Talk:Spectral theorem

message: I think the "Fundamental theorem of Linear Algebra" is normally taken to be the statement that connects kernel, row and column spaces of a matrix. AxelBoldt 02:25 Jan 24, 2003 (UTC)
Added a statement of the spectral theorem for bounded self-adjoint operators.  Hopefully this should leave this article in a more-or-less definitive state.CSTAR 21:09, 13 May 2004 (UTC)

message: The point is that the sentence
: Then—using the fact that A x = \lambda x iff \overline{A} \overline{x} = \overline{\lambda} \overline{x}
is completely irrelevant.  Please remove it.--CSTAR 21:12, 25 April 2006 (UTC)
replies: text:  Message to archelon (who seems to refer to her/himself in the third person). I did. --CSTAR 21:41, 25 April 2006 (UTC)
replies: text:  Message from Archelon (which does indeed refer to itself in the third person).  As you like.  Archelon 22:55, 25 April 2006 (UTC)
thread_title:  Recent edit 

message: Can anyone explain this line, it wasn't entirely clear for me:
This is finite-dimensional, and A has the property that it maps every vector w in K into K
How do we know that anything in K is mapped into K? Why can't it be mapped into Span(e)? Can anyone clarify, please? 216.7.201.43 13:47, 10 August 2006 (UTC)
replies: text: Is that paragraph clearer now? --CSTAR 14:59, 10 August 2006 (UTC)
replies: text:  Perfect, that makes sense now. It was hard to follow before. 216.7.201.43 15:32, 10 August 2006 (UTC)
thread_title:  Proof 

message: Hi, 
I am trying to understand the spectral theorem for normal matrices. I understand the thing about the Schur decomposition and that using that on the normality criterion implies that the upper triangular matrix in the Schur decomposed factorization of the normal matrix has to obey the relation \mathbf{TT}^* = \mathbf{T}^*\mathbf{T}, i.e., the upper triangular matrix has to be be normal as well. Then the article says: Therefore T must be diagonal (the whole point of the theorem). My problem is that I do not see very easily why that is so. I can easily understand that if it is diagonal the normality criterion is met, but I do not understand that it has to be diagonal. To try and understand it I then wrote down the expression for the diagonal elements of (\mathbf{TT}^*)_{i,i} and found that this implies that 
:\sum_{k=1}^n\left|T_{i,k}\right|^2 = \sum_{k=1}^n\left|T_{k,i}\right|^2
where the span of the sums can be further reduced by taking advantage of the fact that T is upper triangular.
However, still, it is not evident for me why T has to be diagonal. Evidently, I am missing some (probably trivial) point in the line of argumentation here. Could this section be elaborated a bit in the article to make it more understandable? -- Slaunger (talk) 13:57, 12 February 2009 (UTC)
replies: text: if you write down an arbitrary, say, 2 by 2 upper-triangular matrix T and impose the condition T*T = TT*, you will see immediately that the off-diagonal entry must be zero. Mct mht (talk) 04:50, 13 February 2009 (UTC)
replies: text: Actually, I did just that as well prior to posting, and I do see that clearly for the n=2 case. I just find that somewhat primitive, and figured there should be a smarter argument valid for the general n-dimensional case. -- Slaunger (talk) 10:57, 13 February 2009 (UTC)
replies: text: the same 2 by 2 argument, applied to an appropriate 2 by 2 operator matrix, yields the general case. Mct mht (talk) 10:48, 15 February 2009 (UTC)
replies: text:  which I think is becoming such a complicated argument that only a small fraction of readers would get that by themselves, which makes me think it could be a good thing to extend the explanation about this crucial point in the line of argumentation;-) I would rather not like to do it myself thaough, as I am not a native writer and my linear algebra is rusty... --Slaunger (talk) 20:10, 16 February 2009 (UTC)
thread_title:  Normal matrix - a detail I do not understand 

message: The article doesn't say why it's called the spectral theorem. ᛭ LokiClock (talk) 02:17, 5 September 2011 (UTC)
thread_title:  Why the name? 

message: Principal axes of a Matrix redirects here from the principle axis disambiguation link. But principal axes aren't mentioned anywhere? Warrickball (talk) 09:04, 21 May 2012 (UTC)
thread_title:  "Principal axes of a Matrix"? 

message: This article refers to:
: a unitary operator 
But the article titled unitary operator says a unitary operator is a certain kind of operator from a space into itself.  Should another definition be used here instead?  If so, it should be stated here. 160.94.28.231 (talk) 01:20, 30 April 2016 (UTC)
thread_title:  Definition of "unitary operator" 

message: The article claims the following:
"An equivalent condition is that A∗ = A, where A∗ is the hermitian conjugate of A", refering to the conjugate of the operator A. 
Is this in general true? With the usual inner product it indeed is, but for any? Is the definition of a scalar product so tight that really ANY inner product has the hermitian conjugate as the conjugate operator?  — Preceding unsigned comment added by 77.0.166.225 (talk) 02:13, 27 January 2019 (UTC)
thread_title:  Hermitian condition and hermitian conjugate 

message: I thought that perhaps the most compehensive spectral theorem was the one that applies to normal operators, at least on a Hilbert space. But that particular version of the spectral theorem seems to be missing from the article (except in the finite-dimensional case).50.205.142.50 (talk) 20:30, 13 March 2020 (UTC)
thread_title:  What about the spectral theorem for normal operators