can the product of 2 non-square matrices be invertible (without rank)

Question

If I have 2 matrices $A$ and $B$, such that $A$ is of dimensions $n$ x $m$ ($n$ lines/rows and $m$ columns ) and $B$ is $m$ x $n$, when $m>n$. Is it possible for $AB$ or $BA$ to be invertible? I know how to prove this using rank, but I would like an explanations that does not use rank.

Answer 1

It is possible for $AB$ to be invertible. For instance take $A = \left(\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right)$ and $B = \left(\begin{array}{rr} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right)$. The product $AB$ is the $2\times 2$ identity matrix (clearly invertible).

It is not possible for $BA$ to be invertible. I'll explain the "moral" reason for this, then I'll give a more concrete proof. The two matrices $A$ and $B$ represent linear transformations between vector spaces. $A$ represents a linear transformation from a larger vector space to a smaller one, and $B$ represents a linear transformation from a smaller space to a larger one. Thus, the product $BA$ represents a linear transformation from the large space to the large space that goes through a smaller space (we read the linear transformations from right to left). Imagine vector spaces as cotton candy. You can easily squish cotton candy, but once it's been squished, you cannot get it to expand again. You start with a big box of cotton candy. $A$ puts the cotton candy in a small box, and in doing so, it must squish the cotton candy. Then $B$ puts the cotton candy back into the big box. There is no way for it to expand to fill the big box again.

For a formal proof, recall a portion of the invertible matrix theorem: a square matrix is invertible if and only if its null space is $\left\{\vec{0}\right\}$. Since $A$ is $n\times m$ ($m > n$), the null space of $A$ is nontrivial, i.e., there is a nonzero vector $\vec{v}$ such that $A\vec{v} = \vec{0}$. But now $BA\vec{v} = B\vec{0} = \vec{0}$ so $\vec{v}$ is in the null space of $BA$. Thus, $\mathrm{Null}(BA)\neq \left\{\vec{0}\right\}$ so $BA$ is not invertible.

Answer 2

I assume that a $n\times m$ matrix has $n$ lines and $m$ columns

$AB$ can be invertible (find an example) but not $BA$ since $A$ represents a linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$ whose kernel is not trivial.

Answer 3

With $n = 1$, and $m = 2$ we have:

$$(1, 0) \times \left(\matrix{ 1 \\ 0 }\right) = 1$$

Geometrically, the matrix representation of any embedding of $\Bbb{R}^n$ into $\Bbb{R}^m$ composed with a projection back on to $\Bbb{R}^n$ gives an example. The above example corresponds to the embedding of the $x$-axis into the plane and the orthogonal projection onto the $x$-axis.