Replace each element in a matrix with a diagonal matrix

Question

Say I have a matrix A of dimension NxV. I want to create a larger matrix of size NTxVT, i.e. I want to replace each element e of the matrix A(e) with diag(T)*A(e)., while keeping the general orientation of the matrix (for instance, A(e) is to the left of A(e-1), so diag(T)*A(e) is to the left of diag(T)*A(e-1).

Is there a trick to accomplish this in matlab? (making each diagonal matrix and concatenating them will take forever).

Many thanks ^^

Answer 1

A = magic(3);
T = diag([-1 1]);
kron(A,T)

gives

-8     0    -1     0    -6     0
 0     8     0     1     0     6
-3     0    -5     0    -7     0
 0     3     0     5     0     7
-4     0    -9     0    -2     0
 0     4     0     9     0     2

ps. I copied the idea from this example

Answer 2

Here is a solution using bsxfun

A = magic(3);
T = [-1 1]
T = diag(T);
M=bsxfun(@times,permute(A,[3,1,4,2]),permute(T,[1,3,2,4]));
M=reshape(M,size(T).*size(A));

It creates a 4D-Matrix where the individual blocks are M(:,i,:,j), then this is reshaped to a 2D-Matrix.

The image processing toolbox provides another solution which is very short but slow:

A = magic(3);
T = [-1 1]
T = diag(T);
M=blockproc(A,[1 1],@(x) x.data.*T);

And finally a implementation which generates a sparse matrix, which might be helpful for large T as your matrix will contain many zeros:

T=[-1 1];
A=magic(3);
%p and q hold the positions where the first element element is stored. Check sparse(p(:),q(:),A(:)) to understand this intermediate step
[p,q]=ndgrid(1:numel(T):numel(T)*size(A,1),1:numel(T):numel(T)*size(A,2));
%now p and q are extended to hold the indices for all elements
tP=bsxfun(@plus,p(:),0:numel(T)-1);
tQ=bsxfun(@plus,q(:),0:numel(T)-1);
%
tA=bsxfun(@times,A(:),T);
M=sparse(tP,tQ,tA);

When T is of size nx1 the sparse solution cuts your memory usage by a factor of roughly n/1.55.

Answer 3

The simplest way I can think of is to combine arrayfun and cell2mat functions:

B = cell2mat(arrayfun((@(x) T .* x), A, 'UniformOutput', false));

First, I transformed matrix A into a cell array of matrices T .* x where x is an element of A (a assumed that T is a matrix).

Then I used cell2mat to transform in back into a matrix.

Here is a complete example (execute online):

A = magic(3);
T = diag([-1 1]);
B = cell2mat(arrayfun((@(x) T .* x), A, 'UniformOutput', false));

resulting in:

B =

  -8   0  -1   0  -6   0
   0   8   0   1   0   6
  -3   0  -5   0  -7   0
   0   3   0   5   0   7
  -4   0  -9   0  -2   0
   0   4   0   9   0   2

Answer 4

Using just indexing:

A = magic(3); 
T = diag([-1 1]); %// example data from Daniel's answer
[a1, a2] = size(A);
[t1, t2] = size(T);
M = A(ceil(1/t1:1/t1:a1), ceil(1/t2:1/t2:a2)).*T(repmat(1:t1,1,a1), repmat(1:t2,1,a2));

Answer 5

Using good old-fashioned matrix multiplication -

M = reshape(diag(T)*A(:).',[size(A,1)*size(T,1) size(A,2)])

---

Sample run -

A = magic(4)
T = magic(3)
M = reshape(diag(T)*A(:).',[size(A,1)*size(T,1) size(A,2)])

will result in -

A =
    16     2     3    13
     5    11    10     8
     9     7     6    12
     4    14    15     1
T =  %// Notice that only diag(T) elements would be used to calculate M 
     8     1     6
     3     5     7
     4     9     2
M =
   128    16    24   104
    80    10    15    65
    32     4     6    26
    40    88    80    64
    25    55    50    40
    10    22    20    16
    72    56    48    96
    45    35    30    60
    18    14    12    24
    32   112   120     8
    20    70    75     5
     8    28    30     2