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Supose $v_1,...,v_n$ is a basis of a inner product space $V$. Prove that there exists a unique vector $u$ of $V$ such that $\langle u,v_i\rangle=c_i$ for all $i=1,...,n$ (each $c_i$ is constant).

I'm not sure how to prove this. I tried proving it by contradiction. Suppose there's $w$ in $V$ such that $\langle w,v_i\rangle=\langle u,v_i\rangle=c_i$ then $\langle w-u,v_i\rangle=0$. But this doesn't get me very far, as it does not imply $w=u$? How do I prove this? Also, how to prove existence in here?

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  • $\begingroup$ I suppose the $c_i$'s are supposed to be fixed? $\endgroup$ – user137731 Oct 31 '16 at 19:03
  • $\begingroup$ Yes, each $c_i$ is a constant. $\endgroup$ – GoetheGrimm Oct 31 '16 at 19:06
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Correction: Let $\{e_k\}$ be an orthonormal basis. Expand the $\{e_k\}$ in terms of $\{v_i\}$, $$ e_k=\sum_ka_{ki}v_i $$ Then if $w=\sum_k b_ke_k=\sum_{ki}b_ka_{ki}v_i$ we can find the $b_k$ by solving: $$ c=A^tb $$ where $c$ is the vector of $c_i$'s, $A$ the matrix of the coefficients of expansion $a_{ki}$ and $b$ the vector of $b_k$'s

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  • $\begingroup$ Yes - you're right. Uniqueness is only for orthonormal basis. $\endgroup$ – am301 Oct 31 '16 at 19:14
  • $\begingroup$ I corrected it now $\endgroup$ – am301 Oct 31 '16 at 19:30

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