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SamIam

  • 4 years ago

Help Please :D

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  1. SAMIAM
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    What happens if you multiply by P from the left?

  3. SAMIAM
    • 4 years ago
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    IAP=DP A^-1=P

  4. SAMIAM
    • 4 years ago
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    Is that correct?

  5. SAMIAM
    • 4 years ago
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    uhhh i thought P*P^-1=Identity matrix

  6. SAMIAM
    • 4 years ago
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    oh ya maybe I am wrong

  7. SAMIAM
    • 4 years ago
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    No it equals Identity matrix

  8. anonymous
    • 4 years ago
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    yes

  9. anonymous
    • 4 years ago
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    \[A^{-1} P A = D\] \[AA^{-1} P A = PA = A D\] \[P A A^{-1} = P = A D A^{-1} \]

  10. anonymous
    • 4 years ago
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    Oh crap I used the wrong variables. Just substitute P for A ;)

  11. SAMIAM
    • 4 years ago
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    noooooooo it was P^-1AP=D

  12. SAMIAM
    • 4 years ago
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    lol ok

  13. anonymous
    • 4 years ago
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    For the second part \(A \neq D \)

  14. anonymous
    • 4 years ago
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    This is known as http;//en.wikipedia.org/wiki/Diagonalizable_matrix

  15. SAMIAM
    • 4 years ago
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    I so dont get what he did

  16. anonymous
    • 4 years ago
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    Here D is the diagonal matrix.

  17. SAMIAM
    • 4 years ago
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    ya? how did u get that?

  18. anonymous
    • 4 years ago
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    \[ P^{-1}AP = D \implies A = PDP^{-1} \]

  19. anonymous
    • 4 years ago
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    Using the right variables this time... \[P^{-1} A P = D\] We can multiply by P from the left to get \[ P P^{-1} A P = PD\] but \[P P^{-1} = I\] so that means \[AP = PD\] Do the same thing on the right side with the inverse: \[AP P^{-1} = PD P^{-1}\] but since \[P P^{-1} = I\] then \[A = P D P^{-1} \]

  20. SAMIAM
    • 4 years ago
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    So how are we able to ignore the I?

  21. anonymous
    • 4 years ago
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    It's just the matrix equivalent of the number one.

  22. anonymous
    • 4 years ago
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    Because \( I \) is the identity matrix.

  23. anonymous
    • 4 years ago
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    with regard to multiplication.

  24. SAMIAM
    • 4 years ago
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    oh ok ya

  25. SAMIAM
    • 4 years ago
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    ohhhh ok

  26. anonymous
    • 4 years ago
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    ^ Identity, not inverse :)

  27. SAMIAM
    • 4 years ago
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    LOL Thanks foolformath and jemurray

  28. SAMIAM
    • 4 years ago
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    That was clear

  29. anonymous
    • 4 years ago
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    Oh yes Identity not inverse.

  30. SAMIAM
    • 4 years ago
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    So from what i see A=d

  31. SAMIAM
    • 4 years ago
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    A=D

  32. anonymous
    • 4 years ago
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    A is not equal to D.

  33. SAMIAM
    • 4 years ago
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    oh no????

  34. anonymous
    • 4 years ago
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    It could be, but it doesn't have to be.

  35. anonymous
    • 4 years ago
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    D is a diagonal matrix, A may be not.

  36. SAMIAM
    • 4 years ago
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    huh? not getting it lol

  37. SAMIAM
    • 4 years ago
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    -_-

  38. anonymous
    • 4 years ago
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    What makes you think A = D?

  39. SAMIAM
    • 4 years ago
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    cuz I plugged this into the original equation

  40. SAMIAM
    • 4 years ago
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    |dw:1327223222985:dw|

  41. SAMIAM
    • 4 years ago
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    LOL I dont think that made sense so i take back what i said

  42. SAMIAM
    • 4 years ago
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    So is this wrong?

  43. SAMIAM
    • 4 years ago
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    -_-

  44. anonymous
    • 4 years ago
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    There is no evidence that those two matrices are equal.

  45. SAMIAM
    • 4 years ago
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    hey thanks jemurray

  46. SAMIAM
    • 4 years ago
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    That was really kinf o fyou to sit on my problem that long

  47. SAMIAM
    • 4 years ago
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    I am serious. I know it is late and .......

  48. SAMIAM
    • 4 years ago
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    Thanks i really appreciate it

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