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Two squared matrices $A$ and $B$, with $B\neq0$, give $AB=0$. Prove that $\det(A)=0$.

After trying with some examples, I believe that $A$ needs to have lines that are equal or can be made equal by scalar multiplication, B needs to have columns that are equal or can be made equal by scalar multiplication, like $$ A= \begin{bmatrix} 1 & 2 \\ 2 & 4 \\ \end{bmatrix} $$ and $$ B= \begin{bmatrix} 2 & 4 \\ -1 & -2 \\ \end{bmatrix} $$

which would mean that $\det(A)=0$ and $\det(B)=0$. But this is still far from being a proof of anything. Am I on the right track? What would be my next step?

Michael Hardy
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Anna Sarah
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  • Do you know that det$(AB)=$det$(A)$det$(B)$. – HarshCurious Nov 02 '16 at 17:01
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    I don't know whether this helps, but $\det(0)=0$ and $\det(AB)=\det(A)\det(B)$. So $\det(A)\det(B)=0$, hence $\det(A)=0$ or $\det(B)=0$. Unfortunately, $B \neq 0$ doesn't imply $\det(B)\neq0$ – wythagoras Nov 02 '16 at 17:01
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    Let $b$ be any nonzero column of $B$. Then $AB = 0$ implies $Ab = 0$. What does this tell you about $A$? –  Nov 02 '16 at 17:02

2 Answers2

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By contradiction: suppose $\det A \ne 0$, than $A$ is invertible and, left multiplying $AB=0$ by $A^{-1}$, we have: $$ A^{-1}(AB)=A^{-1}\cdot 0 \iff (A^{-1}A)B=0 \iff I B= 0 \iff B=0 $$

Emilio Novati
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If $AB=0$ then $Ab_i=0$ $\forall 1\leq i\leq n$ (where $b_i$ denotes the i-th column of B. Let $b_j$ be any non-zero column of B then we have: $Ab_j=0$ which means that $b_j$ is an eigenvector of A with corresponding eigenvalue 0. Since 0 is an eigenvalue of A then $det(A)=0$

Aggelos Bessis
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  • Sorry, didn't see that B was a sqaure matrix. – Aggelos Bessis Nov 02 '16 at 17:38
  • Forget I said anything – Aggelos Bessis Nov 02 '16 at 17:38
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    You should be able to delete your answer. – wythagoras Nov 02 '16 at 17:41
  • Your argument works if you let $b$ be any nonzero column of $B$. Then $b$ is an eigenvector for eigenvalue zero. –  Nov 02 '16 at 17:42
  • So if $AB=0$ then $ Ab_i=0 \forall i $ where $b_i$ is the i-th column of B. Then B is made of eigenvectors that correspond to the eigenvalue 0. Is this better? – Aggelos Bessis Nov 02 '16 at 17:51
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    Some of the columns of $B$ may be zero, so those columns won't count as eigenvectors. However, $B$ is nonzero, so at least one of its columns must be nonzero, and that column is an eigenvector which corresponds to the eigenvalue zero. –  Nov 02 '16 at 22:48