How to use matlab to find the optimized matrix with regarding to minimization of Frobenius norm

I'm trying to solve a minimization problem whose purpose is to optimize a matrix whose square is close to another given matrix. but I have some problems to execute it by using matlab. The problem is illustrated as follows:

First question is how should I define the such an undetermined matrix when using optimization function in matlab;

Second question is if there is a specified toolbox to solve this problem?

[Han Lin]

Have you found the answer?

[li]

Yes. The answer provided by Torsten is very good. However, I found the answer provided by Johan Löfberg is simple (See the answer here https://math.stackexchange.com/a/1672198/171352). The code was used in the paper A Weightedly Uniform Detectability for Sensor Networks

Answer 1

[Torsten]

function main

G=zeros(4,4);
G(1,1)=0.48;
G(1,2)=0.24;
G(1,3)=0.16;
G(1,4)=0.12;
G(2,:)=G(1,:);
G(3,:)=G(1,:);
G(4,:)=G(1,:);

Aeq=zeros(4,12);
Aeq(1,1)=1.0;
Aeq(1,2)=1.0;
Aeq(1,3)=1.0;
Aeq(2,4)=1.0;
Aeq(2,5)=1.0;
Aeq(2,6)=1.0;
Aeq(3,7)=1.0;
Aeq(3,8)=1.0;
Aeq(3,9)=1.0;
Aeq(4,10)=1.0;
Aeq(4,11)=1.0;
Aeq(4,12)=1.0;
beq=ones(4,1);

lb=zeros(1,12);
ub=ones(1,12); 

x0=ones(1,12)/3;

x = fmincon(@(x)obj(x,G),x0,[],[],Aeq,beq,lb,ub);

function val=obj(x,G)
Q(1,1)=x(1);
Q(1,2)=x(2);
Q(1,3)=0.0;
Q(1,4)=x(3);
Q(2,1)=x(4);
Q(2,2)=x(5);
Q(2,3)=x(6);
Q(2,4)=0.0;
Q(3,1)=0.0;
Q(3,2)=x(7);
Q(3,3)=x(8);
Q(3,4)=x(9);
Q(4,1)=x(10);
Q(4,2)=0.0;
Q(4,3)=x(11);
Q(4,4)=x(12);

val=sum(reshape((Q^2-G).^2,16,1));
end

Best wishes

Torsten.

[li]

Thanks a lot. But when I run it with Matlab, it happens with errors, how could I run this code correctly?

[Torsten]

Please post the complete error message.

Best wishes

Torsten.

[li]

I warped up your codes (without end) in a .m file, it outputs the following message,

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the default value of the function tolerance,
and constraints are satisfied to within the default value of the constraint tolerance.

<stopping criteria details>

The result is promising, but when I type val and Q in the command window try to get the true values of val and Q, it pops up the following error messages:

>> val
Undefined function or variable 'val'.

>> Q
Undefined function or variable 'Q'.

How to solve this problem?

Thanks a lot.

[Torsten]

Add the lines

Q=[x(1) x(2) 0 x(3); x(4) x(5) x(6) 0; 0 x(7) x(8) x(9); x(10) 0 x(11) x(12)];
val = sum(reshape((Q^2-G).^2,16,1));

after the call to fmincon.

Best wishes

Torsten.

[li]

Thanks, but I tried with same problems, does it mean I put the new lines of codes in the wrong place?

The .m file I execute is as follows:

function main
G=zeros(4,4);
G(1,1)=0.48;
G(1,2)=0.24;
G(1,3)=0.16;
G(1,4)=0.12;
G(2,:)=G(1,:);
G(3,:)=G(1,:);
G(4,:)=G(1,:);
Aeq=zeros(4,12);
Aeq(1,1)=1.0;
Aeq(1,2)=1.0;
Aeq(1,3)=1.0;
Aeq(2,4)=1.0;
Aeq(2,5)=1.0;
Aeq(2,6)=1.0;
Aeq(3,7)=1.0;
Aeq(3,8)=1.0;
Aeq(3,9)=1.0;
Aeq(4,10)=1.0;
Aeq(4,11)=1.0;
Aeq(4,12)=1.0;
beq=ones(4,1);
lb=zeros(1,12);
ub=ones(1,12); 
x0=ones(1,12)/3;
x = fmincon(@(x)obj(x,G),x0,[],[],Aeq,beq,lb,ub);
Q=[x(1) x(2) 0 x(3); x(4) x(5) x(6) 0; 0 x(7) x(8) x(9); x(10) 0 x(11) x(12)];
val = sum(reshape((Q^2-G).^2,16,1));
function val=obj(x,G)
Q(1,1)=x(1);
Q(1,2)=x(2);
Q(1,3)=0.0;
Q(1,4)=x(3);
Q(2,1)=x(4);
Q(2,2)=x(5);
Q(2,3)=x(6);
Q(2,4)=0.0;
Q(3,1)=0.0;
Q(3,2)=x(7);
Q(3,3)=x(8);
Q(3,4)=x(9);
Q(4,1)=x(10);
Q(4,2)=0.0;
Q(4,3)=x(11);
Q(4,4)=x(12);
val=sum(reshape((Q^2-G).^2,16,1));

Thank you so much for your help!

[Torsten]

Add the lines

Q=[x(1) x(2) 0 x(3); x(4) x(5) x(6) 0; 0 x(7) x(8) x(9); x(10) 0 x(11) x(12)];
val = sum(reshape((Q^2-G).^2,16,1));
disp(Q);
disp(val);

after the call to fmincon.

Best wishes

Torsten.

[li]

It works, thank you so much!

[Flora]

I have a similar issue, but could not solve it with the given code. This because, it is not clear to me why these values for Aeq , beq, lb, ub and x0 are chosen. Where is the criteria of "q11+q12+q14=1" given in the code?

[Torsten]

The unknowns are ordered as

x(1)=q11
x(2)=q12
x(3)=q14
x(4)=q21
x(5)=q22
x(6)=q23
x(7)=q32
x(8)=q33
x(9)=q34
x(10)=q41
x(11)=q43
x(12)=q44

Thus the condition that

q11+q12+q14=1

is given as

Aeq(1,1)=1.0;
Aeq(1,2)=1.0;
Aeq(1,3)=1.0;
beq(1)=1;

Best wishes

Torsten.

[CA Hau]

May I know if there are two constraints for the first rows (eg. q11+q12+q14 = 1 & q11-q12-q14 = 0.01) how should I modify the constraint? Thank you in advance for any enlightenment

[Walter Roberson]

 q11+q12+q14 = 1 & q11-q12-q14 = 0.01

would be

Aeq(1,1) = 1;
Aeq(1,2) = 1;
Aeq(1,3) = 1;
b(1,1) = 1;
Aeq(2,1) = 1;
Aeq(2,2) = -1;
Aeq(2,3) = -1;
b(2,1) = 0.01;

But remember that 0.01 does not have an exact representation in binary floating point, so the closest you can get to that second constraint is q11-q12-q14 = 0.01000000000000000020816681711721685132943093776702880859375