kronecker product by number of iteration

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)

Answer 1

[Geoff Hayes]

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

[Akmyrat]

I=[1 0;0 1]
F=[2 1;1 2]

for i=1:3
    B=i+1;
end
    if B<=3
        A(i)=I
     elseif B>3
         A(i)=F
   end
kron(kron(A(1),A(2)),A(3))

the result i want in auto basis, with i-iteration, Kron product by itself, if for example i change i=1:4, then result should be: kron(kron(kron(A(1),A(2)),A(3)),A(4))

please can u try...thanks !

[Geoff Hayes]

Do you mean for your if statement to be in the for loop?

Your use of A(i) is invalid. Try it, and you will observe the In an assignment A(I) = B, the number of elements in B and I must be the same. You could use a 3D matrix instead

 % pre-allocate memory to A
 n = 3;
 A = zeros(2,2,n);

 for k=1:n
    if (i+1)<=3
        A(:,:,k)=I;
     else
        A(:,:,k)=F;
   end
 end

Then the Kron product would be

 B = A(:,:,1);
 for k=1:n
    B = kron(B,A(:,:,k));
 end

[Akmyrat]

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]

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.