## Homogeneous system of polynomial equations

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Hi all,

Previously I asked a question that currently has no satisfactory answer http://mathoverflow.net/questions/55939/least-sum-squares-given-constraints-on-subcomponents

It comes from an engineering problem. I was thinking to formulate it differently and hope that someone becomes interested and/or know how to solve it.

By formulating differently, I will have the following system:

$\mathbf{D} \mathbf{R} \mathbf{\theta} = \mathbf{0}_{2N \times 1}$

D is a $2N \times 6N$ block diagonal matrix that contains unknowns $\mathbf{D} = diag(\mathbf{x}^T - a_1 \mathbf{z}^T, \mathbf{y}^T - b_1 \mathbf{z}^T, \dots, \mathbf{x}^T - a_N \mathbf{z}^T, \mathbf{x}^T - b_N \mathbf{z}^T)$ where $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are 3x1 orthogonal unit vectors that we need to find (ie., $\mathbf{x}^T \mathbf{x} = 1, \mathbf{x}^T \mathbf{y} = 0, \dots$). $a_i, b_i$ are known parameters.

R is a $6N \times M$ matrix that contains only numerical entries (measured and computed). $\mathbf{\theta}$ is a vector of M other unknown parameters.

By doing so, I isolated unknowns in two separate matrices (vectors). However, it is still not trivial.

I tried to solve this, but there is no obvious way. One way is that I tried to find $\mathbf{x}, \mathbf{y}, \mathbf{z}$ such that $\mathbf{D} \mathbf{R}$ is rank-deficient. I'm not sure if this can be solved, either exactly or in least-squared, in closed-form or numerically. Any idea, discussion is appreciated.

edit: I was not clear. If I set determinant of any MxM sub-matrix of $\mathbf{D} \mathbf{R}$ to 0 and express it as a function of elements in $\mathbf{x}, \mathbf{y}, \mathbf{z}$, then I have a number of homogeneous polynomials. That's why the title comes.

edit2: M is much smaller than 2N.

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## 1 Answer

One suggests that we can have a good prior for $\theta$ (but not for $\mathbf{D}$). Then, he proposed that we fix $\theta$, find $\mathbf{R}$ for least square $|| \mathbf{D} \mathbf{R} \theta||$ given the constraints above; then fix $\mathbf{R}$ and find $\theta$ for least square $|| \mathbf{D} \mathbf{R} \theta||$ given $||\theta|| = 1$. Then keep repeating (fix $\theta$ then $\mathbf{R}$) until they converges.

The latter least square is easy, and the former, I believe there is a closed form or numerical solution.

But my question is if such numerical scheme works? Is there any proof for it? What is the name of this method? I could not recall this numerical scheme.

Please enlighten.

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I don't understand why you can vary $R$. Doesn't that represent some measurement which is fixed? From your question I inferred that $D$ and $\theta$ contain all the unknowns you want to solve for. Assuming you meant $D$ rather than $R$, then the iterative procedure you described is simply partial optimization; for instance to minimize $f(x,y)$, we first fix $x$ and minimize $y$ and then minimize $x$ for the fixed $y$. This certainly doesn't converge to the global minimum in general, but will if the graph of $f$ is strictly convex. – John Jiang Mar 14 2012 at 0:55
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