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

- dan815

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- Capnkookcx

http://www.speakpipe.com/voice-recorder/msg/csdlw5sv1ihf8bxl

- Capnkookcx

http://www.speakpipe.com/voice-recorder/msg/csdlw5sv1ihf8bxl

- Capnkookcx

I like your hit song danny

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## More answers

- Capnkookcx

I knew it >.< beach

##### 1 Attachment

- Empty

\[(AB)^T = B^TA^T\]

- dan815

|dw:1443196662053:dw|

- dan815

|dw:1443197127671:dw|

- Empty

To be self-adjoint it must satisfy \(A^\dagger = A\) so: if \(A = T T^\dagger\) then:
\[A^\dagger = (TT^\dagger)^\dagger = (T^\dagger)^\dagger T^\dagger = TT^\dagger = A\]

- Empty

\[(AB)^\dagger =B^\dagger A^\dagger\]
\[A=T\]\[B=T^\dagger\]

- dan815

|dw:1443197401538:dw|

- Empty

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

- dan815

|dw:1443197507163:dw|

- dan815

|dw:1443197836817:dw|

- dan815

|dw:1443198350958:dw|

- Empty

|dw:1443198286335:dw|

- Empty

|dw:1443198442919:dw|

- Empty

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

- Empty

https://en.wikipedia.org/wiki/Normal_operator

- dan815

|dw:1443199313254:dw|

- 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 Aâˆ—A = AAâˆ—. 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 = UTUâˆ—, where U is unitary and T is upper-triangular. If A is normal, one sees that TTâˆ— = 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

- Empty

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

- Empty

\[AA^\dagger = AA = A^\dagger A\]

- dan815

dced

- Empty

|dw:1443200116964:dw|

- Empty

yeah phone died

- Empty

give it a few minutes before it starts back up lol

- dan815

yep okay i got this

- dan815

like if u look at the DFT

- dan815

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

- 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

- dan815

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

- dan815

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

- 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
\[y = A x\]

- dan815

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

- dan815

lel

- Empty

:O

- dan815

i only wan to see a matrix no more expressions

- dan815

expressions are fake, they are made up matrixces

- Empty

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

- dan815

infinite matrix

- dan815

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

- Capnkookcx

http://prntscr.com/8je71c

- dan815

i kinda glanced over the text book so

- dan815

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

- dan815

https://www.youtube.com/watch?v=FEeX_rz7PO0

- dan815

he's our boy

- dan815

lets fly him out to the americas

- Empty

Nice good idea

- Empty

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

- dan815

go charge your phone duudee

- Empty

it's on the charger ok I think it's probably good now though but I'm eating

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