Constructing a matrix from different blocks

Question

Suppose I have a symbolic matrix called MAIN as follows $$ \begin{matrix} H_{00} & H_{01} & \ldots & H_{0n}\\ H_{10} & H_{11}& \ldots & H_{1n}\\ \vdots & \vdots & \ddots & \vdots\\ H_{n0} & \ldots & \ldots & H_{nn}\\ \end{matrix} $$

where each of these matrix elements is a block itself (a matrix or a vector). For example $H_{01}$ is

$$ \begin{matrix} h_1\\ h_2\\ \vdots\\ h_i \end{matrix} $$

or $H_nn$ is $$ \begin{matrix} h_{11} & h_{12} & \ldots & h_{1i}\\ h_{21} & h_{22}& \ldots &h_{2i}\\ \vdots &\vdots & \ddots & \vdots\\ h_{j1}&h_{j2}&\ldots & h_{ij}\\ \end{matrix} $$

I have calculated each of these matrix elements before and now I want to construct the MAIN matrix using them so that each matrix element is placed in the appropiate position of the MAIN matrix and the additional elements which are created due to this replacement are set to zero. How can I do this?

I applied the ArrayFlatten command to the calculated matrix elements and then simply defined the MAIN matrix using this simplified example command

MAIN={{H00,H01,H02},{H10,H11,H12}};

but the result was not my desire one.

A Toy example: assume the initial symbolic MAIN matrix is a 2x2 matrix and H00 is just a number, H01 and H10 are 2x1 matrices, and also H22 is a 2x2 matrix so the final form of the MAIN matrix after substituting must be as follows: $$ \begin{matrix} H_{00}& u_{11}&0\\ 0&u_{21}&0\\ y_{11}&p_{11}&p_{12}\\ y_{21}&p_{21}&p_{22}\\ \end{matrix} $$

where I have called the inner elements of H01,H10 and H22 by $u$,$y$ and $p$ respectively.

Answer 1

Complete answer

I assume 6 matrices of unknown dimensions in a known (2x3) grid, according to the toy example. I construct a list of the matrices, set some basic variables and get their dimensions.

mat1 = ConstantArray[aa, {2, 2}]; 
mat2 = ConstantArray[bb, {3, 3}]; 
mat3 = ConstantArray[c, {4, 4}]; 
mat4 = ConstantArray[d, {5, 5}]; 
mat5 = ConstantArray[f, {6, 6}]; 
mat6 = ConstantArray[g, {7, 7}];
matlist = {mat1, mat2, mat3, mat4, mat5, mat6};(*list of sub-matrices*)
T = 3;(*Row length in MAIN*)
TT = 2; (*Column length in MAIN, T can be used for square matrix *)
nsm = 6; (*number of sub-matrices*)
end = {20, 20};(*coordinates of bottom right point in MAIN*)
ted = Dimensions /@ matlist

I then get the coordinates of the top-left point of each matrix and construct the MAIN matrix mat containing the placeholders.

xx = Accumulate /@ Most /@ Partition[Prepend[Riffle[ted[[All, 1]], 1, T + 1], 1], T + 1];
yy = Table[ConstantArray[Accumulate[Drop[Prepend[Max /@ Partition[ted[[All, 2]], T], 1], -1]]
[[i]], T], {i, 1, TT}];
aux = Join @@ Transpose /@ Thread[{yy, xx}]
mat = Normal[SparseArray[{aux -> ConstantArray[B, nsm], end -> 0}]];

I then perform the replacement.

a = Join @@ Position[mat, B];
b = Join @@ Dimensions /@ matlist;
coor = Flatten /@ Partition[Thread[{a, b + a - 1}], 2];
Table[mat[[coor[[i, 1]] ;; coor[[i, 2]], coor[[i, 3]] ;; coor[[i, 4]]]] = matlist[[i]];, 
{i, 1, nsm}]
mat // MatrixForm

The code works with an equal number of matrices in each row but it is easy to alter xx and yy to accomodate for that (it just needs to count the length of each row instead of having a fixed T). For a 3*3,, 4*3,4 *4 etc matrix you add the matrices in matlist and change T. To see the various steps remove the semicolons.

I am certain that a much simpler version with Grid or SparseArray is possible.