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Let $v_1, v_2,\ldots, v_n$ 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 $(x, v_k) = 0$ for every $k$, then $x = 0$.

c) If $(x, v_k) = (y, v_k)$ for every $k$, then $x = y$.

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a) One appropriate choice of $v$ suffices. Try $v=x$.

b) Write $x=\lambda_1v_1+\ldots+\lambda_nv_n$ and use the bilinearity of the inner product when computing $(x,x)=(x,\lambda_1v_1+\ldots+\lambda_nv_n)$.

c) Apply b) to $x-y$ instead fo $x$.

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  • $\begingroup$ could you possibly be a little more explicit in the explanation? I'm still rather confused on how to solve this problem. $\endgroup$
    – user72195
    Apr 12 '13 at 1:42
  • $\begingroup$ @user72195 How can I be more explicit in (a), to get started? $\endgroup$
    – Julien
    Apr 12 '13 at 1:42
  • $\begingroup$ What does setting v=x actually do? $\endgroup$
    – user72195
    Apr 12 '13 at 1:45
  • $\begingroup$ @user72195 It does $(x,x)=0$. Now $(x,x)=\|x\|^2$. So? $\endgroup$
    – Julien
    Apr 12 '13 at 1:46
  • $\begingroup$ then x must be 0. Does anything need to be considered about v though? Such as, what is the difference between creating (x,x) vs. (v,v)? $\endgroup$
    – user72195
    Apr 12 '13 at 1:49

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