I found a true/false exam. Here is the exercise (other question are here:
2) Let $q(x)=x^7+1$. Set $W=\{p(x)\in \mathbb R_n[X]\mid p(0)=0\}$. Let $r(x)\in W$. Suppose $$\int_0^1 (q(x)-f(x)p(x)dx=0,$$ for all $p(x)\in W$. There is $s(x)\in W\backslash \{r(x)\}$ such that $$\int_0^1 (q(x)-s(x))^2dx< \int_0^1 (q(x)-r(x))^2dx.$$
2) Let $\mathbb C^5$ the space with the weighted euclidian inner product with weight $(2,1,3,1,4)$. Let $S\in \mathcal L(\mathbb C^5)$. Suppose $\det S=2-i$. There is $v=(v_1,...,v_5)\in \mathbb C^5$ such that $$2|v_1|^2+|v_2|+3|v_3|^2+|v_4|^2+4|v_5|^2\neq 2|w_1|^2+|w_2|+3|w_3|^2+|w_4|^2+4|w_5|^2,$$ where $S(v)=(w_1,...,w_5)$.
My Answers:
1) I would say TRUE. But I have problem to prove it rigorously. I thing $\int_0^1 pq$ is a scalar product. Therefore, there is a unique $r$ s.t. $<q-r,p>=0$ for all $p\in W$. Now I tried to prove that $\|r-s\|^2=0$ to get the contradiction $r=s$, but I have problem to do it.
2) Since $\det(S)=2-i\neq 0$, then $S$ invertible. We have that $$\|S(v)\|^2=\|w\|^2=2|w_1|^2+|w_2|+3|w_3|^2+|w_4|^2+4|w_5|^2\neq 2|v_1|^2+|v_2|+3|v_3|^2+|v_4|^2+4|v_5|^2=\|v\|^2,$$ and thus, $S$ is not an isometry. But I don't get any contradiction anywhere. So I need help here.
Any help would be appreciated.