Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to prove if $X$ is a Hilbert space and $M$ and $N$ it's closed: $$ (M+N)^\perp=M^\perp\cap N^\perp $$ thanks

share|improve this question
    
Do you mean $X$ is a Hilbert space, and $M,N\subseteq X$ are closed? It's a bit unclear. Also, what have you tried? Where are you stuck? –  Stahl Mar 27 '13 at 23:16
add comment

1 Answer 1

This is true for every sets $M$ and $N$, in any inner-product space $X$. They don't need to be subspaces, and they don't need to be closed. Note that the orthogonal of any set is always a closed subspace.

In words: a vector $x$ is orthogonal to $M+N$ if and only if it is orthogonal to $M$ and $N$.

Sufficient: assume $(x,m)=0$ for all $m\in M$ and $(x,n)=0$ for all $n\in N$. Then $(x,m+n)=(x,m)+(x,n)=0$ for all $m\in M$ and all $n\in N$. So $(x,k)=0$ for all $k=m+n\in M+N$.

Necessary: assume $(x,k)=0$ for all $k\in M+N$. Since $M\subseteq M+N$,, this implies $(x,m)=0$ for all $m\in M$. Likewise, $(x,n)=0$ for all $n\in N$.

Now a more instructive way for the "necessary" direction: first observe the fundamental related fact $$ S\subset T\quad\Rightarrow\quad T^\perp\subseteq S^\perp. $$ Since $M$ and $N$ are both contained in $M+N$, it folows that $ (M+N)^\perp \subseteq M^\perp$ and $ (M+N)^\perp \subseteq N^\perp$. Hence $ (M+N)^\perp \subseteq M^\perp\cap N^\perp$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.