How to create a matrix with 1 on ij-th position and zeros elsewhere from a lower triangular matrix?

It may be a very simple question For a symmetric matrix A (3x3), say A=[2 4 6;4 8 11;6 11 20], the way to extract its unique elements (on and lower the diagonal) in an output vector B is:

B=(A(tril(A)~=0))
B =  2
     4
     6
     8
    11
    20

How can I create matrices C1,C2,C3,...,C6, such that

B(1)=A.*C1, B(2)=A.*C2, ..., B(6)=A.*C6
C1=[1 0 0;0 0 0;0 0 0];
C2=[0 0 0;1 0 0;0 0 0];
C3=[0 0 0;0 0 0;1 0 0];
C4=[0 0 0;0 1 0;0 0 0];
C5=[0 0 0;0 0 0;0 1 0];
C6=[0 0 0;0 0 0;0 0 1];

[Niveen El Zayat]

I want to add another inquiry, In the above setting B=vech(A), on the relation B(f)=A(i,j), we have the following relation on the indices that f=(j-1)(k-j/2)+i, where k is the order of the matrix A (here k=3) if I know f and K can I know the corresponding i,j ??

[Niveen El Zayat]

I search and I guess this may work

IND=find(tril(A)~=0)'

[I,J] = ind2sub(size(A),IND)

[Stephen23]

Try this:

[I,J] = find(tril(A)~=0)

Note that you can already find this in my answer, on this line:

[R,C] = find(T);

[Niveen El Zayat]

Stephen, my last comment answer my last question I add, the relation compute the index (i,j) corresponds to f. which as you comment it coincide with a part of you answer as well.

Thanks a lot for your prompt reply

Best regards

Answer 1

[Stephen23]

A = [2,4,6;4,8,11;6,11,20] 
S = size(A);
T = tril(true(S));
[R,C] = find(T);
N = nnz(T);
Z = zeros([S,N]);
V = 1:N;
Z(sub2ind([S,N],R,C,V(:))) = 1

Each page of Z (i.e. the third dimension) is one of the requested matrices, which you can access easily using indexing:

>> Z(:,:,1)
ans =
   1   0   0
   0   0   0
   0   0   0
>> Z(:,:,2)
ans =
   0   0   0
   1   0   0
   0   0   0
...
>> Z(:,:,6)
ans =
   0   0   0
   0   0   0
   0   0   1

[Niveen El Zayat]

Thanks a lot it works well