How to code this quadratic minimization problem subject to a linear constraint?

After having performed Factor Analysis (FA), I am now trying to implement the following rotation strategy (Swanson 2017 - details of rotation strategy in appenix A) to interpret the three factors structurally and economically. The strategy is in 3 steps but I will report just the first one as the other steps will follow suit:

1) Quadratic minimization problem subject to a linear constraint:

s.t.

where

is the 80x3 matrix of unrotated factors, (3x1) is the first column of λ which is the 7x3 matrix of factor loadings and is the third column of U which is a 3x3 orthogonal matrix. and are known (find data attached: fa_pre = (80x3) and lambda_1 = (1x3)). Obviously, I would like to solve it for to fully have .

Can you help me code the first minimization problem?

[Bianca Piccirillo]

Hi!! I need to do the exact same computation for a project, did you solve it by any chance?

Thank you very much!

Answer 1

[David Goodmanson]

Hi Armando/Bianca

Suppose lambda is a 1x3 row vector. The code below is based on the fact that the solution abc (which is the column vector [a b c]' ) must meet two conditions:

lambda*abc = 0
[0 0 1]*abc = 1   % forces c=1

There is a vector n that is perpendicular to both lambda and [0 0 1] and you can have an aribitrary amount x*n of that without affecting the two conditions. So x can be optimized to minimize the quadratic quantity.

F = Fpre;
lam = lambda_1';              % row vector

z = [0 0 1];
n = null([lam;z])';
G = [n; lam; z];
invG = inv(G);
c = [1 0 0]*invG'*F'*F*invG;
x0 = -c(3)/c(1);
abc0 = invG*[x0 0 1]'          % the result
S0 = abc0'*F'*F*abc0           % the quantity that is minimized
lam*abc0                       % check, should be zero
    
    abc0 =
    0.0140
   -0.6621
    1.0000
    
    S0 = 35.5975
    
    ans = -4.6838e-17

To check for a minimum you can change both a and b such that the two conditions are still satisfied. Note that for da = +-005, S-S0 has the same positive value in both cases, the right result for the bottom of a quatratic well.

da = .005;
db =  -lam(1)*da/lam(2);     % db s.t. lam*abc is still 0
abc = abc0 + [da db 0]';
S = abc'*F'*F*abc;
S-S0                         % should be positive
lam*abc                      % check, should be 0
   ans = 0.0667
   ans = -6.0715e-17

da = -.005;
db =  -lam(1)*da/lam(2);     % db s.t. lam*abc is still 0
abc = abc0 + [da db 0]';
S = abc'*F'*F*abc;
S-S0                         % should be positive
lam*abc                      % check, should be 0
   ans = 0.0667
   ans = -3.2960e-17