# Talk:Spectral theorem

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Field: Analysis

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)

Request: The section on the Spectral theorem for unbounded operators is very vague - it more or less just mentions its existence. Since for many applications it is precisely unbounded (differential) operators that are of interest, it would be very useful to at the very least state the Spectral theorem for unbounded self-adjoint operators.

## examples

The section "functional analysis" takes some pains to point out the shift operator, and a scaling operator on L^2[0,1] have no eigenvalues, and then immediately states a theorem that none-the-less, these are unitarily equivalent to some multiplication operator on some measure space. It sure would be nice to have a detailed example for these. linas 05:48, 19 March 2006 (UTC)

A good example would be kinda hard to come by. In the proofs I've seen for this theorem, the multiplication operator and the unitary equivalence are defined quite abstractly, so the constructions in the proofs don't lend themselves to practical calculations. I'm not aware of any method for finding "nice" multiplication operators and unitary equivalences that works in any generality.
However, for the scaling operator you mention (I presume you mean the operator A from the article, with $[A \varphi ](t) = t \varphi (t)$), the multiplication and unitary operators are quite simple: The multiplication operator is just the scaling operator itself, and the unitary operator is just the identity operator. This operator has a spectrum, but no eigenvalues. The spectrum is just $[0,\ 1]$, the range of f(t)=t. However, none of these spectral values are eigenvalues, as the eigenvectors they'd lead to would have to be functions that were zero except at one point (for an eigenvalue $\lambda$, that point would be $f^{-1}(\lambda ) = \lambda$). However, such functions would be zero almost everywhere, so are equivalent to the zero function, so can't be eigenvalues.
I hope this explains it. James pic 10:02, 3 July 2007 (UTC)

## Recent edit

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)

the following was written immediately prior to the appearance of the above comment:

To CSTAR and any others who are interested: Archelon does not wish to argue about the notation for adjointness of linear operators (or complex|Hermitian conjugation, more accurately) (the notation was changed for the sake of the appearance when rendered, but this[*] is a notoriously volatile issue), but has returned the other things CSTAR removed from the proof that the eigenvalues are real (only some of which were originally added by Archelon, incidentally). Archelon 21:19, 25 April 2006 (UTC)

the following was written immediately thereafter:

The sentence is not irrelevant; it is explanatory. Also, a perusal of the page history will reveal that Archelon is not responsible for it (merely in favour of its retention). Archelon's advice to CSTAR: If you find the sentence intolerable, remove it yourself.

[*] (i.e., the latter) Archelon 21:25, 25 April 2006 (UTC)

Message to archelon (who seems to refer to her/himself in the third person). I did. --CSTAR 21:41, 25 April 2006 (UTC)
Message from Archelon (which does indeed refer to itself in the third person). As you like. Archelon 22:55, 25 April 2006 (UTC)

## Proof

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)

Is that paragraph clearer now? --CSTAR 14:59, 10 August 2006 (UTC)
Perfect, that makes sense now. It was hard to follow before. 216.7.201.43 15:32, 10 August 2006 (UTC)

## Target audience

I agree. The bra-ket notation makes this page unnecessarily difficult to understand.

Certainly, the statement of the theorem should be possible without bracket notation (and incidentally, I'm not sure that the $\langle \cdot \vert \cdot \rangle$ notation from physics is appropriate in a maths article - in maths literature $\langle \cdot , \cdot \rangle$ is more common), so I've given an alternative characterisation of the theorem in more elementary terms.
However, the proof really does need bracket notation - bracket notation is the best means we have of working with orthonormal bases. It would be possible to rewrite the proof substituting $\langle x \vert y \rangle$ for x*y, but that would make the proof messier, and most likely make the article less comprehensible to more experienced mathematicians
I guess you could put a note at the beginning of the proof for the mathematical layman explaining that this is all the bracket notation means, but the proof is somewhat of a detour from the article as it is - it might make sense to move the proof to Spectral theorem/Proofs, as per Wikipedia:WikiProject_Mathematics/Proofs#/Proof subpages. This also has the added benefit of moving the more technical aspects of the article out of the way - so people don't have to see them if they don't want to. James pic 09:54, 17 August 2007 (UTC)

James said: "and most likely make the article less comprehensible to more experienced mathematicians"

Ummm, is that who the article is for?

At least give the example from basic linear algebra with real matrices. I just took linear algebra and this page was useless when we covered the spectral decomposition. I don't have time now to change it but just for future reference: this page needs heavy work. —Preceding unsigned comment added by 152.16.225.228 (talk) 15:54, 20 February 2008 (UTC)

## Generalization to non-symmetric matrices

I'm looking at this section and it seems to have some shortcomings. For one, it talks about non-symmetric matrices, when non-Hermitian or non-normal matrices would be more relevant. It also talks about orthonormal systems of eigenvectors, which would only possible for normal matrices, such as Hermitian or real-symmetric matrices - the fact that this doesn't hold for non-normal matrices is what makes this theory interesting, and differentiates it from the more general operator valued theory. It also makes much of this section wrong.

I noticed a recently created article called Eigendecomposition (matrix), which seems to address these issues in greater detail, and with fewer of these fairly elementary mistakes. I'm tempted to replace this section with a See:Eigendecomposition (matrix), or at least replace the section with a condensed version of the article, and a See main article. Thoughts? James pic 09:58, 11 October 2007 (UTC)

i agree that that section has problems, so does the article Eigendecomposition (matrix). it's not clear what the section is saying. for instance, i don't see where it talks about orthonormal systems of eigenvectors, as you do. seems to me it states that eigenvectors of a matrix and its transpose are orthogonal in general (which is not true). i suggest it be removed altogether.
in any case, it's not really a generalization of the spectral theorem, which should be concerned with decomposing an operator or matrix into parts. Mct mht 13:44, 11 October 2007 (UTC)
Ahh, yes. I should really have said orthonormal bases rather than orthonormal systems.
You're probably right about removing it altogether. If I'm reading the section right, it's trying to say roughly the same things the Eigendecomposition article says (although not succeeding). This is a generalisation of the matrix results discussed at the beginning of this article, but neither a special case nor a generalisation of the spectral theorem itself.
I feel I ought to stick up for the Eigendecomposition article here though. The article has some style and structure issues, certainly, but it seems to be fairly accurate for a newly created article, and it deals with a notable topic that's not covered elsewhere. James pic 15:22, 12 October 2007 (UTC)

## Normal matrix - a detail I do not understand

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)

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)
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)
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)
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)

## Why the name?

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

## "Principal axes of a Matrix"?

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)