If $A$ is $4\times2$ and $B$ is $2\times4$, prove that $AB$ is not invertible My attempt :
If $A$ is a matrix of size $4\times2$ and $B$ a matrix of size $2\times4$. Then $AB$ is a matrix of size $4\times4$ .
If $(AB)^{(-1)}$ exists then it would be of the form $$B^{-1}A^{-1}.$$ 
We see that $B^{-1}$ doesn't exist as the size of left inverse and right inverse will not be equal, so they are not equal.  Similarly $A^{-1}$ does not exist. Am I correct? 
 A: It is true that $AB$ is not invertible, but your reasoning is not correct. To see that there must be an error in your reasoning, try to find an example of matrices $A$ and $B$ as in your question for which the $2 * 2$-matrix $BA$ is invertible. This is doable.
In order to see why $AB$ is not invertible it helps to think about the linear maps the matrices encode. In order to be invertible the map corresponding to $AB$ must be injective and surjective. But think about what happens to a vector in 4-dimensional space that get fed into $AB$. First it is squeezed into a 2-dimensional space by $B$ (which causes a problem with injectivity), and then the points of this two-dimensional space are mapped into a much bigger 4 dimensional space by $A$, which causes a problem with surjectivity.
I am sure you can fill in the details in this argument.
A: By the rank-nullity theorem we get $\dim ker(B) \ge 2.$  Thus there is $x \in ker(B)$ with $x \ne 0$, then $ABx=0$. This shows that $AB$ is not invertible.
A: Your proof isn't correct, as pointed out by Eric.
The point is that the rank of $AB$ is at most 2 :
Say we are working with matrices with coefficients from a field $K$. $A$ represents a morphism $\varphi : K^4 \mapsto K^2$. Hence, $rank(A) = dim(\operatorname{Im}(\varphi)) \leqslant 2$. Say $B$ represents a morphism $\psi$. The rank of $AB$ is $dim(\operatorname{Im}(\varphi \circ \psi)) \leqslant dim(\operatorname{Im}(\varphi))\leqslant 2$. If $AB$ were invertible, it would have rank 4.
