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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?

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    $\begingroup$ In second case, you need $A=ABB^{-1}=0$ but otherwise this is fine. $\endgroup$ – Angina Seng May 8 '18 at 20:20
  • $\begingroup$ Dose $\circ$ means component-wise matrix multiplication? $\endgroup$ – Mahdi May 9 '18 at 18:52
  • $\begingroup$ @Mahdi Yes! It's also called "Hadamard product". As far as I know, you can only use it on matrices of same size and for this reason the Hadamard product is also always commutative (the normal matrix product isn't). en.wikipedia.org/wiki/Hadamard_product_(matrices) $\endgroup$ – cnmesr May 9 '18 at 21:49
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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.

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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.

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  • $\begingroup$ \det(A) is a defined MathJax command as well as \cdot $\endgroup$ – SK19 May 8 '18 at 23:22
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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 $\det(B)=0$.

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