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dan815
 one year ago
Let T be an operator
show TT* and T*T are self adjoint
selfadjoint and normal,
selfadjoint means A=A* < hemetian conjugate
Normal in this case means ?
@empty
dan815
 one year ago
Let T be an operator show TT* and T*T are self adjoint selfadjoint and normal, selfadjoint means A=A* < hemetian conjugate Normal in this case means ? @empty

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Capnkookcx
 one year ago
Best ResponseYou've already chosen the best response.0http://www.speakpipe.com/voicerecorder/msg/csdlw5sv1ihf8bxl

Capnkookcx
 one year ago
Best ResponseYou've already chosen the best response.0http://www.speakpipe.com/voicerecorder/msg/csdlw5sv1ihf8bxl

Capnkookcx
 one year ago
Best ResponseYou've already chosen the best response.0I like your hit song danny

Capnkookcx
 one year ago
Best ResponseYou've already chosen the best response.0I knew it >.< beach

Empty
 one year ago
Best ResponseYou've already chosen the best response.1To be selfadjoint 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
 one year ago
Best ResponseYou've already chosen the best response.1\[(AB)^\dagger =B^\dagger A^\dagger\] \[A=T\]\[B=T^\dagger\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.1\[AA^\dagger = A^\dagger A\] This is what it means to be normal

Empty
 one year ago
Best ResponseYou've already chosen the best response.1CORRECT: \[(A_{ij}B_{jk})^T = B_{jk}^TA_{ij}^T = B_{kj}A_{ji}\] BADDDDD: \[(A_{ij}B_{jk})^T \ne A_{kj}B_{ji}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0The spectral theorem extends to a more general class of matrices. Let A be an operator on a finitedimensional 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 uppertriangular. 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
 one year ago
Best ResponseYou've already chosen the best response.1\[AA^\dagger = A^\dagger A\] This is what it means to be normal

Empty
 one year ago
Best ResponseYou've already chosen the best response.1\[AA^\dagger = AA = A^\dagger A\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.1give it a few minutes before it starts back up lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.0like if u look at the DFT

dan815
 one year ago
Best ResponseYou've already chosen the best response.0you are inputting a finite number of element of a column vector

dan815
 one year ago
Best ResponseYou've already chosen the best response.0and 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
 one year ago
Best ResponseYou've already chosen the best response.0which is given by the the different roots of 1 in complex space

dan815
 one year ago
Best ResponseYou've already chosen the best response.0then ya applying that inverse will give u back the initial vector u had

Empty
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x) = \sum_{n=1}^\infty c_n \sin(n \pi x) \] 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
 one year ago
Best ResponseYou've already chosen the best response.0dont waste my time with sins and cos ive moved onto e^ix now

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i only wan to see a matrix no more expressions

dan815
 one year ago
Best ResponseYou've already chosen the best response.0expressions are fake, they are made up matrixces

Empty
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x)=\frac{1}{\sqrt{2 \pi}} \int_{\infty}^\infty g(k)e^{i kx}dk\] This is a matrix equation tho

dan815
 one year ago
Best ResponseYou've already chosen the best response.0okay so tensor products, my prof did them last lecture the one i missed apparently

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i kinda glanced over the text book so

dan815
 one year ago
Best ResponseYou've already chosen the best response.0you know that guy that did the arithmetic function stuff, he has a series on this stuff too lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.0lets fly him out to the americas

Empty
 one year ago
Best ResponseYou've already chosen the best response.1So like let's just try to get into a specific tensor product example and just sorta flesh it out

dan815
 one year ago
Best ResponseYou've already chosen the best response.0go charge your phone duudee

Empty
 one year ago
Best ResponseYou've already chosen the best response.1it's on the charger ok I think it's probably good now though but I'm eating
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