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How can I prove that $\operatorname{Col}(A)^{\bot} = \operatorname{Null}(A^T)$, for $A$ an $m\times n$ matrix?

I know that $\operatorname{Null}(A^T) = \{ x \in \mathbb{R}^n : A^Tx = 0 \}$, but I'm not sure about $\operatorname{Col}(A)$, $\operatorname{Row}(A)$, $\operatorname{Col}(A)^{\bot}$, $\operatorname{Row}(A)^{\bot}$.

I do know that, for example, $\operatorname{Col}(A)$ is the span of columns of $A$. I also do know that $\dim\operatorname{Col}(A)^{\bot} = \dim\operatorname{Null}(A^T)$. It is really easy to prove that it is $n-k$; but then I will need to prove $\subseteq$. If it is possible I'd love to know what is the group for $\operatorname{Col}(A)^{\bot}$ like I wrote for $\operatorname{Null}(A)$ in the beginning of the post, and how to prove with $\subseteq$ after I show that dimensions are equal. (So in total, two ways to prove it.)

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  • $\begingroup$ $\operatorname{Col}(A)^\perp = \{ y \in \mathbb{R}^m : (x,y)=0 \text{ for all }x \in \operatorname{Col}(A)\}$ $\endgroup$ – Nigel Overmars Mar 25 '16 at 12:16
  • $\begingroup$ @NigelOvermars I don't know how this helps me. can I use something simpler? like $Col(A)^{\bot} = \{ y \in R^M : y^Tx = 0 \forall x \in Col(A)\}$. I don't know if it is correct, but I don't understand what you wrote either, and how can I use what you wrote to prove the equality. $\endgroup$ – Ilan Aizelman WS Mar 25 '16 at 12:20
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A vector $v\in\mathbb{R}^n$ belongs to $\operatorname{Col}(A)$ if and only if there exists $x\in\mathbb{R}^m$ such that $Ax=v$.

Suppose $v\in\operatorname{Col}(A)$ and $w\in\operatorname{Null}(A^T)$. Then, with $v=Ax$, $$ \langle v,w\rangle=v^Tw=(Ax)^Tw=(x^TA^T)w=x^T(A^Tw)=x^T0=0 $$ Therefore, every vector in $\operatorname{Null}(A^T)$ is orthogonal to every vector in $\operatorname{Col}(A)$. This is the same as saying that $$ \operatorname{Null}(A^T)\subseteq\operatorname{Col}(A)^\perp $$

Now, if $k$ is the rank of $A$, we know that $\dim\operatorname{Col}(A)=k$, so $\dim\operatorname{Col}(A)^\perp=n-k$.

By the rank nullity theorem, $\dim\operatorname{Null}(A^T)=n-k$ as well, so the inclusion above becomes equality.

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  • $\begingroup$ egreg, this is perfect. Thank you! $\endgroup$ – Ilan Aizelman WS Mar 25 '16 at 12:39

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