## 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

1. Capnkookcx
2. Capnkookcx
3. Capnkookcx

I like your hit song danny

4. Capnkookcx

I knew it >.< beach

5. Empty

$(AB)^T = B^TA^T$

6. dan815

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7. dan815

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8. Empty

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

$(AB)^\dagger =B^\dagger A^\dagger$ $A=T$$B=T^\dagger$

10. dan815

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11. Empty

$AA^\dagger = A^\dagger A$ This is what it means to be normal

12. dan815

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13. dan815

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14. dan815

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17. Empty

CORRECT: $(A_{ij}B_{jk})^T = B_{jk}^TA_{ij}^T = B_{kj}A_{ji}$ BADDDDD: $(A_{ij}B_{jk})^T \ne A_{kj}B_{ji}$

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19. dan815

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20. dan815

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

$AA^\dagger = A^\dagger A$ This is what it means to be normal

22. Empty

$AA^\dagger = AA = A^\dagger A$

23. dan815

dced

24. Empty

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25. Empty

yeah phone died

26. Empty

give it a few minutes before it starts back up lol

27. dan815

yep okay i got this

28. dan815

like if u look at the DFT

29. dan815

you are inputting a finite number of element of a column vector

30. dan815

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

which is given by the the different roots of 1 in complex space

32. dan815

then ya applying that inverse will give u back the initial vector u had

33. Empty

$f(x) = \sum_{n=1}^\infty c_n \sin(n \pi x)$ 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

dont waste my time with sins and cos ive moved onto e^ix now

35. dan815

lel

36. Empty

:O

37. dan815

i only wan to see a matrix no more expressions

38. dan815

expressions are fake, they are made up matrixces

39. Empty

$f(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty g(k)e^{i kx}dk$ This is a matrix equation tho

40. dan815

infinite matrix

41. dan815

okay so tensor products, my prof did them last lecture the one i missed apparently

42. Capnkookcx
43. dan815

i kinda glanced over the text book so

44. dan815

you know that guy that did the arithmetic function stuff, he has a series on this stuff too lol

45. dan815
46. dan815

he's our boy

47. dan815

lets fly him out to the americas

48. Empty

Nice good idea

49. Empty

So like let's just try to get into a specific tensor product example and just sorta flesh it out

50. dan815