How can I create a set of N diagnal matrices range from diag(1, 1, ... , 1) to diag(-1, -1, ... , -1) for testing purposes?

I am needing to create 16 4x4 diagnal matrices consisting of all of the combinations of "1" and "-1" and have them be able to be called for use in a loop. Besides creating each individual matrix, how can I go about accomplishing this?

for instance:

Z1 = diag(1,1,1,1);

.

Z8 = diag(1,-1,-1,1);

.

Z16 = diag(-1,-1,-1,-1);

Thank you for any guidance.

[Simon Chan]

Put them in a cell with index from 1 to 16 instead of assigning 16 different variable names

[Zac Minson]

I ended up doing this, but was wondering if there is a more general method in the case of 2^N matrices.

[Stephen23]

@Zac Minson: using a cell array is a very general method, which works for any number of matrices.

Answer 1

[Matt J]

Do you really need all Z{i} at the same time?

[C{1:N}]=ndgrid([-1,1]);
C=reshape( cat(N+1,C{:})  ,[],N);

Z=zeros(N^2,2^N);  %EDITED - removed loops
Z(1:N+1:N^2,:)=C;
Z=reshape(Z, N,N,[])

[the cyclist]

Generalizing this elegant solution to 2^N:

N = 4; % Set this value 

[C{1:N}]=ndgrid([-1,1]);
C=reshape( cat(N+1,C{:})  ,[],N);

Z=cell(2^N,1);
for i=1:2^N

    Z{i}=diag(C(i,:));
    
end

[Matt J]

Here's a loop-free version:

N = 4; % Set this value 

[C{1:N}]=ndgrid([-1,1]);
C=reshape( cat(N+1,C{:})  ,[],N).';


Z=zeros(N^2,2^N);

Z(1:N+1:N^2,:)=C;

Z=reshape(Z, N,N,[])

Z = 
Z(:,:,1) =

    -1     0     0     0
     0    -1     0     0
     0     0    -1     0
     0     0     0    -1


Z(:,:,2) =

     1     0     0     0
     0    -1     0     0
     0     0    -1     0
     0     0     0    -1


Z(:,:,3) =

    -1     0     0     0
     0     1     0     0
     0     0    -1     0
     0     0     0    -1


Z(:,:,4) =

     1     0     0     0
     0     1     0     0
     0     0    -1     0
     0     0     0    -1


Z(:,:,5) =

    -1     0     0     0
     0    -1     0     0
     0     0     1     0
     0     0     0    -1


Z(:,:,6) =

     1     0     0     0
     0    -1     0     0
     0     0     1     0
     0     0     0    -1


Z(:,:,7) =

    -1     0     0     0
     0     1     0     0
     0     0     1     0
     0     0     0    -1


Z(:,:,8) =

     1     0     0     0
     0     1     0     0
     0     0     1     0
     0     0     0    -1


Z(:,:,9) =

    -1     0     0     0
     0    -1     0     0
     0     0    -1     0
     0     0     0     1


Z(:,:,10) =

     1     0     0     0
     0    -1     0     0
     0     0    -1     0
     0     0     0     1


Z(:,:,11) =

    -1     0     0     0
     0     1     0     0
     0     0    -1     0
     0     0     0     1


Z(:,:,12) =

     1     0     0     0
     0     1     0     0
     0     0    -1     0
     0     0     0     1


Z(:,:,13) =

    -1     0     0     0
     0    -1     0     0
     0     0     1     0
     0     0     0     1


Z(:,:,14) =

     1     0     0     0
     0    -1     0     0
     0     0     1     0
     0     0     0     1


Z(:,:,15) =

    -1     0     0     0
     0     1     0     0
     0     0     1     0
     0     0     0     1


Z(:,:,16) =

     1     0     0     0
     0     1     0     0
     0     0     1     0
     0     0     0     1

[Zac Minson]

Thank you so much. I was really wracking my brain over how to generalize this problem. I appreciate your time.

Answer 2

[Fangjun Jiang]

a=(dec2bin(15:-1:0)-48)*2-1;
z=zeros(4,4,16);
for k=1:16
z(:,:,k)=diag(a(k,:));
end

[the cyclist]

Here is a generalization of this solution:

N = 4; % Set this value

a=(dec2bin((2^N-1):-1:0)-48)*2-1;
z=zeros(N,N,2^N);
for k=1:2^N
    z(:,:,k)=diag(a(k,:));
end

Answer 3

[Jan]

z       = repmat(eye(4), 1, 1, 16);
z(z==1) = 1 - rem(floor((0:15) ./ [1; 2; 4; 8]), 2) * 2;  % Inlined DEC2BIN

Generalized:

n       = 3;
z       = repmat(eye(n), 1, 1, 2^n);
z(z==1) = 1 - rem(floor((0:2^n-1) ./ pow2(0:n-1).'), 2) * 2;

Now z(:,:,i) is the i.th matrix. This is more efficient than storing them in a cell. But if you want this:

zC = num2cell(z, [1,2]);