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.

Is it true that if $A,B$ are closed subsets of a Hilbert space $H$, such that $A\perp B$, we have $A+B+(A\cup B)^{\perp} =H$ ? What if $A,B$ are closed subspaces ?$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

share|improve this question
I guess you meant subspaces (as in your title), not just arbitrary subsets (as in your question test). –  celtschk Aug 22 '12 at 15:14
@celtschk Not necessarily... –  user38178 Aug 22 '12 at 15:26
In that case, how do you define $A+B$? $\{a+b:a\in A \land b\in B\}$? And how exactly do you define $M^\perp$? The set of all vectors which are orthogonal to all vectors in $M$? If I guessed right for both definitions, it's trivial to give a counterexample. –  celtschk Aug 22 '12 at 15:29
@celtschk yes, I would have defined them like you did. Ok, so $A$ and $B$ have to be subspace for the equation to be more meaningful... –  user38178 Aug 22 '12 at 15:34

1 Answer 1

up vote 2 down vote accepted

If $A$ and $B$ are arbitrary sets, it is not hard to construct a counterexample:

Let $\{e_i\}$ be an orthogonal basis of the Hilbert space. Let $A=\{e_1\}$ (that is, the set containing nothing but $e_1$, and $B=\{e_2\}$. Then $A\cup B=\{e_1,e_2\}$, and thus $(A\cup B)^\perp$ consists of all vectors whose first two components are $0$ (because if they aren't, they are either not orthogonal on $e_1$ or not orthogonal on $e_2$). Furthermore, $A+B+(A\cup B)^\perp$ consists of all vectors where the first two components are $1$ (because they are the sum of the only element of $A$, $e_1$, the only element of $B$, $e_2$, and any vector orthogonal to then). Note that not even $e_1$ itself fulfils that condition (because its $e_2$ component is $0$),thus clearly $A+B+(A\cup B)^\perp$ is not the full Hilbert space.

Now things are different if $A$ and $B$ are subspaces. In that case, be $P_A$ the orthogonal projector onto $A$ and $P_B$ the orthogonal projector onto $B$. Since $A$ and $B$ are orthogonal, $P_AP_B=P_BP_A=1$ and thus $P_A+P_B$ is also an orthogonal projector: $(P_A+P_B)^2 = P_A^2+P_AP_B+P_BP_A+P_B^2=P_A+P_B$. And so is $P_C:=1-P_A-P_B$. By construction, $P_A+P_B+P_C=1$.

Now consider an arbitrary vector $v\in H$. Then we can define $v_A=P_Av$, $v_B=P_Bv$ and $v_C=P_Cv$. Clearly $v_A+v_B+v_C = (P_A+P_B+P_C)v = v$. Also $v_A\in A$ and $v_B\in B$ because by definition, $P_A$ and $P_B$ are the orthogonal projectors to $A$ and $B$. So all left to prove is that $v_C\in(A\cup B)^\perp$.

Consider an arbitrary vector $w\in(A\cup B)$. Then either $w\in A$ or $w\in B$. Assume without loss of generality that $w\in A$. Then $P_Aw=w$. Now we calculate the scalar product with $v_C$: $$\langle w,v_C\rangle = \langle P_Aw,P_Cv\rangle = \langle P_CP_Aw,v\rangle = \langle(1-P_A-P_B)P_Aw,w\rangle = \langle(P_A-P_A)w,v\rangle=0.$$ Therefore $v_C$ is orthogonal to $w$, and thus, since $w$ was chosen arbitrary from $(A\cup B)$, $v_C\in(A\cup B)^\perp$.

Thus we can decompose an arbitrary vector $v\in H$ into a sum of a vector $v_A\in A$, a vector $v_B\in B$ and a vector $v_C\in (A\cup B)^\perp$, and therefore $$A+B+(A\cup B)^\perp=H.$$

share|improve this answer

Your Answer


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.