$\sum\limits_{j,k} {{a_{jk}}} {\bar u_j}{u_k}\mathop = \limits^? \int_0^1 | \sum\limits_j {{u_j}} {x^j}{|^2}x{\mkern 1mu} dx$

Question

Let $A\in M_n$ and $U\in \mathbb{C}^n$ and $a_{jk}=\int_0^1 x^{j+k-1}\,dx$

Is this true that

$\sum_{j,k} a_{jk}\bar u_ju_k=\int_0^1\Bigl|\sum_j u_jx^j\Bigr|^2x\,dx$

Answer 1

This is not true because $$ \left| \sum_{j=1}^nu_jx^j \right|^2 = \overline{ \left( \sum_{j=1}^nu_jx^j \right)} \left( \sum_{j=1}^nu_jx^j \right) = \left( \sum_{j=1}^n\bar{u}_jx^j \right) \left( \sum_{j=1}^nu_jx^j \right) = \sum_{j=1}^n \sum_{k=1}^n \bar{u}_j u_k x^{j+k} $$ so $$ \int_0^1\left| \sum_{j=1}^nu_jx^j \right|^2x\,\mathrm{d}x = \sum_{j=1}^n \sum_{k=1}^n \bar{u}_j u_k \int_0^1x^{j+k+1}\,\mathrm{d}x. $$