extract the j-th component of the vector resulting from the matrix product.

I cannot code the fact that I have to extract the j-th component from the vector that results from a matrix product in a constrained minimization problem.

Minimize ∑(RC(Fj)/σP−1/M)^2 for j=1:M

sub ∑x=1

where RC(Fj) = (A⊤x)j⋅ (A+Σx/√x⊤Σx)j

where

• A is a loadings matrix
• A+ is Moore-Penrose inverse of A
• Σ is a covariance matrix
• x is a weights vector

I should therefore extract the j-th component of the following matrix products

(A⊤x)j

(A+Σx/√x⊤Σx)j

Answer 1

[Swastik Sarkar]

Hi @Tommaso Delicato,

In this case, it may be more convenient to compute the full vectors for both (A⊤x) & (A+Σx/√x⊤Σx)and then access the required component by indexing

The following example illustrates this method using randomly generated values:

n = 40;
M = 30; 
j = 20;  

A = randn(n, M);
R = randn(n);
Sigma = R' * R;

x = rand(n, 1);
x = x / sum(x);

v1 = A' * x;
v2 = (pinv(A) * Sigma * x) / sqrt(x' * Sigma * x);

component1 = v1(j);
component2 = v2(j);

RC_Fj = component1 * component2;

% Display results
fprintf('Component (A^T x)_%d = %.4f\n', j, component1);

Component (A^T x)_20 = 0.1311

fprintf('Component (A^+ Σ x / sqrt(x^T Σ x))_%d = %.4f\n', j, component2);

Component (A^+ Σ x / sqrt(x^T Σ x))_20 = 0.4803

fprintf('RC(F_%d) = %.4f\n', j, RC_Fj);

RC(F_20) = 0.0630

This method also makes it easier to compute all RC(Fj)RC(Fj​) values in a loop if needed.

I hope this helps !