Prove : $\rm{col}(B) \subseteq \rm{null}(A)$

Let $A$ be an $m\times n$ matrix and $B$ be a $n\times m$ matrix.

Prove that:

$$AB = 0 \iff \rm{col}(B) \subseteq \rm{null}(A)$$

Here is the problem in my textbook. I just have a simple solution if $B$ is transpose of $A$. But at general case, I don't have result yet.

Thanks :)

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I'll assume your field of scalars is $\mathbb{R}$ (not that it matters).

\begin{align*} &AB = 0 \\ \Leftrightarrow & (AB)x = 0 \quad \forall \, x \in \mathbb{R}^m \\ \Leftrightarrow & A (Bx) = 0 \quad \forall \, x \in \mathbb{R}^m \\ \Leftrightarrow & A y = 0 \quad \forall \, y \in \textbf{col}(B) \\ \Leftrightarrow & \textbf{col}(B) \subset \textbf{null}(A). \end{align*}

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Hints:

1. If $AB=0$ and $v \in \mbox{col}(B)$ then $v=Bx$ for some $x$, so $Av=???$.
2. If $\mbox{col}(B) \subseteq \mbox{null}(A)$ then for all $x$ we have $Bx \in ???$, which means that $ABx=???$.
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… 3. ??? 4. PROFIT! (Sorry, couldn't resist.) –  k.stm Nov 4 '12 at 9:11

You can split a matrix product like $AB$ according to the columns of the right factor $B$: each column of $AB$ is equal to $A$ applied to the corresponding column of $B$ (this is how matrix multiplication works; you can also split the product according to the rows of $A$, but that is not helpful here).

So saying $AB=0$ means precisely that for each column $c$ of $B$ you have $Ac=0$, or $c\in\operatorname{null}A$. So you have to prove the column space of $B$ is contained in $\operatorname{null}A$ if and only if each individual column of $B$ lies in this same space $\operatorname{null}A$. You should be able to prove that yourself.

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