Is it that only with normal matrices, the transition matrix to its [del: inherent] [ins: own] basis is unitary?

Does this even make sense what I translated into english?

PS. I am probably gonna delete this question eventually

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I would prefer if you didn't delete it (if that matters to you). –  S. Carnahan Nov 7 '09 at 17:20

I am not sure I understand the question, but if you are talking about linear algebra over the complex numbers, then it is true that normal matrices (those which commute with their hermitian adjoint) are precisely those which can be diagonalised by a unitary transformation. (This is proven in Herstein's Topics in algebra: §10 of the 1964 edition.)

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I don't really know what I'm asking... A physics friend of mine asked me this (in another language though) and I'm just re-asking it here (translated) :P –  person Nov 7 '09 at 14:44
Then it's difficult to say whether my answer makes any sense. Which language is this, by the way? Perhaps I can understand the original question... –  José Figueroa-O'Farrill Nov 7 '09 at 14:48
Serbian: Da li je samo kod normalnih matrica matrica prelaska na svojstvenu bazu unitarna? –  person Nov 7 '09 at 14:52
Sorry, not among my languages (I have always regretted not studying any Slavic language...) But at any rate, the only theorem i know with the words "matrix", "normal", "unitary" and "basis" is the one I quoted. What throws me off is "inherent": is this a basis where the matrix is diagonal? –  José Figueroa-O'Farrill Nov 7 '09 at 14:58

Jose has already given the answer, but here's a quick proof of it. First, note that any diagonal matrix D is normal, since its adjoint is also diagonal and diagonal matrices commute. Now suppose you have some matrix A that is UDU* for U unitary and D diagonal, i.e. it is conjugate to a diagonal matrix by a unitary change of basis. Then A*=(UDU*)*=UD*U*, so AA*=UDU*UD*U*=UDD*U*=UD*DU*=UD*U*UDU*=A*A, so A is normal.

Here's a more conceptual explanation of what's going on. Normality is a property of a linear transformation on an inner product space. If you pick an orthonormal basis, any transformation that is diagonal with respect to that basis is easily seen to be normal. To say that A is conjugate to a diagonal matrix by a unitary matrix is exactly saying that A is diagonal with respect to some orthonormal basis, since a unitary matrix is just changing from the standard orthonormal basis to some other orthonormal basis.

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I think I understand the question in Serbian (my native language is from the same Slav family) and he is really asking about normal matrices and their diagonalization by a unitary transformation.

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Good. Then the theorem in Herstein applies. So the answer is "Yes", or maybe "Da" :) –  José Figueroa-O'Farrill Nov 7 '09 at 15:05
Hey thanks guys (both of you) :) But I have to delete the question eventually since I can't let my friend see I suck at math :PPP –  person Nov 7 '09 at 15:09

Other people have already answered the mathematical question. I'll point out a possible source of linguistic confusion: the word "eigen", in eigenvector and eigenbasis, is German for something like "own" or "self". I would guess that the poster may have been translating "eigen" into English rather than leaving it alone.

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Possibly, but in some languages they do not use "eigen-". For instance in Spanish they use "auto-" and in French "propre". I have no idea what they use in Serbian. –  José Figueroa-O'Farrill Nov 7 '09 at 19:43
For the sake of completeness: the proper word in Serbian is "sopstveni" (it's an adjective, not a prefix) and it literally means "own". Same term is used in Bosnian and Montenegrin, while Croatian term for it is "svojstveni". –  Harun Šiljak Nov 13 '10 at 7:30