How can I show that a real symmetric matrix has real eigenvalues?
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See too http://math.stackexchange.com/questions/746223/besides-being-symmetric-when-will-a-matrix-have-only-real-eigenvalues – Ellie Kesselman Dec 10 '14 at 22:17
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1It's funny how none of these showed up when I was typing up the question. – sodiumnitrate Dec 10 '14 at 23:10
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Hint: Think about the complex inner product $(u,Au)$ when $u$ is an eigenvector.
Gyu Eun Lee
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$A\mathbf{u} = \lambda\mathbf{u}\Rightarrow A\mathbf{u}^* = \lambda^*\mathbf{u}^* \Rightarrow -\mathbf{u}^T A\mathbf{u}^* =- \lambda^*\mathbf{u}^T\mathbf{u}^* \Rightarrow \mathbf{u}^H A\mathbf{u} - \mathbf{u}^T A\mathbf{u}^* =\lambda \mathbf{u}^H\mathbf{u} - \lambda^*\mathbf{u}^T\mathbf{u}^* \Rightarrow 0 = (\lambda -\lambda^*)\mathbf{u}^H\mathbf{u} \Rightarrow\lambda =\lambda^*. $
Alex Silva
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