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$A$ and $B$ are nonsquare and $AB$ is square. How do you go about checking the requirements of whether $(AB)^{-1}$ exists when the only thing you know is that $B$ has linearly independent columns and $A$ linearly independent rows? Are there any other things relevant in this case for invertibility?

From numerical analysis I have a feeling that $AB$ is non-singular in such case, but can not prove it.

Note: the only relevant proving techniques I found relate to $A^TA$ and $AA^T$, but I can not use these techniques since $A \ne B$.

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    Lots of answers, but they get deleted? – 2017-02-23
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    Two answers were self-deleted, but you're not missing anything. – 2017-02-23

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Suppose $A$ is $m \times n$ and $B$ is $n \times m$, so $AB$ is $m \times m$. If $n < m$, then the rank of $AB$ is at most $n$, so it's certainly not invertible. If $n \ge m$, you do have a chance for $AB$ to be invertible, but it's not always true, even if $A$ and $B$ both have rank $m$. Note that $\ker(AB) = \{x: Bx \in \ker(A)\}$. $\ker(A)$ is a subspace of dimension $n-m$ while $\text{Ran}(B)$ has dimension $m$, and what you need is that these intersect only at $0$.

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    By $Ran(B)$ you mean $Rank(B)$? – 2017-02-23
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    Thank you for the input. The $\ker$ approach gives me more insight into the problem. Anyhow, seems more complex than I thought it was. – 2017-02-23
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    No, the range of $B$, i.e. $\{Bx: x \in \mathbb F^m\}$ where this is over field $\mathbb F$. – 2017-02-23
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    Ok, it makes sense. Thanks! – 2017-02-23