Solving Optimisation Problem with Rank Constraint in MATLAB
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I have a typical least squares problem, i.e I have to find the value of x that minimizes norm of C∗x(:)−d. C is 180x16 matrix(C is rank deficient,i.e rank(C)=7), x is 4x4 matrix & d is 180x1 vector. However, I have a constraint that rank(x)=1. If x was a 16x1 vector and didn't have rank constraint, this problem could be easily solved by using y = pinv(C)*d in MATLAB. But since x is a matrix and has rank constraint, I am not able to proceed further. I would be grateful if someone provides me hint or suggestion to tackle this problem.
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Torsten
2015-11-20
2 个投票
1. Solve y=pinv(C)*d
2. Determine the best rank-1 - approximation x to y as discussed in the previous thread:
My guess is that x solves your original problem, but I'm not 100% certain.
Best wishes
Torsten.
10 个评论
rihab
2015-11-20
Torsten
2015-11-20
To be sure, you could proceed as follows:
Write x as u*v' for two unknown 4x1-vectors u=(u1,u2,u3,u4) and v=(v1,v2,v3,v4).
Then use fminsearch to minimize |C*u*v'-d|_2 in the 8 unknowns u1,u2,u3,u4,v1,v2,v3,v4.
Best wishes
Torsten.
rihab
2015-11-20
Torsten
2015-11-20
function main
A=...;
b=...;
x0=ones(8,1);
x=fminsearch(@)(x)optim(x,A,b),x0);
function fun=optim(x,A,b)
u(1:4,1) = x(1:4);
v(1:4,1) = x(5:8);
rank1 = u*v';
y = reshape(rank1,[16,1]);
residuals = A*y-b;
fun = sum(residuals.^2);
Best wishes
Torsten.
rihab
2015-11-20
Torsten
2015-11-20
Try A and b real-valued first.
For complex A and b, the program has to be adapted.
Best wishes
Torsten.
rihab
2015-11-20
I did not test it, but maybe something like
function main
A=...;
b=...;
x0=ones(16,1);
x=fminsearch(@)(x)optim(x,A,b),x0);
function fun=optim(x,A,b)
realu(1:4,1) = x(1:4);
imagu(1:4,1) = x(5:8);
realv(1:4,1) = x(9:12);
imagv(1:4,1) = x(13:16);
u=complex(realu,imagu);
v=complex(realv,imagv);
rank1 = u*v';
y = reshape(rank1,[16,1]);
residuals = A*y-b;
fun = abs(residuals);
Best wishes
Torsten.
rihab
2015-11-20
Torsten
2015-11-23
In the last line of the code, you will have to replace
fun = abs(residuals)
by
fun = residuals'*residuals
or
fun = norm(residuals)
Best wishes
Torsten.
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