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I have to (numerically) compute $$\int_{0}^{\infty} f(z)(zK+\sigma^2I)^{-1}u\,dz$$ where $u$ is a constant vector, $z$ is a scalar and $f(\cdot)$ is a known function.

The problem that I have is for each increment of $z$ I have to keep recalculating the Cholesky of $zK+\sigma^2I$.

If the Cholesky of $K$ is known is there a method of not continuously doing this for each new value of $z$?

I should mention that $K$ is a positive definite symmetric matrix, and usually dense.

Pantelis Sopasakis
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sachinruk
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1 Answers1

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This is not an answer that directly addresses your question, but if you could use eigendecomposition instead of Cholesky decomposition, so that $K=QDQ^\ast$ (where $Q$ is unitary and $D$ is a nonnegative diagonal matrix), then \begin{align*} \int_0^\infty f(z) (zK+\sigma^2 I)^{-1} u dz &= \int_0^\infty f(z) Q (zD+\sigma^2 I)^{-1} Q^\ast u dz. \end{align*} If $f(z)$ is a real-valued function, the RHS can be further rewritten as $$ Q \left(\int_0^\infty f(z) (zD+\sigma^2 I)^{-1} dz\right) Q^\ast u. $$

user1551
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  • but doesn't $K=QDQ^*$ make it a diagonal matrix? K is a dense matrix :( – sachinruk Apr 16 '13 at 02:08
  • Yes, $D$ is diagonal. Why is that a problem? – Michael Grant Apr 16 '13 at 02:23
  • @user72388 $D$ is a diagonal matrix, but $K$ is not necessarily one. – user1551 Apr 16 '13 at 03:21
  • Actually never mind. You are right, and this is certainly what I am looking for. Thank you so much :D – sachinruk Apr 16 '13 at 03:26
  • Ah! This only works if the $\sigma^2 I$ term is spherical, so that it can also be diagonalized by $Q$. I'm also working on some Gaussian process code, but have non-uniform measurement variances, so alas cannot use this approach (and I fear there may be none). – MRule Feb 17 '23 at 11:16