I would to solve $X=T^TT$ ,I assume that the matrices are real, we know $ X $ and we seek
a solution in $ T $.
Case 1: $\operatorname{rank}(X)>n $
There are no solutions in $ T $.
Case 2: $ \operatorname{rank}(X)\leq n $
There is an orthogonal matrix in $ P$ and a diagonal matrix
$$ D=\operatorname{diag}(\lambda_1,\cdots,\lambda_n,0,\cdots,0) $$ 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=\operatorname{diag}(\lambda_1,\cdots,\lambda_n) $ , for instance ,$ S=\operatorname{diag}(\sqrt{\lambda_1},\cdots,\sqrt{\lambda_n}) $
I just want someone to confirm me if it's a true.
Thank you very much.
