Solve linear equation in matrix form

2023 5 6
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更新时间：2023 5 12
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I need to find pi^(0) from this linear equation
pi^(0) is a row vector, e is column vector of 1
where A_0, R, B, A, and C are square matrix.
Matrix A_0, B, A, and C are given.
R can be found by using successive substitutions method
After i found R, i use this code to find pi^(0) (q is the same as pi^(0))
q=sym('q',[1 mm]) %row vector
I=eye(mm);
e=ones(mm,1); %column vector of 1
a=q*(inv(I-R))*e %size 1 1
aa=a-1
j=solve(aa,sym('q1')) %j is the same as q1=
w=q*(A_0+R*B) %size 1 row mm column
k=subs(w,sym('q1'),j) %subs q1 with j
[C,S]=equationsToMatrix(k,q)
rank(S)
rank(C) %if rank(C)=rank([C,S]) the system can be solved
rank([C,S])
ji=C\S %ji is pi^(0) which i need to find and must fulfill all value less than 1 and non negative.
But, after i run, rank(C)/=rank([C,S]) so the system can't be solved
Is there any code that is wrong? or there is code that more simple than this code?

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回答（1 个）

Torsten
Torsten 2023-5-6
编辑：Torsten 2023-5-6

0 个投票

Why don't you use
pi0 = null((A0+R*B).');
if ~isempty(pi0)
pi0 = pi0.'/(pi0.'*inv(I-R)*e)
end

9 个评论
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DoinK
DoinK 2023-5-6
@Torsten It shows
pi0 =
3×0 empty double matrix
It means no result right, Sir?
Torsten
Torsten 2023-5-6
编辑：Torsten 2023-5-6
Yes, it means (A0+R*B).' has full rank and thus null((A0+R*B).') is empty.
Thus obviously the same result as with your code.
DoinK
DoinK 2023-5-6
Oh i see, your code is the same as my code but shortened.
Thankyou Sir.
So there is one question left, how to find pi^(0).
I'm sure that matrix R is right, so it must be the code that is wrong (?)
Torsten
Torsten 2023-5-6
编辑：Torsten 2023-5-6
pi^0 does not exist for the matrices supplied.
I thought you had understood this because of the empty nullspace of (A0+R*B).' (or in your code: because rank(C)<rank([C,S])).
DoinK
DoinK 2023-5-7
Oh i get it, so the code already good.
The problems are the matrix (A_0, B, A, C, or R).
Thankyou so much Sir.
DoinK
DoinK 2023-5-9
I change my matrix, but still cannot find the pi^(0).
It is continuous time markov chain (sum of every row equal to zero),
Q=[A_0 C 0 0 0 ...; B A C 0 0 ...; 0 B A C 0 ...; ...] and the matrix only on tridiagonal of matrix Q.
(the sum of upper diagonal) / (the sum of lower diagonal) <1.
example the second row 3/12 <1
example the third row 3/(7+5)<1
all have the same value that is 1/4 < 1
Here are my matrix and matrix R by successive substitution.
A_0=[-3 3 0;12 -15 3;7 5 -15]
B=[1 6 5;0 4 4;0 0 7]
A=[-15 3 0;4 -15 3;3 2 -15]
C=[0 0 0;0 0 0;3 0 0]
mm=3;
R=zeros(mm);
X=0;
Z=0;
Err=0;
for i=1:500;
fprintf('i=%g',i);
R=((-C)-((R)^2)*B)*A^-1;
num2str(R,'%7.3f');
Err=R-X;
if isequal(R,X)
break
end
era=abs(Err);
erb=abs(X);
mera=max(era);
merax=max(max(era));
merb=max(erb);
merbx=max(max(erb));
ert=merax/merbx
if ert<=10^(-5)
break %iterasi i= 58
end
X=R;
num2str(X,'%7.3f');
end
From the linear equation pi^(0) * (A_0+R*B)=0,
because pi^(0) is not zero, so (A_0+R*B) must be zero.
Torsten
Torsten 2023-5-9
For that pi^0*M = 0 for a matrix M and a nonzreo vector pi^0, M does not need to be zero. pi^0 has to lie in the nullspace of M.'.
DoinK
DoinK 2023-5-12
Thankyou for your explanation Sir, it really help me.
One more question, from my matrix A_0, B, A, C, and R
If I compute manual
[a b c] * (A_0+RB)=[0 0 0] , I get 3 linear equation name 1, 2, and 3.
[a b c]*(inv(I-R))*[1 1 1].'=1 , I get 1 linear equation name 4.
From my calculation, If I use eq. 1 2 3 and 4, no solution exist.
If I use eq. 1 2 and 3, a=0, b=0, c=0.
If I use eq. 4 and 2 more equation between (1,2,or 3) , there are solutions for a, b, c but the value is not fix. Have a litte different if I use another equation to compare.
Example if I use eq. 4 1 2 compare to eq. 4 2 3, there is a difference in the 6th number after comma.
Torsten
Torsten 2023-5-12
编辑：Torsten 2023-5-12
I don't understand why you experiment above with solutions to 3 equations out of the 4.
For that your problem has a solution, there must exist a vector different from the null vector that satisfies pi^0*(A_0+R*B) = 0 or (A_0+R*B).'*pi^0.' = 0. If such vector(s) exist, you'll get a basis of the vector space spanned by these vectors by the command null((A_0+R*B).'). If the result of this command is empty, your problem has no solution.
Equivalently you can check whether rank((A_0+R*B).') < 3. If this is the case, you'll also be able to solve your problem.

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DoinK
DoinK
2023-5-6

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Torsten
Torsten
2023-5-12

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