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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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    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.2012-09-19

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

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    @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).2012-09-19