If I have an array of 4 dimensions say A=complex(​rand(2,2,2​,2),rand(2​,2,2,2)). If I need to calculate the inverse of this matrix, as defined beow , how should I do it?

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vishav Rattu
vishav Rattu 2016-10-21
评论： vishav Rattu 2016-10-21
I need to compute B(x,y,z,w) such that if I multiply the terms of A and B , I should get an identity matrix I(a,b,c,d):
I=zeros(a,b,c,d);
for a=1:2
for b=1:2
for c=1:2
for d=1:2
for i=1:2
for j=1:2
I(a,b,c,d)=A(a,b,i,j)*B(i,j,c,d)+I(a,b,c,d);
end
end
end
end
end
end
The I(a,b,c,d)= 1 only when a=b=c=d
  2 个评论
Matt J
Matt J 2016-10-21
It is puzzling that you are organizing your data in 4D form, when you appear to want to do simple 2D matrix algebra with it. Are you sure it wouldn't be better just to reshape your data into 2D form
A=reshape(A,4,4)
and keep it that way?
vishav Rattu
vishav Rattu 2016-10-21
编辑：vishav Rattu 2016-10-21
But if I reshape it, will I get the above B matrix that is required? That is, will B satisfy the above condition, that is I(a,b,c,d)=1 only if a=b=c=d? If I am able to get such a B, then I don't need to keep it in a 4d form.

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采纳的回答

Matt J
Matt J 2016-10-21
[a,b,c,d]=ndgrid(1:2);
I=reshape(a==b & b==c & c==d, 4,4);
A=reshape(A,4,4);
B=reshape( A\I , 2,2,2,2);

更多回答（1 个）

KSSV
KSSV 2016-10-21
Are you looking for something like this?
A = rand(2,2,2,2) ;
B = zeros(2,2,2,2) ;
for i = 1:2
for j = 1:2
B(:,:,i,j) = inv(A(:,:,i,j)) ;
end
end
% check
for i = 1:2
for j = 1:2
A(:,:,i,j)*B(:,:,i,j)
end
end

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