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Let $M$ be a positive semi-definite matrix, symmetric with real entries. Then $M$ can be written as $X X^T$. One way is by a Cholesky decomposition (unique for positive definite but not necessarily for positive semi-definite $M$). Also note that for any $X X^T$ decomposition, $Y Y^T$ is also a decomposition where $Y = X R$ with $R$ orthogonal.

I was wondering about the (possibly non-unique) factorization where $X$ is $U D^\frac{1}{2}$ and columns of $U$ are orthonormal eigenvectors of $M$ and $D$ is the diagonal matrix of sorted eigenvalues of $M$. Is this called something? I've been informally calling this $X$ the 'square root' of $M$ but I know that this is wrong and I would like to know if there is a correct word for it.

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The true square root, of course, is $UD^{\frac{1}{2}}U^T$, assuming the eigenvectors have been chosen to be orthonormal (as they always can be). – Tracy Hall Oct 11 '10 at 17:14
spectral decomposition? – ronaf Oct 11 '10 at 21:47
I don't think I've ever seen it named, it gets mentioned/used in a bunch of numerical linear algebra books, but nobody has seemed to bother naming it. The non-uniqueness of $X$ is due to the fact that you can premultiply $U$ by a diagonal matrix with entries of either 1 or -1, and it would still be a valid matrix of eigenvectors. – J. M. Oct 11 '10 at 22:49
This object has to flaws, and therefore must not have received a name. 1- It his non-unique (in a stronger way than mentionned by J.M.), because you may permute the $d_(jj}$ arbitrarily. 2- Contrary to Choleski factorization, it is not computable in finitely many algebraic operations. – Denis Serre Oct 12 '10 at 5:11
@ronaf You're correct, it's basically the spectral decomposition UDU^T. Since the matrices are the matrices of the spectral decomposition, it won't get a separate name. Could you post your remark as an answer so that the question can be "answered". – Greg Kuperberg Oct 12 '10 at 20:05

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