1
$\begingroup$

$A$ 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

$\endgroup$
3
  • $\begingroup$ 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 $\endgroup$ Apr 15, 2012 at 20:24
  • $\begingroup$ 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 $O^T O = I$. $\endgroup$ Apr 15, 2012 at 23:39
  • $\begingroup$ 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 ) $\endgroup$ Apr 16, 2012 at 1:02

1 Answer 1

0
$\begingroup$

Your last comment confirms the tentative answer I gave in the comment above. All you need to check is the scalar relation $y^T A^{-1} y = z^T z$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.