Show that: If $A \circ B =0$ but $A \neq 0$ and $B\neq 0$ then you have that $\det(A)=0=\det(B)$ 
Show that: If $A$ and $B$ are real $n \times n$ matrices with $A \circ
B=0$ but $A \neq 0$ and $B \neq 0$, then you have that
  $\det(A)=0=\det(B)$
($0$ is the matrix which only has zeroes)

I'm not sure how to do this correctly but something tells me that a proof by contradiction would be very useful here.
Let $\det(A) \neq 0 \Rightarrow B = Id \circ B = A^{-1} AB = 0 \Rightarrow B=0 
\,\,\,\unicode{x21af}$
Let $\det(B) \neq 0 \Rightarrow A = Id \circ A = B^{-1} BA = 0 \Rightarrow A=0 \,\,\,\unicode{x21af}$
And that's why we must have that $\det(A)=0=\det(B)$
I think this should work because we just showed a contradiction for every possibility and this means that the negation of our assumption must be correct.
But will this also cover that both determinants must necessarily be zero? I think it only shows that at least one determinant must be zero... : /
Maybe there is a better way to prove this?
 A: Your proof using a property of determinant is OK. Here is a second proof:
We have $A\circ B=0$ so for every $X\in\mathcal M_{n,1}(\Bbb R)$:
$$A(BX)=0\implies Im(B)\subset \ker A$$
If $B$ is invertible then $Im(B)=\mathcal M_{n,1}(\Bbb R)=\ker A\implies A=0$ which is a contradiction so $B$ isn't invertible. Moreover, $B\ne0$ so $Im B\ne\{0\}$ and so $\ker A\ne\{0\}$ which means that $A$ isn't invertible. Ofcourse you can translate the non invertibility in term of determinant.
A: You can use the formule of determinants $$Det(AB)=Det(A).Det(B)$$, thus suppose that $$Det(A)=0$$ and $$Det(B)\neq 0$$ thus $$B$$ is invertible and $$AB=0$$ Multiplying by inverse B concluded that $$A=0$$ absurd.
A: Yes, your proof is okay (aside from Lord Shark the Unknown's comment), because you do it like this:

We want to show that $\det(A)=0$ and $\det(B)=0$. Now for proof by contradiction, assume $\det(A)\neq 0$ or $\det(B)\neq 0$.
Case 1: $\det(A)\neq 0$. Then ... hence contradiction.
Case 2: $\det(B)\neq 0$. Then ... hence contradiction.
Since all cases lead to a contradiction, our assumption "$\det(A)\neq 0$ or $\det(B)\neq 0$" must be wrong, hence $\det(A)=0$ and $\det(B)=0$.

