- dan815

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

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.