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Given a full-rank matrix $X$, and assume that the eigen-decomposition of $X$ is known as $X=V \cdot D \cdot V^{-1}$, where $D$ is a diagonal matrix.

Now let $C$ be a full-rank diagonal matrix, now I want to calucate the eigen-decomposition of $C \cdot X$, that is to find a matrix $V_c$ and a diagonal matrix $D_c$ such that $C \cdot X =V_c \cdot D_c \cdot V_c^{-1}$. Since the eigen-decomposition of $X$ is known, how can we obtain $V_c$ and $D_c$ from $V$ and $D$, respectively? Thanks!

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    Sorry, $X$ is not symmetric, I have rectified the problem.2012-05-08

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There is no simple relation between the eigen-decompositions of $C$, $X$ and $C X$. In fact, $C X$ does not even have to be diagonalizable. About all you can say is that $\text{det}(CX) = \det(C) \det(X)$, so the product of the eigenvalues for $CX$ (counted by algebraic multiplicity) is the product for $C$ times the product for $X$.

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    @JohnSmith: You still think that? e.g. C = \pmatrix{a & 0\cr 0 & b\cr} with $a, b \ne 0$, X = \pmatrix{1/a & 1/a\cr 0 & 1/b\cr} (diagonalizable with eigenvalues $1/a$ and $1/b$), CX = \pmatrix{1 & 1\cr 0 & 1\cr} (not diagonalizable).2012-05-08