# How to show $u=q$ for the following inner product

$$\langle u, v\rangle=\langle q,v \rangle$$ where $$u,q,v$$ are elements of the inner product space, not fixed. Does this imply $$q=u$$?

Attempt: $$\langle 0,v \rangle = 0$$ therefore $$\langle u,v \rangle = \langle q,v\rangle$$ iff. $$u=q$$ since $$\langle u-q,v \rangle =0$$ implies $$u-q=0$$

why my idea fails to complete the proof:

$$u-q$$ can just be an orthogonal vector to $$v$$

• Welcome to MSE. This question may make sense to you, but it makes no sense at all to me, and I've been learning/teaching linear algebra since the early 1970s. If you'd like folks to answer your question, you should click "edit" below your question and take the time to rewrite it clearly, in complete sentences, identifying one of them as the question you want answered (preferably through the use of a question-mark). Then you should add what you've tried so far, and where you got stuck; that way we can help you more effectively. – John Hughes Apr 14 at 2:29
• fixed took a while sorry – Rick Apr 14 at 2:32
• still doesn't make sense – Ariel Serranoni Apr 14 at 2:34
• please inform me if it still doesn't make sense, I'm trying my best to formulate my thought here – Rick Apr 14 at 2:39
• The statement $\langle u - q, v \rangle = 0$ implies that $u - q = 0$ is false, for the exact reason you stated: this only implies that $u - q$ is perpindicular to $v$. If your space has more than one dimension, or if $v = 0$, there are plenty of nonzero perpindicular vectors to $v$. – user571438 Apr 14 at 2:43

Fix $$u,q$$. If $$\left=\left$$ for all $$v$$ in your inner product space, then you're good. We can prove this, as it implies that $$\left=0$$ by multilinearity, hence $$q-u=0$$. Then we have $$q=u.$$ If $$V$$ is our inner product space, this property can also be used to those that the map $$V\to V^*$$ by $$v\mapsto\left$$ is a monomorphism.