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.

  • 4
    $\begingroup$ If $\det A = 5$, then $A$ is nonsingular. So, $AB=0$ implies $B=0$. $\endgroup$ – Sungjin Kim Dec 5 '16 at 22:01
  • 3
    $\begingroup$ 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
  • 3
    $\begingroup$ If you assume $\det(A)\ne 0$, then $B$ must be $0$. $\endgroup$ – user251257 Dec 5 '16 at 22:02
  • 2
    $\begingroup$ $\det(AB)=\det(A)\det(B)$ 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


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 $B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.