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Say I have a symmetric matrix $A$ and a symmetric matrix $B$ such that $B$ is congruent with $A$, i.e. there exists a non-singular matrix $X$ such that $B = X^TAX$. Is there a general relation between the eigenvectors and eigenvalues of $A$ and $B$ if $X$ is known?

lcortesh
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    If $A$ is diagonalizable you have $A=PDP^T$ for $D$ diagonal and $P$ orthogonal, so $B=X^T(PDP^T)X=(X^TP)D(P^TX)$ – Emilio Novati Jun 21 '15 at 14:43
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    the congruent matrices has the same signature meaning that the number of positive/negative/zero eigenvalues are the same. i don't think you can say anything about the eigenvectors. – abel Jun 21 '15 at 14:56
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    Indeed @abel is correct. For example, every positive definite symmetric matrix is congruent to the identity matrix. – JamesM Jun 21 '15 at 15:00
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    I do not think so. For example, take $A = id$ then $B$ can be any positive definite matrix. Of course if you know $X$ then $B = X^{\top}X$ and you can compute eigenvalues and eigenvectors of $B$. For example taking $X$ diagonal then you can arrange the eigenvalues of $B$ as positive but arbitrary numbers. – Holonomia Jun 21 '15 at 15:00

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