When is $R \, A^{-1} \, R^t$ invertible?

Question

In the context of a Gaussian model, I came across a matrix product $R \, A^{-1} \, R^t$ where $R$ is a $m \times n$ rectangular matrix and as implied $A$ is $n \times n$ and invertible.

On which properties of $R$ does the existence of $(R \, A^{-1} \, R^t)^{-1}$ depend?

My idea was to use the rank: $\operatorname{rank} R , A^{-1} , R^t \leq \min(\operatorname{rank} R, \operatorname{rank} A) = \min(\operatorname{rank} R, n)$, so if $R$ does not have full rank or $m > n$, it cannot be invertible. That's not enough, though. — , Jul 02 '14 at 10:50

daw · Accepted Answer · 2014-07-02T11:34:53.617

4

First, if $m=n$ and $R$ is invertible, then $RA^{-1}R^T$ is invertible as well.

Second, if $A$ is in addition symmetric and positive definite on the range of $R$, then $RA^{-1}R^T$ is invertible if $R$ has rank $m$. Actually, in this case $RA^{-1}R^T$ is symmetric and positive definite as well.

This does not work without definiteness of $A$: $$ R=\begin{pmatrix}0&1\end{pmatrix}, \quad A^{-1}=\begin{pmatrix}1 & 1 \\ 1 &0\end{pmatrix}, \quad RA^{-1}R^T = 0, $$ which provides an example, where $R$ has full rank, but $RA^{-1}R^T$ is not invertible.

edited Jul 02 '14 at 11:34

answered Jul 02 '14 at 11:20

daw

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You might slightly relax the restrictions on $A$ by requiring $A$ to be SPD on the range of $R^T$. – Algebraic Pavel Jul 02 '14 at 11:26
Yes, you are right. – daw Jul 02 '14 at 11:35
Thank you for the answer. This is what I expected. Do you know a theorem from which the answer follows, or sketch a proof? – Jul 02 '14 at 11:46

When is $R \, A^{-1} \, R^t$ invertible?

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