I would like to find the inverse of the sum of a Kronecker product and a diagonal matrix. I found this answered here, but I don't see how the last step is valid.

Because of the fact that $C$ is diagonal, the spectral decomposition of $D^{−1}C$ itself is easy to obtain from that of $D^{−1}$.

It says that if we know the spectral decomposition of a symmetric matrix $D^{-1}$, we can easily find the same for $D^{−1}C$, where $C$ is diagonal. Can someone explain how to do this?

Since $D^{−1}C$ is not symmetric and not guaranteed to be orthogonally diagonalizable anymore, I don't quite see how this solution would work.



1 Answer 1


If $D^{-1}$ is symmetric, and $C$ is diagonal, then you're right that their product is not necessarily symmetric, and as a result, any spectral decomposition will not be "easy to obtain."

For example, say $$ D^{-1} = \left[ \begin{array}{cc} 2 & 1\\ 1 & 3 \end{array}\right] $$ and $C = \text{diag}(2,3)$. Then $$ D^{-1}C = \left[ \begin{array}{cc} 4 & 3\\ 2 & 9 \end{array}\right] \neq (D^{-1}C)'. $$

  • $\begingroup$ Thank you. But since this has the form $UDV'$ instead of $UDU'$, how would I simplify the next step? ie, $(I + UDU')^{-1} = U(I + D) ^{-1}U'$, does a similar simplification exist for $(I + UDV')^{-1}$ ? $\endgroup$
    – Max
    Mar 13, 2017 at 15:52

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