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$A$ is symmetric positive definite matrix and $S$ is such that $A=SS^{T}$. Further

$y=Sz$
Does there exist a simple ( or any verifiable) relation exist only involving $A$,$y$ and $z$ ?

Thanks

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Yes, but it does not define y or z in terms of the other, nor A. What woould you want of such a relation? Gerhard "Ask Me About System Design" Paseman, 2012.04.15 –  Gerhard Paseman Apr 15 '12 at 20:24
    
Multiply both sides by $S^{-1}$ and square them: $y^T A^{-1} y = z^T z$. On the other hand, starting from the last expression, there exists some factorization of $A=SS^T$ such that $y=Sz$. So that's the best one can do, since $A$ does not determine $S$ uniquely, only up to orthogonal transformation $S\to SO$, where $O^T O = I$. –  Igor Khavkine Apr 15 '12 at 23:39
    
As Igor pointed , there can be a lot of $S$ possible , which also say that there can be a number of $y$ and $S$ exist for pair of $A$ and $z$. What I'm trying to do is, suppose by some contraption I generated a $y$, (without explicitly finding out $S$ ) , I want to verify that for such $y$, there indeed exist some $S$ satisfying $A=SS^{T}$ ( again I'm not interested in calculation of S,existence is sufficient ) –  zimbra314 Apr 16 '12 at 1:02
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Your last comment confirms the tentative answer I gave in the comment above. All you need to check is the scalar relation $y^T A^{-1} y = z^T z$.

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