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I have a matrix equation of the form: $Sx=x^TKxp$, while $ S = \begin{pmatrix} 0 & a & a \\ b & 0 & b \\ c & c & 0 \\ \end{pmatrix} $, $a$ and $b$ are real, known numbers, $K = \begin{pmatrix} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 \end{pmatrix} $, $p = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and $x = \begin{pmatrix} r \\ s \\ t \end{pmatrix}$ with $r$, $s$ and $t$ as the unknows. Is it possible to solve this equation system analytically or only numerically? How can a approach be?

Doran Strehnisch
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  • You know $x$ has the form $\lambda S^{-1}p$ for some scalar $\lambda$. Substitute this into $S x = (x^T K x)p$, you get a equation in $\lambda$.... – achille hui Sep 08 '19 at 09:45
  • Is $c$ also given? If $c=0$, then $x^TKx=0$ necessarily and $S$ doesn't have an inverse. – amsmath Sep 08 '19 at 09:58
  • Thanks a lot to achille hui and amsmath: The substitution with $\lambda$ was a great idea and I was able to solve the equation. Also sorry I forgot to mention, that $a$, $b$ and also $c$ are real, known numbers (and (a,b,c) > 0). – Doran Strehnisch Sep 08 '19 at 12:49

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Let $u=x^TKx=rs+st+tr$ (1).

$Sx=up$ is then a straightforward equation which gives $r,s,t$ in terms of $u$.

Substitute into (1) and you have a quadratic equation for $u$.

  • I think also because the r.h.s has just this one entry one can get a necessary condition for solvability in terms of what (a,b,c) is, if I didn't do a mistake at the calculation actually one need $\begin{pmatrix} -c & a-c & a \ -b & a& a-b \ b -c & -c & b\end{pmatrix} x = 0$. – hal4math Sep 08 '19 at 10:55