# Dot Product and Orthogonal Complement

Let V be the vector space of all real-valued bounded sequences. Then for $a,b \in V$ $\langle a,b \rangle :=\sum _{n=1}^{\infty } \frac{a(n) b(n)}{n^2}$ defines a dot product. Find a subspace $U \subset V$ with $U \neq 0, U \neq V, U^\bot=0$.

I couldn't find anything that works, thank you in advance. $U^\bot$ is the orthogonal complement of $U$, in case the notation is confusing.

-
Try the subspace $U$ of all sequences with only finitely many non-zero terms. – t.b. Jan 18 '11 at 13:47
I see, if you have a sequence that has only finitely many zero-terms you can always find a corresponding sequence with finitely many non-zero terms where some $a(n_0) \cdot b(n_0) \neq 0$ right? – Listing Jan 18 '11 at 13:52
@Theo: you should make that hint an answer. – Willie Wong Jan 18 '11 at 13:56
@user3123: that is not quite what you want. You want to show $U^\perp = 0$, which means that you want to show for any given non-zero sequence $a\in V$, there exists some $b$ with only finitely many non-zero terms such that $\langle a,b\rangle\neq 0$. Then you want to show that there exists some element in $V$ that is not in $U$. – Willie Wong Jan 18 '11 at 14:00
Ok so thats not hard to show? If $a \in V$ is a non-zero sequence there is a $n_0 \in \mathbb{N}$ such that $a(n_0) \neq 0$. Then you take $b(n) := \delta_{n,n_0}$ and then $\langle a,b\rangle\neq 0$, $b$ has only 1 non-zero term. And the sequence $c = \{0,1,1,1,\ldots \}$ is not in $U$. ( $\delta$ is en.wikipedia.org/wiki/Kronecker_delta ) – Listing Jan 18 '11 at 14:05

The sequences with only finitely many non-zero terms form a proper subspace $U$ of $V$. For $k \in \mathbb{N}$ let $\delta_{k} \in U$ be the sequence for which $\delta_{k}(k) = 1$ is the only non-zero entry. Now compute the scalar product (= dot product) $\langle \delta_{k}, b \rangle$ for all $k$ in order to see that $U^{\perp} = 0$. In other words: if $b$ is orthogonal to all $\delta_{k}$'s then $b$ must be zero.