solve linear equation system with partially unknown coefficient matrix

Hello everybody,

I have a question concerning solving linear equation systems with matlab. Concidering I have an equation system: A*x=y.

x and y are column vectors, where all entrys are known. The data for x und y are measured values.

A is a square matrix, which is diagonal symmetric: . Also the matrix is linearly independent. I know, that some coefficients have to be 0.

Is there any way to solve this equation system in Matlab to get the missing coefficients of A?

In this example there are 8 unknown coefficients, but only 4 rows.

I have to say, that for x and y there are different measurements available, so I could expand these vectors to matrices:

I thought about a solution approach like least absolute deviation, but I don't know, how to consider in matlab the conditions:

• zero elements
• symmetric matrix

Is there any way to solve this problem? At the moment I'm not sure, if this is possible at all?

Thank you for your help!

[John D'Errico]

Knowing that some coefficients must be zero is not what linearly independent means or implies.

Regardless, the solution posed by Bruno, which uses kron to essentially unroll the elements of A is the approach I would use. I believe I could also have solved the problem using symbolic algebra, to create a linear system in those unknowns. The result would be converted to a system of linear equations, solved using linear algebra. But used properly, kron is a better solution, because it never requires you to go into the symbolic domain - that would be much slower.

[J_Automotive]

Hello John,

the information that it has to be linearly independent can be shown by using the real system, it's not an information, which you can get out of my example.

The solution from Bruno seems to work quick and fine, so I also prefer this one. For me it's also important that the solution approach works with the main functions of matlab, because I don't have access to many Toolboxes.

Answer 1

[Bruno Luong]

Just using linear algebra, no extra tollbox is needed, of course n==1 is underdetermined problem

% Generate random matrix 
n = 10;
L = rand(4);
A = L+L.';
A(3,1)=0;
A(4,2)=0;
A(2,4)=0;
A(1,3)=0;
A
% And X/Y data compatible with equation
X = rand(4,n)
Y = A*X;
clear A

% Engine

% Enforce symmetric
ij = nchoosek(1:4,2);
i = ij(:,1);
j = ij(:,2);
ku = sub2ind([4 4],i,j);
kl = sub2ind([4 4],j,i);
p = size(ij,1);
R = (1:p)';
M = kron(X.',eye(4));
sz = [p size(M,2)];
C = accumarray([R ku(:)],1,sz) + accumarray([R kl(:)],-1,sz);

% Enforce A(3,1) & ... A(4,2) == 0
K = sub2ind([4 4],[3, 4], ...
                  [1, 2]);
sz = [2 size(M,2)];
C0 = accumarray([(1:2)', K(:)],1, sz);
C = [C; C0];
p = size(C,1);

% Solve least squares with linear constraints
z = [M'*M, C';
     C, zeros(p)] \ [M'*Y(:); zeros(p,1)];
A = reshape(z(1:16),4,4)

[J_Automotive]

Hello Bruno,

can you explain what M is?

Thank you for your answer.

Best regards

J

[Bruno Luong]

Ops sorry, I fix the code and initialize now M

[J_Automotive]

Hello Bruno,

thank you quite a lot, this is exactly what I was looking for.

I will need a little bit time to understand the solving approach, but that's ok :)

Best regards

J

[J_Automotive]

Hello Bruno, i checked your solution, but i have difficulties to understand it completely.

Would it be possible to consider one more condition, which would be every element has to negative, so only have to be positive values?

Thank you

[Bruno Luong]

Least_squares with extra constraints

A(i,j) <= 0 for i~=j

A(i,i)>=0

% Generate random matrix 
n = 10;
L = rand(4);
A = L+L.';
A = A.*(2*eye(4)-1);
A(3,1)=0;
A(4,2)=0;
A(2,4)=0;
A(1,3)=0;
A
% And X/Y data compatible with equation
X = rand(4,n)
Y = A*X;
% Y = Y+0.1*randn(size(Y));
clear A

% Engine
M = kron(X.',eye(4));
[I,J] = ndgrid(1:4);
I = I(:);
J = J(:);
b = I<=J & ~ismember([I,J], [1 3; ...
                             2 4], 'rows');
I = I(b);
J = J(b);
ku = sub2ind([4 4],I,J);
kl = sub2ind([4 4],J,I);
p = size(I,1);
R = (1:p)';
sz = [size(M,2) p];
isd = I==J;
P = accumarray([ku, R; ...
                kl, R; ...
                ku, R], [isd-1; isd-1; isd], sz);

z = lsqnonneg(M*P, Y(:));
A = reshape(P*z,4,4)

(Note: This is the last time I answered to a modified question.)

Answer 2

[Ameer Hamza]

One approach is to find a least square solution using fmincon()

x = rand(4, 1);
y = rand(4, 1);

A = @(a) [a(1) a(2) 0 a(3);
          a(2) a(4) a(5) 0;
          0 a(5) a(6) a(7);
          a(3) 0 a(7) a(8)];
objFun = @(a) sum((A(a)*x-y).^2, 'all');
sol = fmincon(objFun, rand(8,1))

[J_Automotive]

Thank you for your fast reply.

I have to check if I have an Optimization Toolbox License available.

After checking the result I will answer, whether this was useful.

Best regards

J

[J_Automotive]

Hello Ameer,

sorry, I don't have a license for the Optimiztion Toolbox, so I wasn't able to check your solution.