# If matrices $B$ and $AB$ have the same rank, show that they have the same null spaces.

Not sure how to do this one. A rather big hint was given, but I can't figure out how to actually do the second half of the hint:

First show that $null(B)\subseteq null(AB)$, then use the rank+nullity theorem.

So I figure I got the first part done:

\begin{align} \vec{x}\in null(B)&\Longrightarrow B\vec{x}=\vec{0}\\ &\Longrightarrow A(B\vec{x})=\vec{0}\\ &\Longrightarrow (AB)\vec{x}=\vec{0}\\ &\Longrightarrow \vec{x}\in null(AB)\\ &\Longrightarrow null(B)\subseteq null(AB) \end{align}

Not certain how to proceed from here... Any tips would be great.

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This is not correct: $Bx = 0 \iff A(Bx) = 0$. It should be $Bx = 0 \implies A(Bx) = 0$. –  Vedran Šego Oct 3 '13 at 13:22
Thanks, that makes perfect sense. –  agent154 Oct 3 '13 at 13:24
Now, what does rank+nullity theorem say about $B$? Combined with the fact that $B$ and $AB$ have the same rank, what does this tell you about the nullity of $AB$? –  Vedran Šego Oct 3 '13 at 13:27
You should have other hypotheses on the matrices $A$ and $B$. Are they square? –  egreg Oct 3 '13 at 13:31
@egreg Sadly, no. This is all I was given. –  agent154 Oct 3 '13 at 13:43

Based on the suggestions of Vedran, I think the following might be a solution:

If $\vec{x}$ is in the null space of $B$, then: \begin{align} \vec{x}\in null(B)&\Longrightarrow B\vec{x}=\vec{0}\\ &\Longrightarrow A(B\vec{x})=\vec{0}\\ &\Longrightarrow(AB)\vec{x}=\vec{0}\\ &\Longrightarrow\vec{x}\in null(AB)\\ &\Longrightarrow null(B)\subseteq null(AB) \end{align}

Since $null(B)$ is a subspace of $null(AB)$, we can conclude that the nullity of $B$ is less than or equal to the nullity of $AB$.

$A$ is an $m\times n$ matrix while $B$ is an $n\times p$ matrix, $AB$ is an $m\times p$ matrix. Therefore, $B$ and $AB$ both have the same number of columns. Let $p$ equal the number of columns in both $B$ and $AB$.

Let $k$ equal the rank of both $B$ and $AB$. Let $l$ equal the nullity of $B$ and $o$ equal the nullity of $AB$. Given that $B$ and $AB$ have the same number of columns: \begin{align*} k+l&=p\\ k+o&=p\\ l&=o \end{align*} Therefore, the nullity of $B$ is equal to the nullity of $AB$, and we can then conclude that $null(B)=null(AB)$. $\square$

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Close, but not quite. The rank doesn't refer to the nullspace of a matrix. –  john Oct 3 '13 at 13:53
@john Hmm.. Yes. I should have said that the nullity of $B$ is less than or equal to the nullity of $AB$. But I can't figure out how to combine that with the rank now, given that I don't know what the sizes of $B$ and $AB$ are. I don't know if they have the same number of columns or not. –  agent154 Oct 3 '13 at 14:01
Think through the matrix multiplication. Suppose that $B$ is an $n\times m$ matrix. What are the conditions on $A$ for the matrix product $AB$ to exist? What will be dimensions of $AB$? –  john Oct 3 '13 at 14:04
@john I think I have it now... –  agent154 Oct 3 '13 at 14:22
Excellent! You could probably stand to improve your writing a bit (not a trivial statement, though this often comes with experience) but the important parts of the proof are all there. –  john Oct 3 '13 at 14:26

Assume that $A$ is $m\times n$ and $B$ is $n\times p$. What you know from the rank-nullity theorem is that \begin{gather} \operatorname{rk}(AB)+\operatorname{nl}(AB)=p\\ \operatorname{rk}(B)+\operatorname{nl}(B)=p \end{gather} (where $\operatorname{nl}(X)$ denotes the dimension of $\operatorname{null}(X)$). You also know that $\operatorname{null}(B)\subseteq\operatorname{null}(AB)$, because $x\in\operatorname{null}(B)$ means $Bx=0$, so that also $ABx=0$.

Subtract the equalities above, to get …

If $U_1\subseteq U_2$ are subspaces of $V$ and their dimensions …, then $U_1=U_2$.

(Fill in the blanks)

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HINT:

The rank-nullity theorem tells us that

rank + nullity = no. of columns.

How do the number of columns of $B$ and $AB$ compare? (or do we not know?)

Can we use this to find the nullity of $B$ and/or $AB$?

How does the nullity relate to the nullspace?

.

That should hopefully be quite a lot to go on.

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