Show that AB is singular if A is singular Actually I need to show that $\det(AB) = \det(A)\det(B)$ if $A$ is a singular matrix.
The determinant of $A$ is $0$ if $A$ is singular, so $\det(AB)$ has to be $0$ as well, but I have problems showing that $AB$ is singular if $A$ is singular. How can I show that?
 A: One approach is this: That a matrix $C$ is singular gives us in particular that its null space is non-trivial, that is, for some vector $x\ne0$ we have $Cx=0$. That $C$ is nonsingular, on the other hand, gives us in particular that the column space of $C$ has full rank, that is, for any vector $b$ there is a vector $a$ such that $Ca=b$.
Now, suppose $A$ is singular. If $B$ is also singular, then for some $x\ne 0$ we have $Bx=0$, but then $(AB)x=A(Bx)=A0=0$, and we conclude that $AB$ is also singular.
If, on the other hand, $B$ is nonsingular, use that $A$ is singular to find $b\ne 0$ such that $Ab=0$. Now, use that $B$ is nonsingular to find $a$ such that $Ba=b$. Clearly $a\ne0$ since $b\ne0$. But now we have that $(AB)a=A(Ba)=Ab=0$, and we conclude (again) that $AB$ is singular. 
This completes the proof. Notice, by the way, that we also showed that 1) $AB$ is singular if $B$ is the one assumed singular. On the other hand, since $A,B$ being nonsingular gives us that $AB$ is nonsingular, then we also have that 2) if $AB$ is singular, then at least one of $A$ and $B$ must be singular as well.
A: If $A$ is singular then it isn't injective: there's $y\ne0$ such that
$$Ay=0$$
Now 


*

*if $B$ is invertible then let $x$ such that $Bx=y$ and then 
$$ABx=Ay=0$$
and

*if $B$ is also singular then there's $z\ne0$ such that $Bz=0$ and then
$$ABz=0$$
so we prove that $AB$ isn't injective which's equivalent to $AB$ is singular.

A: Hint: if $x^TA=0$ for some nonzero vector $x$, then ...
A: Let $A$ be a singular matrix. Suppose by way of contradiction that $AB$, for some matrix $B$, has an inverse. Then, there exists some matrix C such that $(AB)C=C(AB)=I$, where I is the identity matrix. But then, $(AB)C=A(BC)$ by associativity. Then imposing $A(BC)=I$ leads us to $BC=A^{-1}$. Which is a contradiction since A was singular by the problem. 
A: Contrapositive: If AB is not singular, then A is not singular. If AB is not singular, then it has an inverse. Its inverse is $B^{-1}A^{-1}$ which implies that B and A are not singular.
Expansion of my answer: if AB is not singular, then A is not singular because if AB is not singular, then AB has an inverse. AB's inverse is $B^{-1}A^{-1}$ which implies that B and A are not singular which implies A is not singular.
