How to covert a vector of lower triangular matrix factors back to the original matrix?

Hi guys

Could you pleas offer me some help with how to covert a vector to a matrix?

for example, the vector I have is [ 1 2 2 3 5 3 4 9 8 4]

And the desirable result is:

1 2 3 4

2 2 5 9

3 5 3 8

4 9 8 4

The vector I have is actually the factors of the lower triangular matrix of a covariance matrix, right now I want to convert it back to the original covariance matrix. The actual length of the vector I have is like 70,000+ ...

Thx a lot!!

Answer 1

[Andrei Bobrov]

p = [ 1 2 2 3 5 3 4 9 8 4];
a = triu(ones(4));
a(a > 0) = p;
out = (a + a')./(eye(4)+1);

[William]

This is a nice solution. Thank you for sharing it!

I wonder whether it's possible to go row-wise to have the final output as below after the third line of your code?

1 2 2 3

0 5 3 4

0 0 9 8

0 0 0 4

[William]

Ah, just figured out that we could use tril instead.

Answer 2

[Yu Jiang]

May not be the optimal solution, by you can try

p = [ 1 2 2 3 5 3 4 9 8 4];

n = 4;
P = zeros(n,n);
k = 1;
for i = 1:n
  for j = 1:i
    P(i,j) = p(k);
    if i~=j
     P(j,i) = p(k);
    end
    k = k +1;
  end
end

Answer 3

[Roger Stafford]

I assume we have n where the desired matrix is to be of size n x n and the vector v which must be of length n*(n+1)/2.

   A = triu(ones(n));
   A(A~=0) = 1:n*(n+1)/2;
   A = A + triu(A,1).';
   A = reshape(v(A),n,n);

A should be the requested matrix.

[Jason]

Hi Roger, it seems good, One question, somehow, my matlab does not have solve function, so I can not use solve function to calculate n(size of matrix) based on v (length of the vector), is there any way around it? Thank you very much!

[Roger Stafford]

You don't need 'solve' for a simple problem like that. If your vector is of length m, then you need to solve the equation m = n*(n+1)/2 for n. Its solution would be:

   n = (-1+sqrt(1+8*m))/2

If your vector is of valid length corresponding to the number of elements in the upper part of a square matrix, then 1+8*m must be a perfect square of an odd integer, so the n you derive will indeed be an integer. Otherwise you don't have a valid vector length.