kronecker product by number of iteration
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lets say A=[1 2;3 4], i want for i=1:5 times, to multiply A itself kronecker product. in this case ,manual will be kron(kron(kron(kron(kron(A,A),A),A),A),A)
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Geoff Hayes
le 8 Juil 2014
Unless you are looking for a way to do this in one line of code, a simple for loop would do the trick
A = [1 2;3 4];
B = A;
for k=1:5
B = kron(B,A);
end
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Geoff Hayes
le 9 Juil 2014
Akymrat - Please format the above code so that it is readable. Highlight the code and press the {}Code button.
There is a bug in the code with the
else
B==2;
R=F2
end
I suspect you meant
elseif B==2
Please make the correction and address the case where B is neither 1 nor 2.
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