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How to assume 'symmatrix' variable as 'real'

5 vues (au cours des 30 derniers jours)
Nishanth Rao
Nishanth Rao le 14 Juil 2022
Commenté : Walter Roberson le 14 Juil 2022
I have created a few symbolic matrices:
syms x [6 1] matrix;
syms k [3 1] matrix;
syms C [3 6] matrix;
b_x = norm(C*x - k)^2;
I am creating a function and want to differentiate it w.r.t x.
d_b_x = diff(b_x, x);
But this generates complex operators like 'conj' which I don't want. I want to assume that all the matrices in this example contain real entries.
How can I do this? The documentation did not have any such commands.
  2 commentaires
Nishanth Rao
Nishanth Rao le 14 Juil 2022
Modifié(e) : Nishanth Rao le 14 Juil 2022
My question is how to assume a symmatrix as real. I just used the differentiation as an example of why I require it. Moreover, for bigger matrices the 'syms' makes the expression excessively long and difficult to interpret / use.
Anyways, thanks for your insight.
Nishanth Rao
Nishanth Rao le 14 Juil 2022
Modifié(e) : Nishanth Rao le 14 Juil 2022
Please read the question title. I have asked about the variable type 'symmatrix'. The code in your question will create 6x1 vector but of variable type 'sym'. The code in my question will create a 6x1 vector (or matrix, however you see it) of variable type 'symmatrix'.
A lot of matrix calculus becomes easier with 'symmatrix' variable types.

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Walter Roberson
Walter Roberson le 14 Juil 2022
Matrix variables do not support assumptions. There does not appear to be any way to do what you would like to do.
  2 commentaires
Nishanth Rao
Nishanth Rao le 14 Juil 2022
Modifié(e) : Nishanth Rao le 14 Juil 2022
Hmm, I see.
Thanks a lot for the response. Would it be too much to ask for, if you can enable assumptions on symbolic matrix variables in future releases?
Walter Roberson
Walter Roberson le 14 Juil 2022
I do not work for Mathworks, so that is not something that I can do.

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