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$A, B$ subspaces of $V$, a finite-dimensional inner product space.

SHOW

[note oc = orthogonal complement (which is defined below)]

1) $oc(A+B) = oc(A) \cap oc(B)$

2) $oc(A \cap B) = oc(A) + oc(B)$

user128616
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1 Answers1

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For (1), can you prove that $$x\cdot (c_1a+c_2b) \hspace{3mm}\forall c_1,c_2\in\text{R} $$ implies that $x\in oc(A)$ and $x\in oc(B)$, and that conversely, $x\in oc(A)$ and $x\in oc(B)$ would imply that $$x\cdot (c_1a+c_2b)=c_1(x\cdot a)+c_2(x\cdot b)=0\hspace{3mm}\forall c_1,c_2\in\text{R}$$ and so the two sets involved are both subsets of each other.

(2) Try a similar approach to the one given above. Alternatively, you could try to derive (2) from (1) using what you know. As in, $$oc(A+B)=oc(A)\cap oc(B)$$ so $$oc(oc(A+B))=oc(oc(A)\cap oc(B))$$ or $$A+B=oc(oc(A)\cap oc(B))$$ By setting $A'=oc(A)$ and $B'=oc(B)$, can you prove the statement?

user140943
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  • for part (1) i don't get how x⋅(c1 a+c2 b) implies x belongs to oc(A) and x belongs to oc(B). could you elaborate please? – user128616 Apr 08 '14 at 02:44
  • It holds for all constants $c_1,c_2$, so take (for example) $c_1=1,c_2=0$ and see what you get. Then do the opposite. – user140943 Apr 08 '14 at 02:45
  • ah that makes sense. i'm curious how doing what you did proves statement (1) though. you basically showed that if we take a vector x that belongs to both oc(A) and oc(B) then both A and B are the same? – user128616 Apr 08 '14 at 02:54
  • If a vector is in oc(A) and oc(B) then it must be in oc(A+B). That is what you are trying to show. – user140943 Apr 08 '14 at 02:55