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In proving the spectral theorem for self-adjoint operators, the first step is to show that an eigenvalue exists (and then you do induction).

Over $\mathbb{C}$, this is easy, since it's an algebraicly closed field.

Over $\mathbb{R}$, the books I've seen use a sort've long proof. But if you have a self-adjoint real matrix, then it is also a self-adjoint complex matrix. Therefore you can find eigenvalues, and you know those eigenvalues will be real because it's self-adjoint. Done. What am I overlooking?

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You might want to give some details about the long proof yu have in mind... – Mariano Suárez-Alvarez Sep 19 '12 at 2:29
BTW, the spectral theorem for real self-adjoint matrices is actually equivalent to the existence of singular value decompositions, which in most proofs uses the Weierstrass extreme value theorem. – lhf Sep 19 '12 at 2:39
up vote 5 down vote accepted

The eigenvalues will be real but you still have to prove that they have real eigenvectors.

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I can't see how is this related to the OP...perhaps you misread? Or perhaps I did, of course. – DonAntonio Sep 19 '12 at 2:34
You're right. But suppose you have a complex eigenvector and the matrix and eigenvalue are real. Then the real (and imaginary) parts of the eigenvector will again be eigenvectors. Just weird that I've checked 2 books and they have super complicated proofs. – StuartHa Sep 19 '12 at 2:38
@user19192, you're right. And Lang does exactly that in his Linear Algebra (chapter XI in both 1st and 2d ed; the 3rd ed is different). – lhf Sep 19 '12 at 3:15

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