solve a symbolic matrix equation

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marcel hendrix 2018-9-1

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https://ww2.mathworks.cn/matlabcentral/answers/417113-solve-a-symbolic-matrix-equation

评论： marcel hendrix 2022-10-22

I wonder if it is at all possible to solve the following equation with the Symbolic Toolbox. Matlab was working for more than 5 hours on this when I terminated the process.

The problem could be that the matrices Aa and Ya are not square. Actually I don't need a symbolic answer, numeric would be fine. The main reason to try the Symbolic Toolbox is that I don't see how to approach 'Ya*D' = Aa*Ya', as Ya is being both post- and pre-multiplied. I guess fine points of vector-matrix algebra are needed here.

-marcel

%%Differentiation script 3
syms e Aa Ya D
syms e1  e2 e3 e4 
syms a11 a12 a13  a21 a22 a23  a31 a32 a33
syms y11 y12 y13 y14  y21 y22 y23 y24  y31 y32 y33 y34 
syms d11 d12 d13 d14  d21 d22 d23 d24  d31 d32 d33 d34  d41 d42 d43 d44
e = [e1 e2 e2 e4];
Aa = [a11 a12 a13
      a21 a22 a23
      a31 a32 a33];
Ya = [y11 y12 y13 y14
      y21 y22 y23 y24
      y31 y32 y33 y34];
D = [d11 d12 d13 d14
     d21 d22 d23 d24
     d31 d32 d33 d34
     d41 d42 d43 d44];
solve(Ya*D'- Aa*Ya==0, Ya)
%EOF

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David Goodmanson 2022-5-7

在 MATLAB Online 中打开

Hi marcel,

I don't believe a nontrivial solution is even possible in general without some conditions on Aa and D. For example,

take

Aa = [1 0 0
2 0
0 3];
D = [5 0 0 0
6 0 0
0 7 0
0 0 8]

then If you solve

Ya*D' = Aa*Ya

all coefficients of Ya are forced to be zero.

marcel hendrix 2022-10-22

Hi David,

It is very likely that you are right. The problem is from circuit simulation, and it assumes the (possibly non-linear) network has a '(periodic) steady-state'. Even when non-periodic, there can be multiple (or no) operating points.

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