How to assume 'symmatrix' variable as 'real'

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Nishanth Rao
Nishanth Rao el 14 de Jul. de 2022
Comentada: Walter Roberson el 14 de Jul. de 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 comentarios
Nishanth Rao
Nishanth Rao el 14 de Jul. de 2022
Editada: Nishanth Rao el 14 de Jul. de 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 el 14 de Jul. de 2022
Editada: Nishanth Rao el 14 de Jul. de 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 el 14 de Jul. de 2022
Matrix variables do not support assumptions. There does not appear to be any way to do what you would like to do.
  2 comentarios
Nishanth Rao
Nishanth Rao el 14 de Jul. de 2022
Editada: Nishanth Rao el 14 de Jul. de 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 el 14 de Jul. de 2022
I do not work for Mathworks, so that is not something that I can do.

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