Matrices with Polynomial Entries functions

Question

Maria 2015-3-6

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/181862-matrices-with-polynomial-entries-functions

编辑： Andrew Newell 2015-3-10

Hi,

I was wondering if Matlab has routines in order to compute the inverse of a Matrix with Polynomial Entries. I have not found any functions till now.

Does anyone know?

Best,

Maria

3 个评论
显示 1更早的评论隐藏 1更早的评论

Maria 2015-3-10

编辑：Maria 2015-3-10

For example, I have a matrix that is in this form

B=[b1 b2 b3; c1 c2 c3; d1 d2 d3];

where the entries form the polynomials, in the form

b1*s^2+b2*s+b3

Right now, I am not using a symbolic expression, but my algorithms starts to suffer in case of big matrices. So, if a symbolic expression would be better, it would be fine for me.

Andrew Newell 2015-3-10

A symbolic expression would probably not be better if that's not your goal. I was just asking for clarification.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

John D'Errico 2015-3-10

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/181862-matrices-with-polynomial-entries-functions#answer_170782

在 MATLAB Online 中打开

People are always looking for the magic solution. Sadly, magic only happens in Harry Potter books.

You say that your algorithm "suffers" now. Your algorithm will also suffer if you try to compute symbolic inverse matrices. Expect a HUGE mess of terms, so it will be quite slow to do anything.

However, in theory, the symbolic toolbox can do what you want. Why not try it? Just don't expect to be terribly happy with the result, but then nothing you will do here will be wonderful.

syms x a1 a2 a3 a4 a5 a6 a7 a8 a9
A = [a1*x + a2,a3*x^2 + a4*x + a5;a6*x^2 + a7,a8*x + a9];
inv(A)
ans =
[  -(a9 + a8*x)/(a5*a7 - a2*a9 - a1*a9*x - a2*a8*x + a4*a7*x - a1*a8*x^2 + a3*a7*x^2 + a3*a6*x^4 + a4*a6*x^3 + a5*a6*x^2), (a3*x^2 + a4*x + a5)/(a5*a7 - a2*a9 - a1*a9*x - a2*a8*x + a4*a7*x - a1*a8*x^2 + a3*a7*x^2 + a3*a6*x^4 + a4*a6*x^3 + a5*a6*x^2)]
[ (a6*x^2 + a7)/(a5*a7 - a2*a9 - a1*a9*x - a2*a8*x + a4*a7*x - a1*a8*x^2 + a3*a7*x^2 + a3*a6*x^4 + a4*a6*x^3 + a5*a6*x^2),         -(a2 + a1*x)/(a5*a7 - a2*a9 - a1*a9*x - a2*a8*x + a4*a7*x - a1*a8*x^2 + a3*a7*x^2 + a3*a6*x^4 + a4*a6*x^3 + a5*a6*x^2)]

And that was only a 2x2 matrix. A 3x3 problem will be much longer.

3 个评论
显示 1更早的评论隐藏 1更早的评论

Maria 2015-3-10

在 MATLAB Online 中打开

I am trying now to do another thing, since the symbolic toolbox is suggested. I explain it with an example.

I have a cell structure called M, that has as entries arrays of numbers, that are the coefficient of a polynomial, for example

M{1,1}=[m1 m2 m3 m4];

I want to convert each entry in a symbolic polynomial and then I want to compute the determinant of this new symbolic polynomial matrix. For example, if M is 2 x 2, I put

p1=poly2sym(M{1,1}); 
p2=poly2sym(M{1,2}); 
p3=poly2sym(M{2,1}); 
p4=poly2sym(M{2,2});

I will have a final symbolic polynomial matrix as

P=[p1 p2; p3 p4];

Then

D=det(P);
D_new=sym2poly(D);

Now, in order to extent this way for any matrices M of generic dimension n x n, I tried to do

Mtest=cellfun(@poly2sym,M);

and I get the error

Error using cellfun sym output type is not supported.

I tried to do

Mtest=cellfun(@poly2sym,M,'UniformOutput',0);

but then, when I want to compute the determinant, I get the error

Dtest=det(Mtest);
Undefined function 'det' for input arguments of type 'cell'.

This way seems not sooo bad to me if I fix this error, do you have any suggestions?

Best,

Maria

Maria 2015-3-10

I need to add that now I am trying to compute only the determinant, and later I will compute the adjoint matrix

John D'Errico 2015-3-10

编辑：John D'Errico 2015-3-10

You need to understand that a cell array is NOT something that the symbolic toolbox can work with. In general, cell arrays are never supported for any computations, since they can contain anything.

I'm not sure why are you trying to use a cell array, when I showed you how to do this with a regular array? If you insist on the use of cell arrays here, then you will need to convert them to a normal symbolic array. Sadly, cell2mat seems not to work here, nor does arrayfun work with uniformoutput set to true. When all else fails, god gave us the loop.

请先登录，再进行评论。

Answer 2

Andrew Newell 2015-3-10

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/181862-matrices-with-polynomial-entries-functions#answer_170786

编辑：Andrew Newell 2015-3-10

在 MATLAB Online 中打开

Maria, is the matrix inverse your goal, or are you using it for some other purpose? A common mistake is to solve a linear equation like A*x = y, where x and y are vectors and A is a matrix, by computing the inverse Ainv and multiplying:

x = Ainv * y

This is very inefficient. In MATLAB, a far more efficient method is to do this:

x = A\y

This tells MATLAB to solve A*x = y using whatever method is most appropriate (see mldivide). Here is an example of the performance:

y = rand(3,1); A = rand(3);
tic % tic/toc does the timing
Ainv = inv(A);
x = Ainv*y;
toc
tic
x = A\y;
toc

On my computer, the first calculation took 1.33 seconds. The second took only 0.0003 seconds, so it was about 4000 times faster!

2 个评论
显示无隐藏无

Maria 2015-3-10

I see, and normally I always try to avoid the inv() command, preferring the \ . I don't have this kind of problem now though, and I need the inverse. Or, what I am currently trying right now, the determinant and the adjoin matrix.

Andrew Newell 2015-3-10

编辑：Andrew Newell 2015-3-10

In one of your comments above, you start with the coefficients of a polynomial (which will always be components of a vector), and then you switch to a 2x2 matrix. Is your polynomial the characteristic polynomial of a matrix? What, ultimately, are you trying to do?

请先登录，再进行评论。

Matrices with Polynomial Entries functions

3 个评论
显示 1更早的评论隐藏 1更早的评论

回答（2 个）

3 个评论
显示 1更早的评论隐藏 1更早的评论

2 个评论
显示无隐藏无

另请参阅

类别

标签

产品

Community Treasure Hunt

Matrices with Polynomial Entries functions

3 个评论 显示 1更早的评论隐藏 1更早的评论

回答（2 个）

3 个评论 显示 1更早的评论隐藏 1更早的评论

2 个评论 显示 无隐藏 无

另请参阅

类别

标签

产品

Community Treasure Hunt

3 个评论
显示 1更早的评论隐藏 1更早的评论

3 个评论
显示 1更早的评论隐藏 1更早的评论

2 个评论
显示无隐藏无