kronecker product by number of iteration

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Akmyrat
Akmyrat 2014-7-7
评论： Geoff Hayes 2014-7-9
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
Geoff Hayes 2014-7-8
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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Akmyrat
Akmyrat 2014-7-9
Goeff can You please see this also !!!??
F2=[1 0 0 0;0 0 0 1]; I=[1 0;0 1]; A=[1 0 0 ;1 1 0;0 1 0]; for i=1:3 B=0; for j=1:3 B=B+A(i,j); end if B==1; R=I else B==2; R=F2 end PTM1=R; for k=i-1 PTM1 = kron(PTM1,R) end end
%% this code kron prod by each iteration it self (Kron(I,I), kron(F2,F2) ans so on..) , but i want like this result: kron(kron(I,F2),I) %% whis kron product of all iteration answers.
thanks in advance!!
Geoff Hayes
Geoff Hayes 2014-7-9
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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