dan815
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Capnkookcx
  • Capnkookcx
http://www.speakpipe.com/voice-recorder/msg/csdlw5sv1ihf8bxl
Capnkookcx
  • Capnkookcx
http://www.speakpipe.com/voice-recorder/msg/csdlw5sv1ihf8bxl
Capnkookcx
  • 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
  • Capnkookcx
I knew it >.< beach
1 Attachment
Empty
  • Empty
\[(AB)^T = B^TA^T\]
dan815
  • dan815
|dw:1443196662053:dw|
dan815
  • dan815
|dw:1443197127671:dw|
Empty
  • 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
  • Empty
\[(AB)^\dagger =B^\dagger A^\dagger\] \[A=T\]\[B=T^\dagger\]
dan815
  • dan815
|dw:1443197401538:dw|
Empty
  • Empty
\[AA^\dagger = A^\dagger A\] This is what it means to be normal
dan815
  • dan815
|dw:1443197507163:dw|
dan815
  • dan815
|dw:1443197836817:dw|
dan815
  • dan815
|dw:1443198350958:dw|
Empty
  • Empty
|dw:1443198286335:dw|
Empty
  • Empty
|dw:1443198442919:dw|
Empty
  • 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
  • Empty
https://en.wikipedia.org/wiki/Normal_operator
dan815
  • dan815
|dw:1443199313254:dw|
dan815
  • 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
Empty
  • Empty
\[AA^\dagger = A^\dagger A\] This is what it means to be normal
Empty
  • Empty
\[AA^\dagger = AA = A^\dagger A\]
dan815
  • dan815
dced
Empty
  • Empty
|dw:1443200116964:dw|
Empty
  • Empty
yeah phone died
Empty
  • Empty
give it a few minutes before it starts back up lol
dan815
  • dan815
yep okay i got this
dan815
  • dan815
like if u look at the DFT
dan815
  • dan815
you are inputting a finite number of element of a column vector
dan815
  • 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
  • dan815
which is given by the the different roots of 1 in complex space
dan815
  • dan815
then ya applying that inverse will give u back the initial vector u had
Empty
  • 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
  • dan815
dont waste my time with sins and cos ive moved onto e^ix now
dan815
  • dan815
lel
Empty
  • Empty
:O
dan815
  • dan815
i only wan to see a matrix no more expressions
dan815
  • dan815
expressions are fake, they are made up matrixces
Empty
  • Empty
\[f(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty g(k)e^{i kx}dk\] This is a matrix equation tho
dan815
  • dan815
infinite matrix
dan815
  • dan815
okay so tensor products, my prof did them last lecture the one i missed apparently
Capnkookcx
  • Capnkookcx
http://prntscr.com/8je71c
dan815
  • dan815
i kinda glanced over the text book so
dan815
  • dan815
you know that guy that did the arithmetic function stuff, he has a series on this stuff too lol
dan815
  • dan815
https://www.youtube.com/watch?v=FEeX_rz7PO0
dan815
  • dan815
he's our boy
dan815
  • dan815
lets fly him out to the americas
Empty
  • Empty
Nice good idea
Empty
  • Empty
So like let's just try to get into a specific tensor product example and just sorta flesh it out
dan815
  • dan815
go charge your phone duudee
Empty
  • 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.