Diagonal displacement of matrix into another matrix

Hi,

I have this matrix 2X8:

and from this I need to create 4 matrices of zeros 5X5 where I displace the 2X8 matrix thi way:

Note that I have to do this parametrically so that if for exaple I have a matrix 2X10 it needs to be dsiplaced in 5 matrices 6X6 and so on.

Any suggestions?

[Adam Danz]

"and from this I need to create 5 matrices"

Your example shows 4 matrices.

How would the 2x10 output look?

Will all input matrices be 2xn where n is divisible by 2?

[Walter Roberson]

If you have a 2 x N matrix, then the pattern should seem to be that you would create (N-1) matrices, the first starting from row 1, the second from row 2, the third from row 3, and so on to (N-1) which would start in row N-1 and end in row N

[Adam Danz]

My interpretation is that there would be N/2 matricies.

[Lodovico Morando]

Adam Danz, you are right, It is just a typo, there are 4 matrices. And yes, the output of the first matrix would always be 2Xn where n is divisible by 2. I would have 5 matrices in case I had a 2X10 matrix as input.

[Lodovico Morando]

I am not quite sure I understand Walter but thank you. My problem anyway is that I know how to di this mechanically, but I need to do it parametrically with a loop, and I can't find a way to make it so. Do you guys have some examples when dealiing with matrices like in this case?

[Lodovico Morando]

And yes Adam in the end I will alwasy have N/2 matrices

[Adam Danz]

Got it. Looks like Matt J and I answered at nearly the same time.

His approach puts the output matrices into a cell array. To access output matrix number n in his answer,

outputMatrices{n}

My approach stores the output matricies within a 3D array. To access output matrix number n in my answer,

A(:,:,n)

[Lodovico Morando]

Yea, using cell arrays is a good solution too, but I solved it another way anyway. Assembling together diagonal matrices. Thank you very much anyway.

Answer 1

[Matt J]

Call your matrix A,

A=reshape(1:16,2,[]), %example input

A = 2×8

1 3 5 7 9 11 13 15 2 4 6 8 10 12 14 16

●

N=size(A,2)/2;

outputMatrices=cell(1,N);

z0=num2cell(zeros(1,N));

for i=1:N
  
    z=z0; z{i}=A(:,2*i-1:2*i);
    
    outputMatrices{i}=blkdiag(z{:});
    
end

outputMatrices{:} %the resulting matrices

ans = 5×5

1 3 0 0 0 2 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

●

ans = 5×5

0 0 0 0 0 0 5 7 0 0 0 6 8 0 0 0 0 0 0 0 0 0 0 0 0

ans = 5×5

0 0 0 0 0 0 0 0 0 0 0 0 9 11 0 0 0 10 12 0 0 0 0 0 0

ans = 5×5

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 13 15 0 0 0 14 16

Answer 2

[Adam Danz]

Fast & efficient vectorized method

m is the input matrix size 2xN where N is divisible by 2.

A is the output array containing the N/2 matrices along the third dimension.

% m = reshape(1:16,2,8);
m = reshape(1:20,2,10)

m = 2×10

1 3 5 7 9 11 13 15 17 19 2 4 6 8 10 12 14 16 18 20

●

% determine the linear index for the first 4 values (baseIdx)
nMat = size(m,2)/2; 
matSize = [nMat,nMat]+1; 
baseIdx = [1 2, matSize+[1,2]]; 

% determine the linear index of all values of m into A (Aidx)
A = nan([matSize,nMat]); 
interval = 0: prod(matSize)+matSize(1)+1 : numel(A); 
Aidx = baseIdx' + interval;

% Place values of m into A
A(Aidx) = m

A = 
A(:,:,1) =

     1     3   NaN   NaN   NaN   NaN
     2     4   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN


A(:,:,2) =

   NaN   NaN   NaN   NaN   NaN   NaN
   NaN     5     7   NaN   NaN   NaN
   NaN     6     8   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN


A(:,:,3) =

   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN     9    11   NaN   NaN
   NaN   NaN    10    12   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN


A(:,:,4) =

   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN    13    15   NaN
   NaN   NaN   NaN    14    16   NaN
   NaN   NaN   NaN   NaN   NaN   NaN


A(:,:,5) =

   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN   NaN   NaN
   NaN   NaN   NaN   NaN    17    19
   NaN   NaN   NaN   NaN    18    20

Answer 3

[Bruno Luong]

Simple loop would be the simplest for me

% Test matrix
A=randi(10,2,8)

n = size(A,2)/2
B = zeros(n+1,n+1,n);
for k=1:n
    i = k+(0:1);
    j = 2*k+(-1:0);
    B(i,i,k) = A(:,j);
end

B

Answer 4

[Lodovico Morando]

Thank you very much all.