How to assume 'symmatrix' variable as 'real'

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Nishanth Rao 2022-7-14

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https://ww2.mathworks.cn/matlabcentral/answers/1759600-how-to-assume-symmatrix-variable-as-real

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.

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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 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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