# Determinant of square matrix multiplied on right and left by rectangular matrices

Let $A$ be an $n\times n$ matrix with, say, real entries. Let $B$ be $m\times n$, where $m<n$. If the determinant of $A$ is known, can we say anything about $\det(BAB')$? What if we put conditions of some sort on $B$?

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Arturo: You're right, thanks. I have the inequality backwards. Will fix this in the question. –  Minor Key Feb 28 '12 at 20:53
Is $B'$ the transpose? –  Arturo Magidin Feb 28 '12 at 21:00
If $\mathrm{rank}(B)\lt n$, then $\det(BAB')= 0$; if $\mathrm{rowspace}(B)\cap\mathrm{nullspace}(A)\neq\{\mathbf{0}\}$, then $\det(BAB') = 0$. Don't know off-hand about the remaining case. –  Arturo Magidin Feb 28 '12 at 21:04
Yes, $B'$ is the transpose. –  Minor Key Feb 28 '12 at 21:04
Above, that should be "If $\mathrm{rank}(B)\lt m$..." –  Arturo Magidin Feb 28 '12 at 21:26
Consider the case that $A$ is $0$ outside an $m \times m$ block, say with rows in set $I$, columns in set $J$. Then $B A B' = B_I A_{IJ} (B_J)'$ where $B_I$ and $B_J$ are the $m \times m$ submatrices of $B$ with columns in $I$ and $J$ respectively, and $A_{IJ}$ is the $m \times m$ submatrix of $A$ with rows in $I$ and columns in $J$. Now $\det(A)=0$, while $\det(BAB') = \det(B_I) \det(A_{IJ}) \det(B_J)$. Since $\det(A_{IJ})$ could be anything, the only condition on $B$ that would let you say anything about this is that the result is $0$ if all $m \times m$ blocks of $B$ have determinant $0$, i.e. if $B$ has rank less than $m$.