Gram-Schmidt process without normalization Consider a basis $\{v_1,v_2,...,v_n\}$ of a subspace of $\mathbb{R}^n$.
On these basis vectors, apply the Gram-Schmidt process without the normalization step. Let the orthogonal basis obtained be $\{u_1,u_2,...,u_n\}$. Since $v_1=u_1$, we have $||v_1||=||u_1||$. Is there anything that can be said about the norms of other corresponding pairs of vectors (i.e., $||v_i||$ and $||u_i||$, $i\geq 2$)?
 A: Note that $v_i=u_i+w_i$, where $w_i$ is the projection of $v_i$ on $\mathrm{span}(u_1,\ldots, u_{i-1})$. Also $u_i$ is orthogonal to $w_i$, and hence we have $\|v_i\|^2 = \|u_i\|^2 + \|w_i\|^2$, hence we must have $\|u_i\|\le\|v_i\|$.
A: Running the Gram-Schmidt algorithm without the normalization step you have $$u_{k}=v_{k}-\sum_{j=1}^{k-1}\langle v_{k},u_{j}\rangle u_{j}:=v_{k}-\pi_{v_{k}},\quad 2\leqslant k\leqslant n .
$$
that is, $v_{k}=u_{k}+\pi_{v_{k}}$. But,  $\|v_{k}\|^{2}=\|u_{k}+\pi_{v_{k}}\|^{2}$, give
\begin{align*}\|u_{k}+\pi_{v_{k}}\|^{2}&=\langle u_{k}+\pi_{v_{k}},u_{k}+\pi_{v_{k}}\rangle\\&=\langle u_{k},u_{k}\rangle+\langle \pi_{v_{k}},u_{k}\rangle+\langle u_{k},\pi_{v_{k}}\rangle+\langle \pi_{v_{k}},\pi_{v_{k}}\rangle\\&=\|u_{k}||^{2}+2\langle u_{k},\pi_{v_{k}}\rangle+\|\pi_{v_{k}}\|^{2},\quad u_{k},\pi_{v_{k}}\in {\bf R}\\\end{align*}
Thus $\|v_{k}\|^{2}=\|u_{k}\|^{2}+2\langle u_{k},\pi_{v_{k}}\rangle+\|\pi_{v_{k}}\|^{2}$. But $\langle u_{i},u_{j}\rangle=0$ for $i\not=j$ by construction, then $\langle u_{k},\pi_{v_{k}}\rangle=0$.
Thus, the Pythagorean theorem holds $\|v_{k}\|^{2}=\|u_{k}\|^{2}+\|\pi_{v_{k}}\|^{2}$ with a relation between $\|v_{k}\|$ and $\|u_{k}\|$.
