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dan815

  • one year ago

Let T be an operator show TT* and T*T are self adjoint self-adjoint and normal, self-adjoint means A=A* <--- hemetian conjugate Normal in this case means ? @empty

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  1. Capnkookcx
    • one year ago
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    http://www.speakpipe.com/voice-recorder/msg/csdlw5sv1ihf8bxl

  2. Capnkookcx
    • one year ago
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    http://www.speakpipe.com/voice-recorder/msg/csdlw5sv1ihf8bxl

  3. Capnkookcx
    • one year ago
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    I like your hit song danny

  4. Capnkookcx
    • one year ago
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    I knew it >.< beach

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  5. Empty
    • one year ago
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    \[(AB)^T = B^TA^T\]

  6. dan815
    • one year ago
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    |dw:1443196662053:dw|

  7. dan815
    • one year ago
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    |dw:1443197127671:dw|

  8. Empty
    • one year ago
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    To be self-adjoint it must satisfy \(A^\dagger = A\) so: if \(A = T T^\dagger\) then: \[A^\dagger = (TT^\dagger)^\dagger = (T^\dagger)^\dagger T^\dagger = TT^\dagger = A\]

  9. Empty
    • one year ago
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    \[(AB)^\dagger =B^\dagger A^\dagger\] \[A=T\]\[B=T^\dagger\]

  10. dan815
    • one year ago
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    |dw:1443197401538:dw|

  11. Empty
    • one year ago
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    \[AA^\dagger = A^\dagger A\] This is what it means to be normal

  12. dan815
    • one year ago
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    |dw:1443197507163:dw|

  13. dan815
    • one year ago
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    |dw:1443197836817:dw|

  14. dan815
    • one year ago
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    |dw:1443198350958:dw|

  15. Empty
    • one year ago
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    |dw:1443198286335:dw|

  16. Empty
    • one year ago
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    |dw:1443198442919:dw|

  17. Empty
    • one year ago
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    CORRECT: \[(A_{ij}B_{jk})^T = B_{jk}^TA_{ij}^T = B_{kj}A_{ji}\] BADDDDD: \[(A_{ij}B_{jk})^T \ne A_{kj}B_{ji}\]

  18. Empty
    • one year ago
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    https://en.wikipedia.org/wiki/Normal_operator

  19. dan815
    • one year ago
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    |dw:1443199313254:dw|

  20. dan815
    • one year ago
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    The spectral theorem extends to a more general class of matrices. Let A be an operator on a finite-dimensional inner product space. A is said to be normal if A∗A = AA∗. One can show that A is normal if and only if it is unitarily diagonalizable. Proof: By the Schur decomposition, we can write any matrix as A = UTU∗, where U is unitary and T is upper-triangular. If A is normal, one sees that TT∗ = T*T. Therefore T must be diagonal since a normal upper triangular matrix is diagonal (see normal matrix). The converse is obvious. In other words, A is normal if and only if there exists a unitary matrix U such that

  21. Empty
    • one year ago
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    \[AA^\dagger = A^\dagger A\] This is what it means to be normal

  22. Empty
    • one year ago
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    \[AA^\dagger = AA = A^\dagger A\]

  23. dan815
    • one year ago
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    dced

  24. Empty
    • one year ago
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    |dw:1443200116964:dw|

  25. Empty
    • one year ago
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    yeah phone died

  26. Empty
    • one year ago
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    give it a few minutes before it starts back up lol

  27. dan815
    • one year ago
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    yep okay i got this

  28. dan815
    • one year ago
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    like if u look at the DFT

  29. dan815
    • one year ago
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    you are inputting a finite number of element of a column vector

  30. dan815
    • one year ago
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    and u have that element * element sqquare matrix for the fouriet transform vector which is spitting out the dependancy on each frequency of your e^inx

  31. dan815
    • one year ago
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    which is given by the the different roots of 1 in complex space

  32. dan815
    • one year ago
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    then ya applying that inverse will give u back the initial vector u had

  33. Empty
    • one year ago
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    \[f(x) = \sum_{n=1}^\infty c_n \sin(n \pi x) \] in index notation I guess you could put: \[f_x = \sum_{n=1}^\infty S_{nx}c_n\] or more linear algebra like \[y = A x\]

  34. dan815
    • one year ago
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    dont waste my time with sins and cos ive moved onto e^ix now

  35. dan815
    • one year ago
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    lel

  36. Empty
    • one year ago
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    :O

  37. dan815
    • one year ago
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    i only wan to see a matrix no more expressions

  38. dan815
    • one year ago
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    expressions are fake, they are made up matrixces

  39. Empty
    • one year ago
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    \[f(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty g(k)e^{i kx}dk\] This is a matrix equation tho

  40. dan815
    • one year ago
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    infinite matrix

  41. dan815
    • one year ago
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    okay so tensor products, my prof did them last lecture the one i missed apparently

  42. Capnkookcx
    • one year ago
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    http://prntscr.com/8je71c

  43. dan815
    • one year ago
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    i kinda glanced over the text book so

  44. dan815
    • one year ago
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    you know that guy that did the arithmetic function stuff, he has a series on this stuff too lol

  45. dan815
    • one year ago
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    https://www.youtube.com/watch?v=FEeX_rz7PO0

  46. dan815
    • one year ago
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    he's our boy

  47. dan815
    • one year ago
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    lets fly him out to the americas

  48. Empty
    • one year ago
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    Nice good idea

  49. Empty
    • one year ago
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    So like let's just try to get into a specific tensor product example and just sorta flesh it out

  50. dan815
    • one year ago
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    go charge your phone duudee

  51. Empty
    • one year ago
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    it's on the charger ok I think it's probably good now though but I'm eating

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is replying to Can someone tell me what button the professor is hitting...

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