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let $x_{1},x_{2},\cdots,x_{n}$ be real numbers,and $0=r_{0}\le r_{1}\le r_{2}\le\cdots\le r_{n}$. show that:$$\sum_{i,j}\min{(r_{i},r_{j})}x_{i}x_{j}=\sum_{i}r_{i}x^2_{i}+2\sum_{i<j}r_{i}x_{i}x_{j}=\sum_{i}(r_{i}-r_{i-1})\left( \sum_{j=i}^{n}x_{j}\right)^2$$

a book say this It's clearly? can you someone explain?

I know $$\sum_{i,j}\min{(r_{i},r_{j})}x_{i}x_{j}=\sum_{i}r_{i}x^2_{i}+2\sum_{i<j}r_{i}x_{i}x_{j}$$ is clearly,But $$\sum_{i}r_{i}x^2_{i}+2\sum_{i<j}r_{i}x_{i}x_{j}=\sum_{i}(r_{i}-r_{i-1})\left( \sum_{j=i}^{n}x_{j}\right)^2$$ maybe not clearly.

Thank you

math110
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1 Answers1

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This is an explicit (and invertible) change of variables for quadratic forms. Here it is with $n=4.$ I used $p=r_1, q = r_2, r = r_3, s = r_4. $ The inequalities on these are irrelevant; all that matters is $r_0=0.$

$$ \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{array} \right) \left( \begin{array}{cccc} p & 0 & 0 & 0 \\ 0 & q-p & 0 & 0 \\ 0 & 0 & r-q & 0 \\ 0 & 0 & 0 & s-r \end{array} \right) \left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array} \right) $$

$$ = \; \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{array} \right) \left( \begin{array}{cccc} p & p & p & p \\ 0 & q-p & q-p & q-p \\ 0 & 0 & r-q & r-q \\ 0 & 0 & 0 & s-r \end{array} \right) $$

$$ = \; \left( \begin{array}{cccc} p & p & p & p \\ p & q & q & q \\ p & q & r & r \\ p & q & r & s \end{array} \right). $$

Note that the inverse of $$ \left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array} \right) $$ really is $$ \left( \begin{array}{cccc} 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \end{array} \right), $$ which is how the minus signs show up when reversing the process.

Will Jagy
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