Real symmetric matrices have only real eigenvalues — is this an incorrect proof?

Question

According to my lecturer there are problems within this proof, but I can't for the life of me see what they are. It seems perfectly valid to me. Any help?

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Answer 1

Your lecturer is correct: there is a lot wrong with the proof above. The result (that every eigenvalue of a real symmetric matrix is real) has nothing to do with the Fundamental Theorem of Algebra---if the characteristic polynomial had no complex roots, then there would be no eigenvalues and the claimed result would be trivially true. The quantifier in the proof above is wrong because we are not interested in the existence of a complex root of the characteristic polynomial but in a property of all such roots. The result above also has nothing to do with the characteristic polynomial.

Here is a correct proof: Suppose $A$ is a real symmetric matrix and $\lambda \in \mathbb{C}$ is an eigenvalue of $A$. Then there exists a nonzero vector $x \in \mathbb{C}^n$ such that $Ax = \lambda x$. Now proceed as in the last four lines of the proof as given above.