# If $A$ is $k\times l$ and $B$ is $l\times k$, $l\geq k$ and full-rank, must $AB$ have full-rank?

Let $1\leq k\leq l$, $A \in \mathbb R ^{k\times l}$ and $B \in \mathbb R ^{l\times k}$ with $rank(A)=rank(B)=k$. Is the square matrix $AB$ invertible ?

I've seen this answered positively in some econometrics lecture notes... But I don't see why it should hold. Obviously, $AB$ is invertible iff $\operatorname{im} B\cap \ker A= \{0\}$. By the nullity-rank theorem, this happens iff $\dim (\operatorname{im} B + \ker A) = l$, that is $\operatorname{im} B \oplus \ker A = \mathbb R^l$.

However I fail to see why this should be the case without any further assumption. I'm a little rusty with my linear algebra, can someone provide a counter-example to the claim ?

A counterexample is $$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix} \begin{bmatrix}0 & 0 \\ 1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}.$$ In general, if $k < \ell$, then $A$ will have a nontrivial kernel, so there will be at least one vector $x \in \mathbb R^{\ell}$ such that $Ax=0$. The easiest way to make sure $AB$ is not invertible is to make $x$ one of the columns of $B$: in that case, $x$ is definitely in the image of $B$, so there will be a vector $y\in\mathbb R^k$ such that $x = By$ and therefore $ABy=0$.
Also, in the case $k=1$, the quoted statement is just saying "the dot product of two vectors in $\mathbb R^\ell$ is never zero", which is very false.
Let $n = l-k$. With $A : \mathbb{R}^l \to \mathbb{R}^k$ and $B : \mathbb{R}^k \to \mathbb{R}^l$, let $\mathcal{A} = v_1, \ldots, v_n$ be a basis of $\ker A \cong \mathbb{R}^{n}$. So if $B$ maps $\mathbb{R}^{k}$ to $\operatorname{span}(\mathcal{A})$, then $AB$ maps $\mathbb{R}^k$ to $\mathbf{0}$, i.e, it is not invertible(injective).
• It is $A$, not $B$, which should have a kernel of dimension $\ell-k$ if it has rank $k$. – Misha Lavrov May 6 '18 at 21:54