How to assume 'symmatrix' variable as 'real'

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Nishanth Rao
Nishanth Rao 2022-7-14
评论： Walter Roberson 2022-7-14
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 个评论
Nishanth Rao
Nishanth Rao 2022-7-14
编辑：Nishanth Rao 2022-7-14
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 2022-7-14
编辑：Nishanth Rao 2022-7-14
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 2022-7-14
Matrix variables do not support assumptions. There does not appear to be any way to do what you would like to do.
  2 个评论
Nishanth Rao
Nishanth Rao 2022-7-14
编辑：Nishanth Rao 2022-7-14
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 2022-7-14
I do not work for Mathworks, so that is not something that I can do.

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