Re: complex orthogonal group

To clarify the notation: by a null vector in a vector space with a symmetric bilinear form < , >, I mean a vector x with <x,x>=0. Maybe this was not clear in my first posting.


Any good thorough book on linear algebra (i.e., designed not for
teaching students) will contain the spectral theorem over an
algebraically closed case.

Though I have to admit that I did not check really thoroughly, I doubt that this is true.
A key assumption in the proof of the standard (finite-dim) spectral theorem for normal operators is that the orthogonal complement of a subspace has trivial intersection with that subspace; this fails for the C^n case, since it has null vectors.

Here is an example of a symmetric matrix that is not diagonalizable:

( l + 1/2     i/2   )
(                   )
(   i/2     l - 1/2 )

where l is an arbitrary complex parameter. The matrix has JNF in the basis {(1,i),(1,-i)} with a 2x2-Jordan block to eigenvalue l; this is also how I came up with it.

This is just to show that there can be no general spectral theorem for algebraically closed fields with respect to the standard symmetric bilinear form.


Thanks for your answer,
Tobias

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