# Why the complicated proof for existence of eigenvalue

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