# Rank of composition of non-square transpose matrices

Let $d' < d$ be natural numbers, and let $L$ be a real matrix of dimension $d'\times d$, with maximum rank (that is, with rank $d'$). Let $A$ be a real square matrix of dimension $d$, and let's suppose that $A$ is regular. We consider the product $LAL^T$. This product is a square matrix of dimension $d'$ and I want to see that it is also regular, or equivalently, that it has rank $d'$.

EDIT. Thanks to the answers now I see that the regular condition for $A$ is not enough, but what about if $A$ is positive definite?

I have tried to see the matrices as linear maps and use the rank-nullity theorem to obtain information, but I'm only able to find that the rank of $AL^T$ is $d'$. Any ideas?

Also I'm interested in knowing if the same follows if we replace $L^T$ by any other matrix of maximum rank with the same dimensions. That is, the product $LAM$ will be regular for matrices $L$ and $M$ with dimensions $d'\times d$ and $d\times d'$, respectively, and both ranks $d'$?

The condition of regularity for $A$ is insufficient. For instance, consider the following example: $$L = \pmatrix{1&0}, \quad A = \pmatrix{0&1\\1&0}$$ A sufficient condition would be for $A$ to be positive definite.
If $A$ is positive definite, then there exists a regular matrix $Q$ such that $A = QQ^T$. With that, we note that $$LAL^T = LQQ^TL^T = (LQ)(LQ)^T$$ We note that the rank of this matrix is equal to that of $LQ$, namely $d'$.
The result fails to generally hold if we replace $L$ with another matrix. For instance, take $$L = \pmatrix{1&0}, \quad A = \pmatrix{1&0\\0&1}, \quad M = \pmatrix{0\\1}$$
• I suppose that you wanted to say $LAL^T$ instead of $QAQ^T$, but I understood it, thanks! – gtf Jul 2 '18 at 16:44