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If there are two matrices, lets say A & B such that $$ AB = 0 $$ and A is a non singular matrix and B may or may not be a square matrix . Can we infer anything about nature of B . The book says B is a zero matrix but I am unable to prove.

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  • $\begingroup$ Like you do with numbers, multiply, from the left, both sides by $A^{-1}$. $\endgroup$ – orole Jan 18 '18 at 20:56
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If $A$ is non-singular, there exists $A^{-1}$. So $$AB = 0 \implies A^{-1}(AB) = A^{-1}0 \implies (A^{-1}A)B = 0 \implies {\rm Id}\; B = 0 \implies B = 0.$$The moral of the history is that if $A$ is non-singular, you can """""divide by $A$""""".

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Here's another approach: If the nullity of $AB$ is $n$, the nullity of $A$ plus the nullity of $B$ must be at least $n$. Can you see why?

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  • $\begingroup$ I have read about 'nullity' in context of linear transformations. Not sure what you mean here ! $\endgroup$ – Anuj Jan 18 '18 at 20:59
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    $\begingroup$ A $n \times m$ matrix defines a linear transformation $T_A\colon \Bbb R^m \to \Bbb R^n$, right? The nullity of $A$ is defined as the nullity of $T_A$. $\endgroup$ – Ivo Terek Jan 18 '18 at 21:07
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If $A$ is non-singular, $A^{-1}$ exists.

Premultiply $A^{-1}$ to the equation to obtain the desired result.

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If $A$ is nonsingular then it is invertible. Thus, B = $A^{-1}\cdot 0 = 0$.

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