# Is $A'B$ invertible if both $A$ and $B$ are of full column rank?

Suppose that $A$ is a real matrix (dimension $n\times k$) with full column rank. We know that $A'A$ is invertible: $$A'Av=0\implies 0=v'A'Av=\left|Av\right|^2\implies Av=0\implies v=0.$$ Does this result generalize as follows: if $B$ is another real matrix (also dimension $n\times k$) with full column rank, under what conditions is it true that $A'B$ is invertible? As Pavel pointed out in the comment, $A'B$ isn't invertible in general.

I tried doing a similar trick above but it hadn't worked out so far. Thanks.

Edit: Does it matter if we know that $A=WJ$ where $W$ is $n\times l$ with full column rank and $J$ is $l\times k$ with full column rank? The context of these questions come from my reading on the Generalized Method of Moments: $B$ is the matrix of explanatory variables, $W$ is the matrix of instruments, and $J$ is a weighting matrix.

• If the columns of $A$ and $B$ were orthogonal, then $A'B=0$ even if $A$ and $B$ had full column rank. Mar 2, 2015 at 23:04
• @AlgebraicPavel Good point. Let me add a few details. Edit: I actually summarized the problem correctly. The author must have used some unstated assumption. Mar 2, 2015 at 23:05

I'm not sure if there is a simpler condition than what essentially $A^*Bx=0$ for an $x\neq 0$ says: since $B$ has full rank, $0\neq Bx\in\mathrm{Im}(B)$, so $A^*Bx=0$ for an $x\neq 0$ is equivalent to the existence of a nonzero vector $y\in\mathrm{Im}(B)$ such that $y\in\mathrm{Ker}(A^*)$, that is, $y\perp\mathrm{Im}(A)$. Consequently, $A^*B$ is nonsingular if and only if no nonzero vector from the image of $B$ is orthogonal to the image of $A$ (and equivalently with $A$ and $B$ replaced since $A^*B$ is nonsingular iff $B^*A$ is).

I don't like this condition since it does not look very symmetric. You can, of course, require that for all nonzero vectors $x\in\mathrm{Im}(A)$ and $y\in\mathrm{Im}(B)$, $x^*y\neq 0$ holds, but this gives only a sufficient condition for the invertibility of $A^*B$. For example, if $A$ and $B$ are real and $A^*B$ has both positive and negative real parts of the spectrum, then $z^*(A^*B)z=0$ for some nonzero $z$ without the need of $A^*B$ to have a zero eigenvalue and thus violating the simplified condition.

Not sure if I'm getting the question right.

A matrix product is invertible if both the matrices in the product are invertible, that is, A is invertible and B is invertible. A matrix is invertible when it follows: $$n=rank$$ This can also be showed by determinants. Hope this helped!

With this method $$A^TA$$ is also invertible, when $$A$$ is invertible and because $$rank(A)=rank(A^T)$$ $$A^T$$ is also invertible and then the product is also.

Hope this helped

• The question is not about square matrices. Feb 14, 2019 at 19:21

No. Try $$A = \begin{pmatrix} 1&0\\ 0 &1 \\ 0 &0\end{pmatrix}\quad\text{and}\quad B= I\,,$$ where $$I\in \mathbb{R}^{3\times3}$$ is the identity matrix. Both $$A$$ and $$B$$ have full column rank, but the product is not square and therefore not invertible.

• The OP wants $A'B$ invertible, thus in particular square. Feb 14, 2019 at 19:20