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Matrices have a least two major functions in linear algebra. On one hand, they can represent linear transformations as elements of $\text{Hom}(V,V)$). On the other hand they can represent inner products as elements of $\text{Hom}(V,V^*).$

In the case of linear transformations, there is a clear geometrical interpretation of eigenvalues and eigenvectors - eigenvectors are scaled by eigenvalues. For a matrix representing an inner product, it's not as clear to me. I don't expect the eigenvalues of an inner product matrix to be intrinsic to the inner product since eigenvalues of an inner product matrix are not preserved by a change of basis. Is there however a way to interpret the eigenvalues in relation to a chosen basis?

This seems to be a reasonable hope since there are relationships between an inner product matrix and its eigenvalues. For example, a symmetric inner product matrix (say over $\mathbb{R}$) with all positive eigenvalues represents a positive definite inner product.

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  • $\begingroup$ observation: if $B$ is the matrix got by the change of basis due to the eigenvectors of $Q$, where $Q$ is the matrix of a inner product, then the product $B^{\top}QB$ will be the reinterpretation of the very same inner product given by $Q$ but in the new basis $\endgroup$ – janmarqz Jul 15 '15 at 16:40
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The matrix $A$ which induces an inner product must be symmetric (or hermitian), non-degenerate and positive definite.

Because of its symmetry, we have using the spectral theorem, a full set of eigenvalues and corresponding eigenvectors and therefore $A$ is diagonalizable with positiv eigenvalues. Even under the change of the basis, $A$ has still the same eigenvalues since matrix similarity states that similar matrices have the same eigenvalues.

A matrix $A$ is similar to $B$ if there exists invertible $S$ such that $$ A=S^{-1}BS $$

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