Decompose a symmetric matrix Given a symmetric $30\times 30$ matrix 
$$A:=(0.55^{|i-j|})_{ij}, \quad 1\le i,j\le 30.$$
How to find a real $20\times 30$ matrix $H$ such that 
$$H^TH=A.$$
What makes this problem non-trivial to me is $H^T$ has more rows than columns. 
If it's not feasible to find such a $H$ by naked eye, what would possibly be an efficient numerical algorithm to find it?
Thanks.
 A: If you take your matrix $A$ with 0.55 replaced by $t,$ with the understanding that $0 < t < 1.$ Next, take the matrix $P$ with the main diagonal all $1$ and the immediate upper diagonal all $-t,$ the congruent matrix
$$ M = P^T A P   $$
becomes the diagonal matrix with a single $M_{11} = 1$  and all other $M_{ii} = 1 - t^2.$ 
With $0 < t < 1$ this means that both $M$ and the original $A$ are positive definite.  
Next, take $Q = P^{-1} \; \; .$
Let me typeset the 4 by 4 case. Below I am writing $Q^T M Q = A,$ where $M$ is diagonal positive, and there is a real positive square root matrix $\sqrt M \; .$ If we name $W = \sqrt M \; Q, \;$ we get $W^T W = A.$ 
$$
\left(
\begin{array}{cccc}
1&0&0&0 \\
t&1&0&0 \\
t^2&t&1&0 \\
t^3&t^2&t&1 \\
\end{array}
\right)
\left(
\begin{array}{cccc}
1&0&0&0 \\
0&1-t^2&0&0 \\
0&0&1-t^2&0 \\
0&0&0&1-t^2 \\
\end{array}
\right)
\left(
\begin{array}{cccc}
1&t&t^2&t^3 \\
0&1&t&t^2 \\
0&0&1&t \\
0&0&0&1 \\
\end{array}
\right) =
\left(
\begin{array}{cccc}
1&t&t^2&t^3 \\
t&1&t&t^2 \\
t^2&t&1&t \\
t^3&t^2&t&1 \\
\end{array}
\right)
$$
