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, 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 $Z$.

up vote 1 down vote accepted

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}) $.

  • thanks. I managed to get the answer in the meantime and is similar to the one that you gave me. In my case I actually also needed that T to be with the least amount of zero elements possible. I managed to do that by using an heuristic for rank constrained optimization to find an appropriate X for that to happen. – jaraujo May 31 '14 at 8:51

(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)

  • Hi Dima. I made a typo as $X$ is positive semidefinite. Sorry about that. I just edited my question – jaraujo May 26 '14 at 7:40
  • Thanks Dima. I went for something close to that using SVD and similar to the new answer I got. – jaraujo May 31 '14 at 8:48

Let $A:=\sqrt X$, which is positive definite by the assumption. So, if $T$ solves $T^TT=X$ then $U:=TA^{-1}$ verifies $U^TU=A^{-1}T^TTA^{-1}=A^{-1}(T^TT)A^{-1}=I$, that is, $T$ writes $T=UA$ with $U$ a linear isometry. Conversely, any $T$ of the form $T:=UA$ with a linear isometry $U$, solves the equation.

  • Hi Pietro. Thanks for your answer. But how do you find the matrix $U$ which is $n \times m$, that solves $U^T U = I$? – jaraujo May 26 '14 at 7:51
  • $U$ is any matrix whose columns are orthonormal. – Pietro Majer May 26 '14 at 8:40
  • (I'm not addressing the case of positive semidefinite matrices as in the edited question) – Pietro Majer May 26 '14 at 8:49
  • Hi Pietro. I tried to investigate on how to find $U$ such that $U^T U = I$, when $U$ is $n \times m$ with $m > n$ but that is impossible no? For it to be possible you must have that $n > m$ which is not my case. Am I right? – jaraujo May 26 '14 at 10:36
  • Of course, such $U$ only exist if $n\ge m$. – Pietro Majer May 26 '14 at 10:51

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