matlab: matrix symbolic solution

Question

Thanks much for your time and for all your help. Actually, I made a mistake in the previous post when specifying the problem. Thus, I reformulate my question using a simpler example. I need to solve symbolically the equation Ct = Z/(P-I) or Ct*(P-I) = Z.

I already know the answer => Ct = [sigma, 1-sigma]

How to program "correctly" the code in order to get the solution

syms sigma;
Ct = sym('Ct',[1 2]);
%    
P = [sigma  1-sigma;
     sigma  1-sigma];
I = [1 0;
     0 1];
Z = [0 0];
%    
solve(Ct*(P-I) == Z);

So far, I get :

Z =

0 0

Warning: The solutions are parametrized by the symbols: z = C_

In solve at 190 In test_matrix_sigma at 13

Or with

  solve(Ct == Z/(P-I), Ct);

I get:

Warning: System is rank deficient. Solution is not unique. Warning: 4 equations in 2 variables.

In /opt/MATLAB/R2013a/toolbox/symbolic/symbolic/symengine.p>symengine at 56 In mupadengine.mupadengine>mupadengine.evalin at 97 In mupadengine.mupadengine>mupadengine.feval at 150 In solve at 170 In test_matrix_sigma at 13

 ---------------------------------------------------------------------------------------

Thanks for the answer !

Now I have two issues: 1) When I try to handla a more complicated system:

syms a b P1 P2;
I = [1 0 0 0;
     0 1 0 0;
     0 0 1 0;
     0 0 0 1]; 
%    
P = [a*P1     (1-a)*P1     (1-b)*(1-P1)  b*(1-P1);    
     a*P1     (1-a)*P1     (1-b)*(1-P1)  b*(1-P1);    
     b*(1-P2) (1-b)*(1-P2) (1-b)*P2      b*(1-P2);    
     b*(1-P2) (1-b)*(1-P2) (1-b)*P2      b*(1-P2)];    
%    
assume(a, 'real'); 
assume(b, 'real'); 
assume(P1, 'real'); 
assume(P2, 'real');
%
answer = null((P-I)');
disp(answer);

I get

ans =

[ empty sym ]

as the only answer.

2) If there is a way in maltlab to "solve" the above symbolic matrix P and find the symbolic determinant ?

For instance, if I do eid(P) it works;

when I do det(P) it gives 0 as answer...

Etienne Pellegrini Etienne Pellegrini · Accepted Answer · 2014-08-18T22:14:29

This post is an answer to a different problem, that was first asked by the OP before being edited. I leave the problem and solution here in case someone ever runs into the same problem:

I need to solve symbolically the following matrix equation to find out Ct (a vector ???):

syms a b P1 P2    
%    
P = [a*P1     (1-a)*P1     (1-b)*(1-P1)  b*(1-P1);    
     a*P1     (1-a)*P1     (1-b)*(1-P1)  b*(1-P1);    
     b*(1-P2) (1-b)*(1-P2) (1-b)*P2      b*(1-P2);    
     b*(1-P2) (1-b)*(1-P2) (1-b)*P2      b*(1-P2)];    
%    
solve(Ct*(P-1) == 0, Ct);

How to proceed ?

So far I get:

Undefined function or variable 'Ct'.

Error in matrix_test (line 10) solve(Ct*(P-1) == 0, Ct);

The error you get is because you did not assign Ct before trying to solve for your equation. In the equation Ct*(P-1) == 0, Matlab does not know what Ct is. You could remedy this by creating a symbolic vector (see sym documentation). For instance:

Ct = sym('Ct', [1 4]);

However, using solve on this would not give you the solutions you're looking for: instead, Matlab is going to give you the trivial answer Ct = 0, which of course is a correct answer to your equation.

What you really want to find is the null space of the (P-1)' matrix: the null space is the set of vectors X such that (P-1)'X = 0 (Which is the same thing as X'(P-1) = 0, so Ct = X'). The Matlab function null (see doc) is what you need. Using your code, I get:

null((P-1)')

ans =

[ -1,  0]
[  1,  0]
[  0, -1]
[  0,  1]

This means that any linear combination of the vectors [-1, 1, 0, 0] and [0, 0, -1, 1] belong to the null space of (P-1)', and therefore its transpose is the Ct you were looking for.

N.B.: This result is easily confirmed by observation of your initial matrix P.

matlab: matrix symbolic solution

2 Answers