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This question is in reference to this proof of the Spectral Theorem.

Let $A$ be a symmetric matrix. Take $u$ to be an orthonormal vector (such that $u^Tu=1$). Then if $u$ is an eigenvector of $A$, then $Au=\lambda u$. This implies that $\lambda=u^TAu$.

The article says that this proves that $\lambda$ is real. I don't follow why this is. Both $u$ and $A$ can contain complex entries! Why would $u^TAu$ have to be real?

algebraically_speaking
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  • What's known is that a real symmetric matrix has real eigenvalues and it's diagonalizable over $\Bbb R$. –  Aug 16 '14 at 11:36
  • Also, the result holds for hermitian matrices. – Dmoreno Aug 16 '14 at 11:38
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    That proof is not very well written and that clearly accounts for some of your confusion. I'd suggest a text that's written by a mathematician. – Tim kinsella Aug 16 '14 at 11:44
  • "Both $u$ and $A$ can contain complex entries!" The concept of symmetric matrix is only defined for real matrices. If it is given that $A$ is symmetric, then $A$ is, by definition, real. Since what I've just said has brought some confusion in the past, let me clarify that it makes sense to say that a complex matrix $M$ is symmetric if $M=M^T$, but that simply isn't done anywhere at this level (I've only seen it done once and it wasn't at the undergraduate level). The same applies to orthogonal matrices, it's a 'real' concept. – Git Gud Aug 16 '14 at 12:08

1 Answers1

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Lemma: A hermitian matrix has real eigenvalues.

Proof: Let $A$ be a complex matrix such that $A^*=A$. Let $\lambda$ be one of $A$'s eigenvalues and $u\neq 0$ a corresponding eigenvector. $Au=\lambda u\implies u^*A^*=\lambda^*u^*$. Hence $$\lambda^* u^*u=u^*A^*u=u^*Au=\lambda u^*u$$ and therefore $\lambda^*=\lambda$. Therefore $\lambda$ is real.

Since real symmetric matrices are hermitian, their eigenvalues are real.

anonymous67
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