Simple way to create a block-diagonal matrix from a single vector.

I am working with a vector of the following form

and I would like to transform this vector into a block-diagonal matrix

without resorting to any sort of looping.

If I call the above vector x the following does, in principle, work

Z = reshape(x',3,4)';
[eye(2),zeros(2),zeros(2,4);zeros(2,4),zeros(2),eye(2)]*blkdiag(Z,Z)

It's just that my actual vector is fairly high dimensional and I feel that there must be a more organic way to obtain the same result, i.e., a way that does not rely on the manual construction of a selection matrix and that does not require typing

blkdiag(Z,Z,Z,Z,Z,Z,Z,Z,Z,Z,Z,Z,Z,..)

everytime the dimensionality changes. So essentialy, I would like to take a vector populate the first n coordinates of the first row of a matrix with the first n coodinates of this vector, fill in the rest of this row with zeros, move to the second row, populate the first first n coordinates with the second (!) n coordinates coordinates of my vector and continue until I reached the -st coordinate of my vectore, then I start populating the first n cooridnates of the -st row of my matrix with zeros and the second n cooridnates with the next elements of my vector, and so on...

I'd very much appreciate any pointers.

Thanks!

Answer 1

[Star Strider]

I don’t understand exactly what ‘z’ is or what the desired result is.

One possibility is to replicate the original matrix using repmat, then create it as a cell array using mat2cell, and then providing that result as the argument to blkdiag.

If I understand the problem correctly, this approach appears to provide the automation desired —

d = 3;                                      % Matrix Dimension (For This Simulation)
n = 5;                                      % Desired Number Of Repititions
z = randi([10 99],d)                        % Matrix To Be Replicated

z = 3×3

28 52 96 17 59 94 57 97 85

zm = repmat(z,1,n)

zm = 3×15

28 52 96 28 52 96 28 52 96 28 52 96 28 52 96 17 59 94 17 59 94 17 59 94 17 59 94 17 59 94 57 97 85 57 97 85 57 97 85 57 97 85 57 97 85

zmc = mat2cell(repmat(z,1,n), d, ones(1,n)*d)

zmc = 1×5 cell array
    {3×3 double}    {3×3 double}    {3×3 double}    {3×3 double}    {3×3 double}

zmc{1}                                      % Demonstrate That The Result Is As Desired

ans = 3×3

28 52 96 17 59 94 57 97 85

zmc{end}                                    % Demonstrate That The Result Is As Desired

ans = 3×3

28 52 96 17 59 94 57 97 85

BD = blkdiag(zmc{:})                        % Show Final Result

BD = 15×15

28 52 96 0 0 0 0 0 0 0 0 0 0 0 0 17 59 94 0 0 0 0 0 0 0 0 0 0 0 0 57 97 85 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 28 52 96 0 0 0 0 0 0 0 0 0 0 0 0 17 59 94 0 0 0 0 0 0 0 0 0 0 0 0 57 97 85 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 28 52 96 0 0 0 0 0 0 0 0 0 0 0 0 17 59 94 0 0 0 0 0 0 0 0 0 0 0 0 57 97 85 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 28 52 96 0 0 0

This replicates the original matrix, segments the replicated matrix into a cell array (a comma-separated list, exactly what blkdiag wants for an argument), then creates the bolck diagonal matrix.

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[Jon B.]

This is exactly what I have been looking for. Thank you!

[Star Strider]

As always, my pleasure!

This was an interesting (and challenging) problem!

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