TITLE: Can product of two singular matrices be invertible?
QUESTION [6 upvotes]: Suppose $A,B$ are square matrices of size $n\times n$.  Can $AB$ be invertible, even though both $A$ and $B$ are singular (not invertible)?
And if not, does it follow that if $A_1 \times A_2 \times ... \times A_n$ is the invertible product of square matrices, then all factors $A_i$ are invertible?

REPLY [11 votes]: $$\det(A_1\cdot A_2\cdot\ldots\cdot A_n)\neq0\quad\Leftrightarrow\quad \det A_1\cdot\det A_2\cdot\ldots\cdot\det A_n\neq0\quad\Leftrightarrow\quad\det A_k\neq 0\quad \forall k$$

REPLY [9 votes]: Suppose $A$ and $B$ are such that $(AB)^{-1} = C$. Then $ABC = I$, so $A$ has inverse $BC$. Likewise, $CAB = I$, so $B$ has inverse $CA$.
(This relies on knowing that any left inverse is also a right inverse and vice versa.)

REPLY [8 votes]: If $A$ and $B$ are not invertible, then $AB$ is also not-invertible. More generally, $$\text{rank}(AB) \leq \max(\text{rank}(A), \text{rank}(B))$$
And the answer to your second question is yes as well by inducting the above argument.