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Let $V$ be an inner product space. Show that if $\langle u,v\rangle=0, \forall v\in V$, then $u=0$.

I have some doubts about this statement; for example, if $v$ were to be the zero vector, couldn't $u$ simply be any vector in $V$? It doesn't seem strong enough to be true.

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    It has to be true for all $v\in V$, so you can't make a counter argument with one specific $v$. – 2017-01-17
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    Oh, I realize my mistake now; I have to assume the antecedent always holds. Thank you! – 2017-01-17
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    Let $u = v$ and invoke the inner product axioms. – 2017-01-17

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HINT:

Consider using one of the properties of the dot product:

  • $\langle x, x \rangle > 0$ for $x \not = 0$
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It is said "$\forall v$", so not only for $v = 0$.

Let $u \in V$. By the statement, for all $v \in V$ (including $u$ itself) $\langle u, v \rangle = 0$. Appears that $\langle u, u \rangle = 0$, which means $u = 0$.

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    The assumption that $u\neq0$ was superfluous; you obtain $\langle u, u \rangle = 0$ without it. – 2017-01-17
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    I see, thanks, corrected – 2017-01-17