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I am trying to build two non-zero square matrices $A$ and $B$ whose product will be zero and who will have any fixed determinant value (e.g. det$(A) = 5$).

I can easily think of two non-zero square matrices that satisfy $AB = 0$, but to get them to have a specific determinant is tripping me up.

Would anyone know of a first step? I imagine it would be easy to start with two triangular matrices.

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    $\begingroup$ If $\det A = 5$, then $A$ is nonsingular. So, $AB=0$ implies $B=0$. $\endgroup$ – Sungjin Kim Dec 5 '16 at 22:01
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    $\begingroup$ If $\det A \ne 0$, then $A$ is invertible, so from $AB=O$ we conclude $B=O$. Then getting any determinant you want for $A$ should be easy. $\endgroup$ – GEdgar Dec 5 '16 at 22:02
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    $\begingroup$ If you assume $\det(A)\ne 0$, then $B$ must be $0$. $\endgroup$ – user251257 Dec 5 '16 at 22:02
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    $\begingroup$ $\det(AB)=\det(A)\det(B)$ so do you have an example of $A$ $B$ such that neither has determinant 0 but their product does have determinant 0? $\endgroup$ – Kitter Catter Dec 5 '16 at 22:03
  • $\begingroup$ Thanks very much, all. I now see that I was missing some key ideas when I asked this question. $\endgroup$ – on-pasta Dec 6 '16 at 3:37
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Hint:

If $A \ne 0$ and $B\ne 0$ are such that $AB=0$, than $A$ and $B$ are not invertible and this means that $\det A =0$ and $\det B=0$.

You can prove this by contraposition. Suppose $A$ is invertible, than

$$ AB=0 \Rightarrow A^{-1}(AB)=A^{-1}0 \Rightarrow (A^{-1}A)B=0\Rightarrow B=0 $$

and analogously for $B$.

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