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(For the original question): As $X$ is positive definite, its rank equals to $m$, and there will be no solution $T$ with $n<m$.

On the other hand if your $n>m$ then you can append the corresponding number of 0 rows to $Z$ from the Cholesky decomposition and obtain the matrix $T$ sought.

For the positive semidefinite case (the edited question): you still can compute the Cholesky-like decomposition; see Cholesky decomposition of a positive semi-definite

(I edited my 2nd edition, as it was nonsense, and made my answer community wiki)

(For the original question): As $X$ is positive definite, its rank equals to $m$, and there will be no solution $T$ with $n<m$.

On the other hand if your $n>m$ then you can append the corresponding number of 0 rows to $Z$ from the Cholesky decomposition and obtain the matrix $T$ sought.

For the positive semidefinite case (the edited question): you still can compute the Cholesky-like decomposition; see Cholesky decomposition of a positive semi-definite

(I edited my 2nd edition, as it was nonsense, and made my answer community wiki)

(For the original question): As $X$ is positive definite, its rank equals to $m$, and there will be no solution $T$ with $n<m$.

On the other hand if your $n>m$ then you can append the corresponding number of 0 rows to $Z$ from the Cholesky decomposition and obtain the matrix $T$ sought.

For the positive semidefinite case (the edited question): it is known to be an NP-complete problem, as the case of $n=1$ reduces to MAX-CUT.

(For the original question): As $X$ is positive definite, its rank equals to $m$, and there will be no solution $T$ with $n<m$.

On the other hand if your $n>m$ then you can append the corresponding number of 0 rows to $Z$ from the Cholesky decomposition and obtain the matrix $T$ sought.

For the positive semidefinite case (the edited question): you still can compute the Cholesky-like decomposition; see Cholesky decomposition of a positive semi-definite

(I edited my 2nd edition, as it was nonsense, and made my answer community wiki)

As $X$ is positive definite, its rank equals to $m$, and there will be no solution $T$ with $n<m$.

On the other hand if your $n>m$ then you can append the corresponding number of 0 rows to $Z$ from the Cholesky decomposition and obtain the matrix $T$ sought.

(For the original question): As $X$ is positive definite, its rank equals to $m$, and there will be no solution $T$ with $n<m$.

On the other hand if your $n>m$ then you can append the corresponding number of 0 rows to $Z$ from the Cholesky decomposition and obtain the matrix $T$ sought.

For the positive semidefinite case (the edited question): it is known to be an NP-complete problem, as the case of $n=1$ reduces to MAX-CUT.