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Let $K(x)_{n\times n}$ be a positive definite matrix defined on $x\in D$ and $K_{i,j}(x)\in C^2(D)$ (or generally $C^k$) for any $1\le i,j\le n$. Of course for any $x$, there exists a invertable matrix $A(x)$ such that $A(x)A(x)^t=K(x)$, and $A(x)$ is not unique. What I want to ask is, whether $K(x)$ has $C^2$ (or $C^k$) decomposition. That is, there exists $A(x)\in C^2(D)$ (or $C^k$), such that $A(x)A(x)^t=K(x)$. If not, what conditions can we add to make this true?

Moreover, are there some references about this question? Thanks a lot!

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2 Answers 2

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The Cholesky decomposition does what you want. It depends smoothly on the input matrix, because every step in the algorithm is a smooth function. It's all just basic arithmetic and square roots.

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  • $\begingroup$ Thank you! Your answer helps me a lot. $\endgroup$
    – W.J.
    Commented Mar 11, 2022 at 2:45
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Of course the square root $K(x)^{1/2}\in{\bf SPD}_n$ does the job, because $S\mapsto S^2$ is a diffeomorphism. Thus the regularity oof $x\mapsto K(x)$ transfers to $x\mapsto K(x)^{1/2}$.

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  • $\begingroup$ Thank you very much! Now I understand that. $\endgroup$
    – W.J.
    Commented Mar 11, 2022 at 2:43

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