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I have an expression of the form:

$ACA^{'}$

where $C$ is an invertible, symmetric and positive definite matrix. I'm trying to figure out if the expression above is invertible. The $C$ matrix is $n \times n$ and the $A$ matrix is $k \times n$. Every row of A has exactly one entry =1 and one entry =-1, all other entries are 0.

Any help, especially something pointing to a theorem, would be greatly appreciated.

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    $\begingroup$ I have an LP where my matrix C and A will be changing given the state of the system and I would like to be able to determine whether my system will ever be infeasible. I hope that fits within this group but please let me know $\endgroup$ – evgeny Jan 21 '14 at 20:42
  • $\begingroup$ Is $k$ larger than $n$? Smaller than $n$? $\endgroup$ – Brian Borchers Jan 22 '14 at 4:35
  • $\begingroup$ smaller than n. I think that if A is of full rank (i.e. k) then this should be invertible, otherwise it's probably never invertible? $\endgroup$ – evgeny Jan 22 '14 at 5:04
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If $C\in\mathbb{R}^{n\times n}$ is (symmetric and) positive definite and $A\in\mathbb{R}^{k\times n}$ with $k\leq n$, then $ACA^T\in\mathbb{R}^{k\times k}$ is invertible if and only if $A$ has full rank. (You can think of $ACA^T$ as the projection of $C$ onto the subspace spanned by the rows of $A$, so it makes sense to expect them to be linearly independent.)

To see this, let $x\in\mathbb{R}^k\setminus\{0\}$ be arbitrary. If $A$ has full (row) rank, then $A^T$ has full (column) rank as well and thus is injective. Hence, $y:=A^Tx \in\mathbb{R}^n\setminus\{0\}$, and the positive definiteness of $C$ yields $$ x^T(ACA^T)x = (A^Tx)^TC(A^Tx) = y^TCy > 0, $$ i.e., $ACA^T$ is (symmetric and) positive definite and thus invertible. Conversely, if $A$ does not have full rank, $A^T$ is not injective and there exists a vector $x\in\mathbb{R}^k\setminus\{0\}$ with $A^Tx = 0$ and hence $ACA^Tx = 0$. It follows that $ACA^T$ is not injective and thus not invertible.

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    $\begingroup$ And with one nonzero entry per row, $A$ will have full row rank as long as the nonzero entries are all in different columns of $A$. $\endgroup$ – Brian Borchers Jan 22 '14 at 13:11
  • $\begingroup$ thanks Christian, that makes sense. My A matrix is an incidence matrix for a transmission network so as long as I don't have parallel lines, it should work out. $\endgroup$ – evgeny Jan 22 '14 at 22:00

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