Say $V$ is a finite dimensional inner product space and $B = \{v_1,\dots, v_n\}$ a basis for $V.$ Suppose that we apply the Gram-Schmidt process to this basis in order to obtain the orthogonal basis $\{w_1,\dots, w_n\}.$ Show that $$\|v_1\|^2 +\dots+ \|v_n\|^2 \geq \|w_1\|^2 + . . . + \|w_n\|^2$$
Hello guys. I am trying to prove this. Now i am trying to do induction. Showed for $n$ is $1$ and supposed up to $n-1.$ Also i writed $w_1$ up to $w_n$ applying Gram Schmidt. But at the final step which is showing $\|v_n\|^2\geq\|w_n\|^2,$ i could not find a way to show it. Can you help?
Let $U=\operatorname{span}(v_1,\ldots,v_{n-1})=\operatorname{span}(w_1,\ldots,w_{n-1})$. Then $w_n=v_n+u$ with $u\in U$ and $w_n\perp U$. In particular, $w_n\perp u$. Then $$ \|v_n\|^2=v_n\cdot v_n=(w_n-u)\cdot (w_n- u)=w_n\cdot w_n-2(w_n\cdot u)+u\cdot u=\|w_n\|^2+\|u\|^2\ge \|w_n\|^2.$$