\dfrac{2}{\sqrt{5}}\right]^{T}\) What's the trick?

    • one year ago
  31. KonradZuse Group Title
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    How do we find the normal EV's again? and what trick?

    • one year ago
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