How to solve this vector equation for optical flow

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I am unable to solve for $\textbf{h}$ in the following equation $\sum\limits_{\textbf{x}=1}^n2\partial F(\textbf{x})/\partial\textbf{x}(F(\textbf{x}) + \textbf{h}^{T}\partial F(\textbf{x})/\partial \textbf{x} - G(\textbf{x})) = \textbf{0}$ where $\textbf{h}$ and $\textbf{x}$ are column vectors of same dimension. Vector $\textbf{x}$ takes $n$ instances of values. $G(\textbf{x})$ and $F(\textbf{x})$ are scalar valued fucntions, taking vectors as input and producing saclars as output. The equation belongs to the seminal paper of Lucas-Kanade Optical flow shown here (4th page top left). I tried simplyfying the problem to : solve for $\textbf{x}$ in $\textbf{x}^{T}\textbf{y} = c$ where $\textbf{x}$ and $\textbf{y}$ are column vectors of same dimension and $c$ is a scalar. $c$ and $\textbf{y}$ are known. I guess my simplifaction is not correct.

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After skimming the paper, it appears the the authors are applying the {$F,G$} functions to a finite set of points {$x_k$} in the region of interest. I'll propose a slight change of notation that made thing simpler (to my mind, at least).

Let's denote $$\eqalign{ F_k &= F(x_k) \cr G_k &= G(x_k) \cr g_k &= \frac{\partial F(x_k)}{\partial x} \cr }$$ Where the last definition is the gradient of $F$, which is needed for a Taylor approximation used in the paper.

The Taylor approximation can be recast in the above notation as $$\eqalign{ F(x_k+h) &= F_k + g_k^Th \cr }$$ Now the paper minimizes the function $$\eqalign{ L &= \sum_{k=1}^M \big(g_k^Th + F_k-G_k\big)^2 \cr dL &= \sum_{k=1}^M 2\,\big(g_k^Th + F_k-G_k\big)g_k^Tdh \cr \frac{\partial L}{\partial h} &= \sum_{k=1}^M 2\,\big(g_k^Th + F_k-G_k\big)g_k^T \cr }$$ Setting the derivative to zero, transposing, and solving for $h$ $$\eqalign{ \sum_{k=1}^M\big(g_kg_k^Th\big) &= \sum_{k=1}^M \big(G_k-F_k\big)g_k \cr \Big(\sum_{k=1}^Mg_kg_k^T\Big)\,h &= v \cr Bh &= v \cr h &= B^{+}v \cr }$$ So the solution vector is something like $$\eqalign{ h &= \Big(\sum_{k=1}^Mg_kg_k^T\Big)^{+}\,\Big(\sum_{k=1}^M (G_k-F_k)g_k\Big) }$$