Is it possible to construct matrices $A$ and $B$ such that $AB$ is invertible?

Question

Let $A$ be an $n \times m$ matrix and let $B$ be an $m \times n$ matrix. Also, let $n$ be less than $m$. Is it possible to construct $A$ and $B$ such that $AB$ is invertible?

I know that $AB$ must have dimensions $n \times n$. If both $A$ and $B$ are full rank, then $AB$ should also be full rank. Thus, $AB$ should be invertible. However, I can't find a line of reason for why $AB$ should also be full rank.

Any guidance is greatly appreciated!

Answer 1

If $A$ is $n\times m$ and $B$ is $m\times n$, with $m\lt n$, then $AB$ cannot be invertible. If $n\lt m$, then yes, it can be, but it doesn’t have to be.

If $n\lt m$, one example would be to let $A$ have first $n$ columns be the $n\times n$ identity and then fill in with zero columns; and let $B$ have first $n$ rows be the $n\times n$ identity, and the rest of the rows be zero. i.e. $$\begin{align*} A &= \left(\begin{array}{ccccccc} 1 & 0 &\cdots &0 & 0 & \cdots &0\\ 0 & 1 & \cdots & 0 & 0 & \cdots &0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 1 & 0 & \cdots &0\end{array}\right)\\ B&=\left(\begin{array}{cccc} 1 & 0 & \cdots& 0\\ 0 & 1 & \cdots& 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 1\\ 0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0\end{array}\right). \end{align*}$$ Then $AB$ is the $n\times n$ identity.

So it is possible for $AB$ to be invertible. However, it need not be invertible necessarily. For instance, just move the identity matrix in $A$ one column to the right, and the product of $A$ and $B$ will not be full rank. For a worst case scenario, take $m=2n$, and let $A$ have $n$ columns of zeros followed by the identity matrix, and leave $B$ the same. Then $A$ and $B$ both have full rank, but $AB$ is the zero matrix.

On the other hand, if $m\lt n$, then note that the rank of $B$ is at most $m$; that means that the rank of $AB$ is at most $m$, and hence cannot equal $n$ (which you would need for $AB$ to be invertible). Alternatively, because $B$ has more columns than rows, there exists a nonzero vector $\mathbf{v}$ such that $B\mathbf{v}=\mathbf{0}$. But then $(AB)\mathbf{v}=\mathbf{0}$, with $\mathbf{v}\neq\mathbf{0}$, which again tells you $AB$ cannot be invertible.

Answer 2

Edit: My answer adresses the original version of the question.

A short answer: no. $A$ and $B$ have $m$ columns and $n$ rows respectively, and you let $n < m$ in the question, so the matrix product $AB$ can't even be defined.

Answer 3

$AB$ need not be invertible. To see this, take\begin{equation*} A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix} \end{equation*} \begin{equation*} B = \begin{pmatrix} 1 & 2\\ 2 & 4\\ 3 & 1 \end{pmatrix} \end{equation*} Notice $A$ and $B$ both have full rank, but \begin{equation*} AB = \begin{pmatrix} 1 & 2 \\ 2 & 4 \\ \end{pmatrix} \end{equation*} does not.