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Is there a way to solve the matrix equation $XX^* = A$, where $X$ is a $n\times k$ unknown matrix and $A$ is a $n\times n$ positive-definite Hermite matrix?

Cholesky decomposition may be useful when $n=k$, but how about the case where $n \neq k$?

Thanks so much!

Simon Wong
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  • Isn't $A$ also symmetric? – Gottfried Helms Oct 25 '13 at 14:50
  • Yes, $A$ is also symmetric. – Simon Wong Oct 25 '13 at 14:51
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    Hmm, if you can assume that you compute that $\sqrt 0 =0 $ then why do you think Cholesky wouldn't be useful when $k\lt n$ (if $k \gt n$ we have no problem at all)? (If you cannot assume exactness, I think you just implement approximation below some defined $\varepsilon$. Actually I've implemented this just this way in my matrix-calculator-program) – Gottfried Helms Oct 25 '13 at 14:53
  • I don't quite understand what you mean by saying "$\sqrt 0 =0$". Could you explain it a little bit more? – Simon Wong Oct 25 '13 at 14:59
  • Hmm, when you perform the Cholesky-decomposition in a case where $k \lt n$ with the most simple algorithm then you arrive after $k$ columns/rows at a residue which is ideally $0$ (practically it can be zero only if you have rational numbers so far). That leads to a cholesky-factor matrix of $n \times n$ where only $n \times k$ have "significant"/non-zero entries. Upps- if $A$ is positive definite, then k must equal n and cannot be smaller than n as asked in your question (I've overlooked this condition) – Gottfried Helms Oct 25 '13 at 15:22

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Start with the Cholesky decomposition. That is, for any such $A$, we have an $L$ such that $LL^* = A$. Now, in the case that $k>n$, construct the $n\times k$ block matrix $X$ as follows: $$ X = \begin{bmatrix} L & 0 \end{bmatrix} $$ Where $0$ here is the $n\times(k-n)$ zero matrix. We find that $$ XX^* = \begin{bmatrix} L & 0 \end{bmatrix} \begin{bmatrix} L^* \\ 0^* \end{bmatrix}= LL^* + 00^* = LL^* = A $$ This will not generally be possible in the case that $k<n$. For a given symmetric, positive semidefinite matrix $A$, if $XX^* = A$ and $X$ is $n\times k$, then $k$ must be greater than or equal to the rank of $A$.

Ben Grossmann
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