## Least sum squares given constraints on subcomponents

Hi all,

I recently encounter a difficult problem.

I wish to minimize in $\mathbf{x}$ the sum $\min \sum_{i=1..n} (\mathbf{x}^T \mathbf{A}_i \mathbf{x})^2$ given the constraints on the norms of all $\mathbf{x}$'s subcomponents (let's say three 3-by-1 vectors) $|\mathbf{x}_1| = 1, |\mathbf{x}_2| = 1, |\mathbf{x}_3| = 1$. $\mathbf{A}_i$ may not be positive-definite.

Yes, it's quartic expression that we want to minimize. I'm not sure if any one has worked on this or similar problem in the math community. I search the literature for sometimes but no use. My question may be similar but actually much more difficult than this http://mathoverflow.net/questions/47025/least-square-given-constraint-on-subcomponents

The 4th-order and constraints on all subcomponents makes it really hard for me to handle.

Any idea to a numerical/analytical solution, is greatly appreciated. Thanks for reading.

p/s: $\mathbf{x} = [\mathbf{x}_1^T , \mathbf{x}_2^T, \mathbf{x}_3^T]^T$. By "subcomponents", I mean the subvectors, as shown in the equation.

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 May I ask what the motivation is for making the quadratic expression a quartic one? Is it to avoid concave curvature? (at the expense of introducing multiple solutions) – Gilead Feb 19 2011 at 4:16 I want the quadratic expressions as close to 0 as possible, and since A is not positive-definite, I have to minimize the sum of squares. I'm not sure if it's a good idea. – Tony Feb 19 2011 at 5:51 @Tony: I'm having a little trouble understanding your question. Are the "subcomponents" orthogonal projections of $x$ onto a set of three (for example) mutually orthogonal, complementary subspaces whose sum is $R^{k}$, the space in which $x$ lives? That's my guess, but if you could clarify it might help folks to answer your question. – drbobmeister Feb 19 2011 at 8:14 It is a subvector. My bad word usage. I added a p.s. to clarify. Thanks for your suggestion, drbobmeister. – Tony Feb 19 2011 at 8:30 OK, so it looks like I was right. Thanks for clarifying, Tony. – drbobmeister Feb 19 2011 at 8:40

As a practical matter, a lot depends on $n$ and the dimension of $x$. If the problem is small enough then you might not be in deep trouble. If the problem is large (e.g. $x$ might have thousands of components), then it could be very hard.