1
$\begingroup$

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..

$\endgroup$
2
$\begingroup$

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 $k<i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.