One of my homework questions is:
Let $v_1, v_2,\dots ,v_n$ be a spanning set (in particular, a basis) in an inner product space $V$ . Prove that
a) If$(x,v)=0$ for all $v\in V$, then $x=0$;
b) If$(x,v_k)=0$ $\forall k$, then $x=0$;
c) If $(x,v_k) = (y,v_k)$ $\forall k$, then $x=y$.
I initially thought for part a) to just say do a proof where $v$ is set to $x$, which would ultimately get $(x,x)=0$, so $x$ would also have to be equal to $0$. But upon thinking about it, would that proof work since it is not specified that $x$ is in $V$, or can you still go about it that way? and if not, what step should I take first?
I am very lost on how to start part b, but was thinking if I am set on part a then b should follow more naturally. Does anyone have a recommendation for a starting point?
And for c, what I thought to do was just go by the definitions that $(x,vk)$ is the same as $||x||||v_k||$, and so we then have $||x||||v_k||=||y||||v_k||$. Can I then cancel out the $||v_k||$ to get $||x||=||y||$? Even this would not be super clean because it ends with $\pm x= \pm y$. Any suggestions how to clear this up/ get on the right track?
Sorry this is so long and detailed. Just trying to understand as thoroughly as possible.
For the a)
The inner product is defined as a function $\varphi: V \times V \to k$ where $k$ is the field that makes $V$ a vector space. (In your case, probably $k=\Bbb{R}$ or $k=\Bbb{C}$ ). Therefore $(x,v)$ with $x \notin V$ doesn´t make any sense in this case. So $x \in V$ and your reasoning is perfectly well done.
For b)
Hint: Since $v_1,...,v_n$ is a spanning set, in particular $x=\sum \alpha_i v_i$, therefore $(x,x)=(x,\sum \alpha_i v_i)=\sum \alpha_i^*(x,v_i)$
(here, $\alpha_i^*$ denotes the complex conjugate of $\alpha_i$). Try finishing the proof.
c)
$(x,v_k)=(y,v_k)$ for all $k$ if and only if $(x,v_k)-(y,v_k)=0$ for all $k$ if and only if $(x-y,v_k)=0$ for all $k$ (using linearity of the inner product.) Can you finish the proof using a) and b)?