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Let $\lbrace u_1,\dots,u_k\rbrace\in V$ be some vectors in an inner product space, $\dim V=n\geq k$. if $\forall v\in V, \|v\|^2=\sum_{i=1}^k|\langle v,u_i\rangle |^2$, prove that they span $V$.

I tried various things: CS inequality, writing the inner products as double-sums (a mess) of coordinates of $v$, proving by negation (if there are $\ell>k$ vectors needed to span $V$). Couldn't sort it out. What am I missing?

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Suppose that $U=\{u_1,...,u_k\}$ does not span $V$ and let $u$ be a non zero element of $V$ orthogonal to $U$, $\|u\|^2\neq 0$, but $\sum_{i=1}^{i=k} |\langle u,u_i\rangle|=0$.

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  • $\begingroup$ Thanks! I should have just wrote $v\in U\oplus U^{\perp}$ and then work with either option. $\endgroup$ – gbi1977 Feb 14 '18 at 22:43

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