# Is $tr(B(B^TWB + D)^{-1}B^TW) = tr((I + D(B^TWB)^{-1})^{-1})$?

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.

• 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 – Thomas Lumley Jun 5 '20 at 22:55
• 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. – rcon1 Jun 8 '20 at 11:34
• 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..." – Thomas Lumley Jun 8 '20 at 22:10
• @ThomasLumley, you're definitely right! I made a huge confusion here. Thanks a lot! – rcon1 Jun 9 '20 at 0:13

## 1 Answer

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.

• 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)? – rcon1 Jun 8 '20 at 11:43
• 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 $Q = B^T W B$? – rcon1 Jun 8 '20 at 11:49