Reduce the following quadratic functions to canonical form and find how the new coordinates are expressed through the old one.
i) $\sum \limits_{i=1}^{n}x_i^2+\sum \limits_{i<j}^{n}x_ix_j$ $\quad $ and $\quad$ ii) $\sum \limits_{i<j}^{n}x_ix_j$
I was able to solve the first one and let me show how I solved it.
Proof: i) I will skip the computations because they are quite trivial. $$\sum \limits_{i=1}^{n}x_i^2+\sum \limits_{i<j}^{n}x_ix_j=y_1^2+\frac{3}{4}y_2^2+\frac{4}{6}y_2^2+\dots+\dfrac{n+1}{2n}y_n^2$$ where $y_k=x_k+\dfrac{x_{k+1}+\dots+x_n}{k+1}$ for $1\leq k< n$ and $y_n=x_n$. The reasoning behind this formula is that you have to complete squares.
ii) But the book says that the second part of the problem can be reduced the first one. I was trying to understand how to reduce it but failed.
I would be very grateful if someone can show it, please!
The author probably intended to choose $y_1=x_1+x_2+\ldots+x_n$. Given this choice of $y_1$, if you can find some linear functions $y_2,y_3,\ldots,y_n$ of $x_1,x_2,\ldots,x_n$ such that $\sum_ix_i^2+\sum_{i<j}x_ix_j=\sum_ic_iy_i^2$, then \begin{aligned} \sum_{i<j}x_ix_j &=\left(\sum_ix_i^2+2\sum_{i<j}x_ix_j\right)-\left(\sum_ix_i^2+\sum_{i<j}x_ix_j\right)\\ &=y_1^2-\left(\sum_ic_iy_i^2\right)\\ &=(1-c_1)y_1^2+c_2y_2^2+\cdots+c_ny_n^2. \end{aligned}