Show that the expression does not depend on what base you pick

Question

I got this question on a test today and I had no idea.

Define an inner product on the vector space $\mathbb{P}_n$ consisting of polynomials of grade lower than or equal to $n$, by letting $$ \langle f,g \rangle=\int_{-1}^1f(t)g(t)\,\mathrm{d}t. $$

Given an orthonormal basis $B=(f_1,f_2,\dots,f_n)$ for $\mathbb{P}_n$, consider the expression $$ G(B)=\sum_{i=1}^N|f_i(0)|^2. $$

Show that the value of $G(B)$ does not depend on what orthonormal basis $B$ one chooses.

Answer 1

Let $g_1,\ldots,g_n$ be another orthonormal basis. Thus, $\sum_{j=1}^n\langle f_i,g_j\rangle g_j(x)=f_i(x)$. Notice that the matrix $A=(\langle f_i,g_j\rangle)_{ij}$ is orthogonal.

Let $v^t=(g_1(0),\ldots,g_n(0))$ and $w^t=(f_1(0),\ldots,f_n(0))$. Notice that $Av=w$.

Now $\sum_{i=1}^ng_i(0)^2=v^tv=w^tA^tAw=w^tw=\sum_{i=1}^nf_i(0)^2$.