0
$\begingroup$

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 ?

$\endgroup$
0
4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Damn econometricians ! Thanks for detailing the thought process of the counterexample. $\endgroup$ – Gabriel Romon May 6 '18 at 21:57
1
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ It is $A$, not $B$, which should have a kernel of dimension $\ell-k$ if it has rank $k$. $\endgroup$ – Misha Lavrov May 6 '18 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.