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Let $V$ a vector space with an inner product. and let $(u_1,u_2,\dots,u_k)$ a sequence of linearly-independent vectors, and $(b_1,b_2,\dots,b_k)$ the orthonormal basis that is generated by the Gram–Schmidt process on $(u_1,u_2,\dots,u_k)$.

How to show that $\langle b_i|u_i\rangle> 0$ for every $1\leq i \leq k$?

I tried to represent $b_i$ in terms of the basis $(u_1,u_2,\dots,u_k)$, and to get the desired result by showing the projection on U of the vector, and from there I couldn't continue..

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Recall that for all $i\geqslant 1$, $b_i$ is positively colinear to: $$u_i-\sum_{k=1}^{i-1}\langle u_i,b_k\rangle b_k.$$ Therefore, $\langle b_i,u_i\rangle$ has the sign of $\|u_i\|^2>0$, since $u_i$ is orthogonal to $b_k$ for $k<i$.

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