Let $M$, $N$ be linear sub spaces of a Hilbert space $H$.
How to show that $$(M+N)^\perp= M^\perp\cap N^\perp?$$
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1This is pretty obviously false (look at lines in $\Bbb{R}^2$, say). – Chris Eagle Dec 05 '12 at 11:15
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I'm sorry,there was a mistake.that should be the intersection.I've corrected it. – ccc Dec 05 '12 at 11:17
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@ccc : See this question – Thibaut Dumont Dec 05 '12 at 13:04
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Quick beginner guide for asking a well-received question + please avoid "no clue" questions – Anne Bauval May 21 '23 at 18:57
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If $d\in (M+N)^{\perp}$ then, for $m\in M $ and $n\in N$, $(d,m+n)=0$. Take $n=0$ and then $m=0$ to conclude that $d\in N^\perp\cap M^\perp$.
On the other hand, if $d\in N^\perp\cap M^\perp$ then you have $d\in N^\perp$ and $d\in M^\perp$. This implies that $(d,n)=0$ and $(d,m)=0$ for $m\in M$ and $n\in N$, hence, $(d,n+m)=(d,n)+(d,m)=0$. Therefore $d\in (M+N)^{\perp}$
Tomás
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