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How to reduce the quadratic form $$\sum_{1 \leq k < l \leq n}(k + il)x_kx_l$$ to the canonical form on the $n$-dimensional complex space $\mathbb{C}^n$?

I started with its symmetric matrix \begin{align*} S = \frac{1}{2}\begin{pmatrix} 0 & 1 + 2i & 1 + 3i & \cdots & 1 + ni \\ 1 + 2i & 0 & 2 + 3i & \cdots & 2 + ni \\ 1 + 3i & 2 + 3i & 0 & \cdots & 3 + n i \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 + ni & 2 + ni & 3 + ni & \cdots & 0 \end{pmatrix}, \end{align*} and then tried to reduce it to the diagonal matrix following the algorithmic recipe, i.e., tried reducing it by eliminating the off diagonal elements row/column after row/column: the first step is to clear the first two rows/columns by calculating $P^TSP$, where \begin{align*} P = \begin{pmatrix} 1 & -1 & 0 & \cdots & 0 \\ 1 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & -\frac{1 + 3i}{1 + 2i} & \cdots & -\frac{1 + ni}{1 + 2i} \\ 0 & 1 & \frac{1}{2(1 + 2i)} & \cdots & \frac{1}{2(1 + 2i)} \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix}. \end{align*} The resultant matrix $P^TSP := (s_{ij})$ has already quite complicated entries: \begin{align*} & s_{kk} = \frac{k^2 - 2 - 3ki}{1 + 2i}, \; k = 3, 4, \ldots, n; \\ & s_{kl} = \frac{k - 2l + 2kl - 4 - li}{2(1 + 2i)}, \; l > k, \end{align*} making the subsequent calculation notoriously nasty, so I am still inconclusive.

Is my attempt on the right track to solve this problem? I think there probably should be a better and cleaner way.

Zhanxiong
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  • can you let us know where this problem comes from? The source will give some indication as to whether a 'nice' solution may exist... – user8675309 Aug 24 '20 at 19:08
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    @user8675309 It is an exercise of the section "Symmetric bilinear form and quadratic form" from a linear algebra textbook written in Chinese, which is used as the linear algebra text for the mathematics department of a reputable Chinese university. I think the book is rarely known internationally, but is widely recognized as ''the most difficult" linear algebra book in China. That being said, it is still entirely possible a "nice" solution may not exist. – Zhanxiong Aug 24 '20 at 20:11
  • one idea: if you let $R$ be an upper triangular matrix with 1s on the diagonal and -1's on the super diagonal, e.g. $R= \left[\begin{matrix}1 & -1 & 0 & 0\0 & 1 & -1 & 0\0 & 0 & 1 & -1\0 & 0 & 0 & 1\end{matrix}\right]$ then $\frac{1}{2}R^TSR$ looks a bit nicer – user8675309 Aug 24 '20 at 22:19
  • @user8675309 Thank you for your suggestion. Maybe I have missed something, while I tested your idea for the $n = 4$ case, it seems didn't show me a clearer road map of how to proceed. – Zhanxiong Aug 25 '20 at 19:14
  • @user8675309 Very interesting. I will think about the next step! Thanks for your inputs! – Zhanxiong Aug 25 '20 at 19:25

2 Answers2

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If $n = 1$, the quadratic form is just $0$. This is a trivial case, but it should not be ignored.

If $n > 1$, it can be shown that the determinant of $S$ is nonzero, so the canonical form is $I_n$.

Juliamisto
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-1

This might be helpful

If we define $$ \Phi_1:=\sum_{1\leq k<l\leq n}(k+il)x_kx_l $$ and $$ \epsilon_{k,l}=\left\{\begin{array}{cc} 1\textrm{ , if }k<l\\ i\textrm{ , if }k>l\\ 0\textrm{ , else } \end{array}\right\}\textrm{, }i=\sqrt{-1} $$ then $$ \Phi_1=\sum^{n}_{k,l=1}\epsilon_{kl}kx_kx_l=\frac{1+i}{1-i}\sum_{k<l}(k+l)x_kx_l-\frac{2i}{1-i}\sum^{n}_{k<l}kx_kx_l $$

Nikos Bagis
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