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I need to prove:

Let $V$ be a real linear space with an inner product. Prove that if $A:V\to V$ is a symmetric linear operator, then the null space of $A$ is orhogonal to the image space of $A$.

I only have $N \subseteq R^\perp$: Take a random $x\in N$ and a random $y\in R, y=Au$. Then $(x,y)=(x,Au)=(Ax,u)=(0,u)=0 \implies x\in R^\perp \implies N\subseteq R^\perp$

Edit: But I still need to prove the other side

egreg
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Pieter
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2 Answers2

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To prove the stronger statement you indicate after the question in the highlighted box, note that \begin{align} w \in \mathcal{R}(A)^{\perp} &\iff (Av,w)=0\;\forall v\in V \\ &\iff (v,Aw)=0\;\forall v\in V \\ &\iff Aw=0 \end{align} The last equivalence holds because $Aw=0$ implies $(v,Aw)=0$ for all $v$, and because, if $(v,Aw)=0$ for all $v\in V$, then it holds for $v=Aw$, which implies $Aw=0$. Therefore, $$ \mathcal{R}(A)^{\perp} = \mathcal{N}(A). $$ Hence, $$ V = \mathcal{R}(A)\oplus\mathcal{R}(A)^{\perp}=\mathcal{R}(A)\oplus\mathcal{N}(A). $$

Disintegrating By Parts
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Let $x \in R^{\perp}$, then $(x,Av)=0$, for all $v\in V$.

In particular, $(Ax,Ax)=(x,AAx)=0$, so $Ax=0$ and $x\in N$. Therefore $R^{\perp}\subset N$

egreg
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Nick
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  • Sorry i dont really understand your answer, this is probably because sometimes only 1 side of the $<$ is filled in, so i cant see for sure if this is good or wrong – Pieter Mar 23 '17 at 17:44
  • I edited in order to use the same notation as the OP, which should make this clearer. – egreg Mar 23 '17 at 22:39