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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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    @ccc : See [this question](http://math.stackexchange.com/questions/246967/show-w-1-w-2-perp-w-1-perp-cap-w-2-perp-and-w-1-perp-w-2-perp/247717#247717)2012-12-05

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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 $d\in (M+N)^{\perp}$

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    Thank you very much Tomas!2012-12-05