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We have two matrices $A$ and $B$, and we know that $AB=0$. Does it mean that $A$ is the null space of $B$?

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    $\begingroup$ A claim that is true and rings a tune in the same key as what you said is that the columns of $B$ are in the null space of $A$. Similarly, transposing the tune, the rows of $A$ (seen as columns) are in the null space of $B^t$. $\endgroup$
    – user561348
    Commented May 20, 2018 at 13:26
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    $\begingroup$ Are you supposed to interpret $A $ as a linear transformation from a set of matrices into a set of matrices? In this case $B $ could be said to be in the null space of $A $. If not (which is likely), see the answer by @Henning Makholm $\endgroup$
    – AnyAD
    Commented May 20, 2018 at 13:34

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It implies that each column of $B$ is in the null space of $A$.

Since the null space is a set of vectors and $B$ is a matrix, $B$ itself cannot be in the null space. It is not guaranteed either that the columns of $B$ will span the null space, although that is possible.

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  • $\begingroup$ So we cannot say that each column of $A$ is in the null space of $B$? $\endgroup$
    – eHH
    Commented May 20, 2018 at 14:20
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    $\begingroup$ @KeH: No; consider for example $$ A=\begin{pmatrix}1&0\\1&0\end{pmatrix} \qquad B=\begin{pmatrix}0&0\\1&1\end{pmatrix}$$ Why do you think the columns of $A$ would have be in the null space of $B$? $\endgroup$ Commented May 20, 2018 at 14:30
  • $\begingroup$ Yeah you're right. Generally, we cannot say so. $\endgroup$
    – eHH
    Commented May 20, 2018 at 17:13
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    $\begingroup$ Note that the rows of $A$ are in the left null space (cokernel) of $B$: en.wikipedia.org/wiki/Kernel_(linear_algebra)#Left_null_space $\endgroup$
    – Noiralef
    Commented May 20, 2018 at 18:33
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Assuming that they are both $n\times n$ matrices over a field $F$, the null space of $B$ is a vector subspace of $F^n$, whereas $A$ is not such a subspace. So, the answer is negative.

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  • $\begingroup$ How do you know that "$A$ is not such a subspace"? could you please explain more? $\endgroup$
    – eHH
    Commented May 20, 2018 at 13:45
  • $\begingroup$ @KeH Because $A$ is a matrix, not a vector space. $\endgroup$ Commented May 20, 2018 at 13:47

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