$AB$ is singular $\Leftrightarrow $ $A$ or $B$ is singular Just for fun, I was solving few linear algebra problems. Hence, I came across the fact that- 


*

*$AB$ is singular $\Leftrightarrow $ $A$ or $B$ is(/are) singular.

*$AB$ is not singular $\Leftrightarrow $ $A$ and $B$ both are not singular.


Notation: $A, B \in \mathbb{F}^{n\times n}$
My question is that let say we have proved the $2$nd fact, does it automatically imply the $1$st fact? Or, I have to prove it separately.
On a footnote, I have proved the second fact using Sylvester's rank inequality. Is there any other elegant way to do it?
Just for the preciseness, you can not use the fact that $\det(AB)=\det(A)\det(B)$ because you have to prove it before you use it, and for the proof let say you need the above mentioned facts.
 A: On the logic part of the question...
Let $X$, $Y$, and $Z$ be propositions respectively that $AB$ is singular, $A$ is singular, and $B$ is singular. Statement 2 is:
\begin{equation}\neg X \leftrightarrow \neg Y \wedge \neg Z\end{equation}
Statement 1 is:
\begin{equation} X \leftrightarrow Y \vee Z \end{equation}
These are logically equivalent (simple application of De Morgan's laws).
On the singular question...
Direction 1:
If $B^{-1}$ exists and $A^{-1}$ exists then $B^{-1}A^{-1}$ is an inverse of $AB$.
Direction 2: If $(AB)^{-1}$ is an inverse of AB. Then:

*

*$(AB)^{-1}(AB) = I$
$((AB)^{-1}A)B = I$ hence inverse of $B$ exists (where $B^{-1} = (AB)^{-1}A$)


*$(AB)(AB)^{-1} = I$
$A(B(AB)^{-1}) = I$ hence inverse of $A$ exists
A: $A \in M(n,F)$ is singular iff any of the following equivalent conditions are met.
$A$ is not injective (has non-trivial kernel)
$A$ is not surjective(image rank less than domain rank)
$A$ has zero determinant
$A$ has $0$ as an eigenvalue
$A$ has no inverse
$A$ is a zero divisor
in validating the two logically equivalent statements you present concerning products, it is to some extent a matter of taste which route you take. for example if $AB$ has zero as an eigenvalue, let $v$ be an eigenvector, so $ABv=0$. clearly if $Bv \ne 0$ then $Bv$ is a non-trivial eigenvector for $A$ with the eigenvalue $0$
A: These statements are literally just contrapositives of each other. If you prove one, you get the other free of charge.
A: Assume that $AB$ is singular then there exists $v\in Ker(AB)$ with $v\neq 0$ (since the kernel is not trivial), it follows that $ABv=0$ then either $Bv=0$ and then $B$ is singular either $Bv\neq 0$ but $Bv$ is in the kernel of $A$ so $A$ is singular.
Assume that $B$ is singular, then take $v\in Ker(B)$ which is not trivial, clearly $ABv=0$ and $v$ is a non-trivial element in $Ker(AB)$ whence $AB$ is singular.
Assume that $A$ is singular then $rk(A)<n$ but $rk(AB)\leq rk(A)$ (trivial inequality) so that $rk(AB)<n$ and $AB$ is singular.
