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I want to multiply a matrix by its transposed in order to knwo if it verify QtQ=I.

Yet when I compute it on matlab I get something strange using a concept I'm not aware of: conj(x).

  • Therefore, how to multiply a matrix by its transposed?

Here is the code I tried:

>> syms x
>> A=[cos(x) -sin(x);
sin(x) cos(x)]

A =

[ cos(x), -sin(x)]
[ sin(x),  cos(x)]

>> A'*A

ans =

[ cos(conj(x))*cos(x) + sin(conj(x))*sin(x), sin(conj(x))*cos(x) - cos(conj(x))*sin(x)]
[ cos(conj(x))*sin(x) - sin(conj(x))*cos(x), cos(conj(x))*cos(x) + sin(conj(x))*sin(x)]
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  • Have you already looked in matlab help? What remains unclear? – Johannes Tue Jun 19 '16 at 11:35
  • @Alex.S Yes, I did, but to my mind a<sub>1,1</sub> should have been 1 for instance. And instead it gives me the complex conjugate... – Revolucion for Monica Jun 19 '16 at 11:48
  • users of matlab are often scientists, more scientist than programmer, and they may find that a matlab specific support site is more beneficial e.g. uk.mathworks.com/matlabcentral/answers/… Also people there are more mathematical and scientific and may be better candidates to answer such a mathematical question. Many left tried going from matlab to python but found what scientist described to me as a "lack of support" and switched to matlab so I guess a matlab forum is better for a scientist than SO(and def better than SU). – barlop Jun 19 '16 at 11:57
  • @barlop okay, too bad :( – Revolucion for Monica Jun 19 '16 at 12:03
  • @Marine1 too good you mean. There is a lot of support for scientists stuck with matlab, less so with python. Where that support is shouldn't matter to you. You shouldn't think negatively if you got an answer that makes sense and points you in the right direction. I can assure you that a bunch of PhD scientists on a matlab forum could solve your matlab problem quicker than they can pick their nose or quicker than they can make small talk. – barlop Jun 19 '16 at 12:10

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