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I am wondering if the following vector operation is known/studied, or alternatively if it can be simplified somehow into matrix multiplications:


$x$ is a vector and $A$ is the resulting matrix. The definition of $A$ is such that it is symmetric and the diagonal is $x$.

I am interested in this operation since I would later want to find optimal solutions to problems like $\min_x||A-D||^2_F$ (let's say this is the Frobenius norm) for some given data matrix $D$. In other words, I would like to find a set of weights $x$ such that if I perform this operation it will induce a matrix $A$ that is as "close" as possible to some target matrix $D$. BTW if it helps, I can think of $x$ values as probabilities so they are between 0 and 1, which means $A$ values will also be between 0 and 1.

I am looking for insights into this representation or the optimization problem.

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What property makes you think it might have been studied before? Since the expressions are not linear in $x$, matrix multiplications are not likely to give rise to it. –  Marc van Leeuwen Nov 21 '12 at 16:28
@MarcvanLeeuwen I do not know if this has been studied before. I was hoping it might be studied since the matrix can be described succinctly. I am interested in the problem because it is a model of a system I am studying, and I am interested in later on fitting it to data (for example $\min_x||A-D||_2^2$ of some data matrix $D$). –  Bitwise Nov 21 '12 at 16:36
@Bitwise Just curious: how do you define $A_{ij}$ when $i>j$? Is it the empty product (i.e. $1$) or $A_{ji}$ (hence $A$ is symmetric)? Also, is the diagonal of $A$ equal to $x$? BTW, for matrix norms, $\|\cdot\|_2$ denotes the 2-norm, not the Frobenius norm $\|\cdot\|_F$. –  user1551 Nov 23 '12 at 14:34
@user1551 I define $A$ symmetrically. Regarding the diagonal, I am flexible on how it is defined but I guess the natural definition would be to make it equal to x. I will update the question. Thanks for the clarifications. You are right of course about the norm - I actually solve these optimizations in vectorized form, so that is why I am used to writing the norm as L2. –  Bitwise Nov 25 '12 at 22:10
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