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$$\lambda_i M\mathbf{v}_i=K\mathbf{v}_i$$

$K$ is a real symmetric matrix and $M$ is real and diagonal.

I need to prove that the eigenvalues of $K$ with respect to the metric $M$ are real, that eigenvectors for different eigenvalues are orthogonal with respect to the metric (i.e $\mathbf{v}^T_iM\mathbf{v}_j=0$ if $i\ne j$) and that there exists an orthonormal eigenbasis.

I've managed to prove the orthogonality of eigenvectors with different eigenvalues, but am struggling with the other two proofs.

Hints would be appreciated.

EDIT: Metric is the termonology used in the text I'm reading. To clear up any confusion, let me make it clear that $M$ is a diagonal, real matrix.

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    $\begingroup$ I'm not sure what you mean when you say $M$ is a metric. Do you mean $M$ is a matrix? $\endgroup$ – Omnomnomnom Jan 28 at 20:43
  • $\begingroup$ $M^{-1}K$ is real symmetric and therefore diagonalizable as required, isn't it? $\endgroup$ – Klaas van Aarsen Jan 28 at 20:44
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    $\begingroup$ $M^{-1}K$ might not be symmetric, but $M^{-1/2} K M^{-1/2}$ should be. $\endgroup$ – tch Jan 28 at 21:07
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Let $C=M^{-1/2}$ (the diagonal matrix whose diagonal entries are the reciprocals of the square roots of the diagonal entries of $M$). Then $C=C^*$ and $C^*MC=E$ (the identity matrix), and if we seek the eigenvectors in the form $v_i=Cu_i$, then the problem for the $u_i$ becomes $C^*KCu_i=\lambda_i u_i$. The matrix $C^*KC$ is real and symmetric, and so its eigenvalues are real and there exists an orthonormal eigenbasis $\{u_i\}$ (with respect to the standard inner product on $R^n$). The vectors $v_i=Cu_i$ are then orthogonal in the metric defined by $M$.

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