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I have the following formula to implement $$\Sigma_{ij} = \rho_{ij}\sigma_i\sigma_j$$ where no implicit summation is implied. I have both the $\sigma$ vector and the $\rho$ matrix and I want to calculate the $\Sigma$ matrix.

To be pedantic what i need is a way to find the elements of the matrix $\Sigma$ such that, take for example the element $(1,2)$, this is given by the normal multiplication $$\Sigma_{12} = \rho_{12}\sigma_1\sigma_2$$

Specifically the vector is a $1\times 10$ vector and the matrix is $10\times 10$. At the end I want another $10\times10$ matrix defined as before.

How can I do this?

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  • $\begingroup$ Try matrix KroneckerProduct[vector, vector] or vector matrix.DiagonalMatrix[SparseArray[vector]]. $\endgroup$
    – Henrik Schumacher
    Apr 12, 2021 at 15:43
  • $\begingroup$ @HenrikSchumacher both of them seem to not work. In particular the KroneckerProduct of the two vectors yields a $1\times 100$ column vector which cannot be multiplied by the matrix. In the second case I'm not sure what's going on. $\endgroup$
    – Quiver
    Apr 12, 2021 at 15:53
  • $\begingroup$ @HenrikSchumacher Your answer sparked something in my head: I converted the vector to a diagonal matrix with diagonal elements the elements of the vector and then multiplied row-by-columns diagMatrix.Matrix.diagMatrix and I seem to get a sensible answer! $\endgroup$
    – Quiver
    Apr 12, 2021 at 16:02
  • $\begingroup$ "both of them seem to not work." What are you talking about? They work (see below). KroneckerProduct[vector, vector] is a $10 \times 10$ matrix. $\endgroup$
    – Henrik Schumacher
    Apr 12, 2021 at 16:09
  • $\begingroup$ Ah, you are taking about a $1 \times 10$ vector. Then just use Flatten : matrix KroneckerProduct[Flatten[vector], Flatten[vector]]. $\endgroup$
    – Henrik Schumacher
    Apr 12, 2021 at 16:14

1 Answer 1

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Just too long for a comment.

n = 1000;
matrix = RandomReal[{-1, 1}, {n, n}];
vector = RandomReal[{-1, 1}, {n}];

A0 = Table[matrix[[i, j]] vector[[i]] vector[[j]], {i, 1, n}, {j, 1, n}]; //
   AbsoluteTiming // First
A1 = matrix KroneckerProduct[vector, vector]; // AbsoluteTiming // First
A2 = vector matrix.DiagonalMatrix[SparseArray[vector]]; // 
  AbsoluteTiming // First

1.32595

0.006415

0.006203

This shows that one should avoid Table if possible.

Max[Abs[A0 - A1]]
Max[Abs[A0 - A2]]

1.11022*10^-16

1.11022*10^-16

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  • $\begingroup$ Yes, the first one worked pretty well! It gave the same result as per the solution I gave in the comments. Thank you very much! $\endgroup$
    – Quiver
    Apr 12, 2021 at 16:25

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