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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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If you are able to write $X$ as $VDV^T$, where $D$ is diagonal, then $X$ is actually symmetric. However the matrix $CX$ needn't be symmetric and hence it doesn't permit a decomposition as $V_C D_C V_C^T$. – user17762 May 8 '12 at 7:16
Sorry, $X$ is not symmetric, I have rectified the problem. – John Smith May 8 '12 at 7:30
up vote 2 down vote accepted

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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Suppose $C$ is full rank, is $CX$ diagonalizable ? – John Smith May 8 '12 at 8:04
Just take your favourite non-diagonalizable matrix for $C X$, and multiply it on the left by a diagonal full-rank $C^{-1}$ to get $X$; for most $C$ the result will be diagonalizable. – Robert Israel May 8 '12 at 8:29
@JohnSmith, As Marvis and Robert said no: take $D=I$ the identity in $\mathbb Q^2$ and $C$ the translation to the right by one (a Jordan block with 3 $1$s). $CD=C$ is not diagonalizable. (Robert was quicker. :) – plm May 8 '12 at 8:29
Oh, I've assumed $C$ is a diagonal matrix of full rank, meaning that all the diagonal entries of $C$ is non-zero. I think in this case if $X$ is diagonalizable, $CX$ is also diagonalizable. – John Smith May 8 '12 at 10:49
@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). – Robert Israel May 8 '12 at 16:01

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