I am trying to find an $n \times m$ fat (i.e., $m > n$) matrix $T$ that solves


$$T^T T = X$$


where $X$ is a given $m \times m$ symmetric, positive semidefinite matrix.


I saw [this post](https://mathoverflow.net/questions/78106/solving-a-quadratic-matrix-equation), but unfortunately it seems that no solution was yet found and $T$ in that case is squared, which is not my case. 


If $T$ were square, one could use the Cholesky decomposition and find $Z$ such that $X = Z^T Z$. Unfortunately, I cannot do this since the Cholesky decomposition always produces a square $Z$.



Solving a quadratic matrix equation with fat matrix
I assume that the matrices are real, we know $ X $ and we seek a solution in $ T $.


Case 1.$ rank(X)>n $.There are no solutions in $ T $.


Case 2.$ rank(X)\leq n $.There is an orthogonal matrix $ P$ 


and a diagonal matrix $ D=diag(\lambda\_1,\cdots,\lambda\_n,0,\cdots,0) $ s.t. $ X=PDP^T $ and
$\lambda\_i\geq 0 $ then $ W^TW=D $ with $ W=TP^{-T} $.since $ P$ is known,it suffices to 


obtain a solution in $W$.we choose $ W=[S\_{n,n},0\_{n,m-n}] $ with$ S^TS=diag(\lambda\_1,\cdots,\lambda\_n) $. for instance , $ S=diag(\sqrt{\lambda\_1},\cdots,\sqrt{\lambda\_n}) $.


