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Next: GRADIENT Up: Claerbout: Eigenvectors for missing Previous: INTRODUCTION


There are unknowns in a vector $\bold m$and knowns in a vector $\bold k$.Pack $\bold m$ and $\bold k$into a column vector $\bold d$ which will be known as ``data space''. There is also a known linear operator $\bold A$,and I have programs that will rapidly evaluate $\bold A \bold d$ and $\bold A' \bold r$ for any vectors $\bold d$ and $\bold r$.Here $\bold A'$ denotes the conjugate of $\bold A$.

\max_{\bold m} \ \lambda({\bold m}) \ \ =\ \ 
 ...\ \ \
{\bold d' \bold A'\bold A \bold d \over \bold d' \bold d}\end{displaymath} (1)

In the physical problem, the components of $\bold m$ and $\bold k$ randomly interleave one another and there are huge numbers of components of each. To gain some insight and see how classical mathematics is relevant, forget for a moment the distinction between known and missing data and imagine that all the components of the data field $\bold d$are independent variables. Then, it is well known (and I will repeat the proof below) that the stationary values of $\lambda$ in (1) are the family of eigenvalues. So if components of $\bold k$ happen to be components of the eigenvector of maximum eigenvalue, then maximization of $\bold \lambda$ will bring $\bold m$to the remaining components of that eigenvector of maximum eigenvalue. My conjecture is that in practice we would find that maximization of $\bold \lambda$would lead to estimated missing data $\bold m$that best fits the various components in the known data $\bold k$.

In many of our applications, $\bold A$ is roughly a unitary matrix because it arises from wave propagation problems. Think of $\lambda$ this way: The denominator limits the total energy; In such cases, maximization should quickly exclude evanescent energy, but a huge number of eigenvalues are all the same, so the maximization is extremely nonunique. To obtain uniqueness with unitary operators we need to introduce a weighting function such as $\lambda(\bold m )=(\bold d' \bold A'\bold W\bold A \bold d)/(\bold d'\bold d)$.

The mathematical problem with the $\bold k$ constraints however, was unknown to Michael Saunders. Since it is a nonlinear problem we should be wary of multiple unwanted solutions, but that need not stop us from trying. My proposed method for the optimization (1) follows from what I know about conjugate gradients: First we easily find a gradient and keep a vector representing the previous step. Second, scaling these by $\alpha$ and $\beta$ respectively, $\lambda(\alpha,\beta )$ is a simple ratio of quadratic forms in two variables. After getting the coefficients of these quadratic forms the problem is of such low dimensionality (two) that almost any method should suffice.

previous up next print clean
Next: GRADIENT Up: Claerbout: Eigenvectors for missing Previous: INTRODUCTION
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