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I am currently trying to show that the Row space of $V$ is the orthogonal complement of the Null space of $V$. That is:

$R(V) = N(V)^\perp$.

This seems like a straight-forward proof and my strategy is to prove that $R(V) \subset N(V)^\perp$ and that $N(V)^\perp \subset R(V)$. I think I have the first down, but am not sure how I would approach the second. Would anyone have any ideas? Thank you!

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2 Answers 2

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Let $F_1,\cdots, F_m$ denote the files of $A.$ Then:

$$x\in N(A) \Leftrightarrow Ax=0 \Leftrightarrow x\perp F_i, i=1,\cdots,m \Leftrightarrow x\in R(A)^{\perp}.$$

So, $N(A)=R(A)^{\perp}$ or, equivalently, $R(A)=N(A)^{\perp}.$

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Proof:

For a linear map $V: \mathbb{R}^n \to \mathbb{R}^m$, let's define:

  • Row space of V: $R(V) = \{z\in\mathbb{R}^m:\exists y\in\mathbb{R}^n,z=V^Ty,\}$
  • Null space / kernel of V: $N(V) = \{x\in\mathbb{R}^n:Vx =0\}$

Then, $$\begin{align} & x\in N(V)\\ & \Leftrightarrow Vx=0\\ & \Leftrightarrow \langle y,Vx\rangle_{\mathbb{R}^m}=0, \quad\forall y\in \mathbb{R}^m\\ & \Leftrightarrow \langle V^Ty,x\rangle_{\mathbb{R}^n}=0, \quad\forall y\in \mathbb{R}^m\\ & \Leftrightarrow \langle z,x\rangle_{\mathbb{R}^n}=0, \quad\forall z\in R(V)\\ & \Leftrightarrow x\perp R(V)\\ & \Leftrightarrow x\in R(V)^{\perp} \end{align} $$ Thus, $N(V)=R(V)^{\perp}$ and vice versa.

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  • $\begingroup$ I was confusing range for row space. Corrected it now. $\endgroup$
    – tvo
    Oct 11, 2016 at 16:43
  • $\begingroup$ Notice now that your proof is basically the same as mfl's. $\endgroup$
    – user137731
    Oct 11, 2016 at 16:44
  • $\begingroup$ Of course, it has always been along similar lines of thought. But my proof uses the inner product to proof the equivalence algebraically, whereas mfl's relies on more abstract logical steps. It's personal taste I guess... $\endgroup$
    – tvo
    Oct 11, 2016 at 16:51

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