I am reading Eilers and Marx (1996) and at the beginning of page 94 they write, for $Q = B^TWB$, $D$ a symmetric positive definite matrix and $W$ a diagonal matrix,

\begin{align} tr\left(B(Q + D)^{-1}B^TW\right) &= tr\left((I + Q^{-1/2}DQ^{-1/2})^{-1}\right) \\ &=\sum_{i=1}^{n} \frac{1}{1+\epsilon_i} \end{align}

where $\epsilon_i$ are the eigen values of $Q^{-1/2}DQ^{-1/2}$.

My question is, why don't they write: $tr\left(B(Q + D)^{-1}B^TW\right) = tr\left((I + DQ^{-1})^{-1}\right)$?

Then we could calculate $\epsilon_i$ as the ratio of the eigen values of $D$ by the ones of $Q$, based on this answer, right?

I appreciate if anyone one point what I am missing here.

  • $\begingroup$ While it doesn't answer the question, I think their $D$ is only positive semidefinite: it's a differencing operator that has $k$ zero eigenvalues for degree-$k$ splines $\endgroup$ – Thomas Lumley Jun 5 at 22:55
  • $\begingroup$ Hi @ThomasLumley. I think their $D$ is fact positive semidefinite. The $\mathbf{D}$ I am writting here is what they call $\lambda D^T D$, for $\lambda > 0$, which is positive definite. $\endgroup$ – rcon1 Jun 8 at 11:34
  • $\begingroup$ No, it isn't. At the top of p94 they define $Q_\lambda=\lambdaD^TD$, and the second line after equation (24) starts "Because $k$ eigenvalues of $Q_\lambda$are zero..." $\endgroup$ – Thomas Lumley Jun 8 at 22:10
  • $\begingroup$ @ThomasLumley, you're definitely right! I made a huge confusion here. Thanks a lot! $\endgroup$ – rcon1 Jun 9 at 0:13

There are two possible explanations

  1. It's not obvious to me (and might well not even be true) that $D$ and $Q^{-1}$ commute. The linked answer was for the case where $DQ^{-1}$ is symmetric and $DQ^{-1}=Q^{-1}D$ which means that the eigenvalues of the product are products of the eigenvalues. If they don't, you can't get the eigenvalues from those of $D$ and $Q$ that way.

  2. It doesn't actually make the next step any easier -- the only fact they need about the $\epsilon_i$ is that $k$ of them are zero. And when they actually compute $\mathrm{tr}(H)$ in the next section of the paper, they do it by adding up the diagonal elements, which is faster than finding an eigendecomposition.

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  • $\begingroup$ Thanks for your time. (1.) I didn't note that they had to commute. My fault. I'll check if they commute. (2.) That makes me think, would building matrix $H = B(B^T W B + D)^{-1}B^T W$ and summing up its diagonal entries still be faster than computing the trace as the ratio of the eigen values of $D$ and $Q$ (in case that is valid)? $\endgroup$ – rcon1 Jun 8 at 11:43
  • $\begingroup$ I guess my question wasn't well posed. The main point of my questions is can we write $tr(B(Q+D)^{-1} B^T W)$ as $tr((I + DQ^{-1})^{-1})$ instead of $tr((I + Q^{-1/2} D Q^{-1/2})^{-1})$ for $Q = B^T W B$? $\endgroup$ – rcon1 Jun 8 at 11:49

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