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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.

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  • $\begingroup$ Not every spanning set is a basis... $\endgroup$
    – DonAntonio
    Sep 29 '16 at 21:57
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Hints:

B: Let x = x_i*v_i and compute ||x|| using the triangle inequality. What does this tell you about x?

C: (x,v_k)=(y,v_k) if and only if (x-y,v_k)=0. Then use B.

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  • $\begingroup$ How would you use x=x_iv_i with the triangle inequality? I guess, since the triangle inequality is ||x+y|| <= to ||x|| + ||y|| where would the x_iv_i fit in? $\endgroup$ Sep 29 '16 at 22:38
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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)?

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  • $\begingroup$ I understand the reasoning for part c and see how it follows from a and b- clever. With the explanation of part b, I am still sort of unsure. Is there another way you could explain this same concept? $\endgroup$ Sep 29 '16 at 22:32
  • $\begingroup$ Can you be a little bit more precise in what makes you unsure about part b)? $\endgroup$ Sep 29 '16 at 22:39
  • $\begingroup$ I see that it stats similarly to the proof of a, but I am not following why we are using the complex conjugate of a exactly. $\endgroup$ Sep 29 '16 at 22:42
  • $\begingroup$ Because the inner product is not exactly linear in the second coordinate: the coefficient multiplying the vector in the second coordinate, comes out as it´s complex conjugate. If you are working on an $\Bbb{R}$-vector space then $\alpha = \alpha^*$ for any $\alpha \in \Bbb{R}$. But if you are working on a $\Bbb{C}$-vector space then $\alpha$ and $\alpha^*$ are not neccesairily the same. $\endgroup$ Sep 29 '16 at 22:48
  • $\begingroup$ Ah ok. So then would you prove this the same way as the first except noting that there is a chance of complex. So basically saying x= av_i and then using that v_i point so then it sort of cancels the same way as the first one? If that makes sense. $\endgroup$ Sep 29 '16 at 23:01

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