I have a system of equations that have the following structure. Let $x\in\mathbb{R}^m$ and let $x_k$ be the $k$-th element of $x$. Let $H_k\in\mathbb{R}^{m\times m}$ for $k=1,\ldots, m$. I need to find the $x$ satisfying, \begin{align} x_k = x^\top H_k x \end{align} for $k=1,\ldots,m$. I know that I can solve this system via fixed point iteration or via Newton's method. Are these the best ways? Are there non-iterative methods?
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Do you have any assumption you can do on $H_k$? For example, if it's tridiagonal then you can for sure improve. – PC1 Nov 03 '21 at 14:43
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I would be okay assuming that $H_k$ is invertible but otherwise doesn't have special structure. – 5d41402abc4 Nov 03 '21 at 14:44
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Are you willing to assume that $H_{k}$ is symmetric? positive definite? – Brian Borchers Nov 06 '21 at 21:38
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@BrianBorchers Okay, let's say symmetry but not necessarily positive definite. – 5d41402abc4 Nov 07 '21 at 17:40