I had never really heard of this result, sometimes called the Matrix Determinant Lemma, but it came up in the process of answering a relatively simple question. Suppose I have an 
On the face of it, this seems intuitively easy — diagonalize
Matrix Determinant Lemma. Let
positive definite matrix and
and
be two
matrices. Then
.
To see this, note that
![\left[ \begin{array}{cc} A & -U \\ V^{H} & I \end{array} \right] = \left[ \begin{array}{cc} A & 0 \\ V^{H} & I \end{array} \right] \cdot \left[ \begin{array}{cc} I & - A^{-1} U \\ 0 & I + V^H A^{-1} U \end{array} \right]](https://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Barray%7D%7Bcc%7D+A+%26+-U+%5C%5C+V%5E%7BH%7D+%26+I+%5Cend%7Barray%7D+%5Cright%5D+%3D+%5Cleft%5B+%5Cbegin%7Barray%7D%7Bcc%7D+A+%26+0+%5C%5C+V%5E%7BH%7D+%26+I+%5Cend%7Barray%7D+%5Cright%5D+%5Ccdot+%5Cleft%5B+%5Cbegin%7Barray%7D%7Bcc%7D+I+%26+-+A%5E%7B-1%7D+U+%5C%5C+0+%26+I+%2B+V%5EH+A%5E%7B-1%7D+U+%5Cend%7Barray%7D+%5Cright%5D&bg=ffffff&fg=333333&s=0&c=20201002)
and take determinants on both sides.
So now applying this to our problem,
But the right side is clearly maximized by choosing 
An Ergodic Walk 
Someone, who will go unnamed, quizzed me on this in late 2002. Old times!
Trying to understand here: Via writing $U = A A^{-1} U$, the lemma is equivalent to the “special case $A=1$”, associated with Sylvester. In turn, the special case seems to be an exponentiated form of the easy (and more well-known?) identity $Tr(AB) = Tr(BA)$.
Er, not quite sure I see how commuting under the trace implies the determinant result? Maybe I am slow. You are the real mathematician after all 😉
It’s more that the reverse implication is true: in a horrible notation because things are multivariate here, d/dx det(1+x) = Tr(x), so the determinant result implies the trace result. Perhaps I could have said “integrated” or “antidifferentiated” instead of “exponentiated”, but in some sense they’re the same on a Lie group.
Thanks Jay. For a moment, I was as surprised as Anant.