I'm a physics undergrad and am studying small osciallations from Goldstien's Classical Mecahnics. We end up having to solve a system governed by the eigenvalue equation: $$ \boldsymbol{V \vec{a}} = \omega^2 \boldsymbol{ T \vec{a}}= \lambda \boldsymbol{T \vec{a}} $$ where V and T are real symmetric matrices with real eigenvalues $\lambda_k$ and eigenvectors $\vec{a_k}$.
We are left with the equations,
$$ \boldsymbol{\tilde{A} T A = I} $$ $$ \boldsymbol{\tilde{A} V A = \lambda} $$ where $A$ is a square matrix with the eigenvectors as its columns and $\boldsymbol{\lambda}$ is a diagonal matrix with the eigenvalues as the diagonal elements. Also the components of the eigenvectors are made to be real
The book concludes from the final equation,
Blockquote The final equation states that a congruence transformation of V by A changes it into a diagonal matrix whose elements are the eigenvalues $\lambda_k$ , the equation admits the solutions, $$ | V - \lambda I | = 0 $$ where the matrix containing the eigenvectors diagonalizes V.
But this seems like now we are solving the eigenvalue problem $$ V \vec{a} = \lambda \vec{a} $$I didn't seem to find any sound justifications to support this sudden conclusion . Is this because of some sort of uniqueness of diagonalization, or is there something else I'm missing.