1

The function $f(X) \Rightarrow \log \det X$ is concave as shown here. However, I was wondering if we could simplify the proof suggested.

When we compute : $g(t) = \log\det(Z + tV)$, why not just saying that:

$$\begin{aligned} g(t) &= \log\det(Z)(I+tZ^{-1}V) \\ &= \sum_i \log(1+t\lambda_i). + \log\det Z \end{aligned}$$ with $(\lambda_i)$ the eigenvalues of $Z^{-1}V$?

guhur
  • 147
  • 6
  • Are your matrices positive semidefinite or something? And yes, I agree that the introduction of $Z^{1/2}$ is unneeded. – darij grinberg Oct 10 '18 at 16:37
  • OK thanks. I think that the problem is not defined if the matrices are not positive definite. – guhur Oct 10 '18 at 16:48

1 Answers1

0

Actually, we can not simplify the proof. Indeed, $Z^{-1}V$ is not necessarily symmetric, and thus, the decomposition to the eigenvalues is not guaranteed.

This explains the use of the trick with $Z^{1/2}$

guhur
  • 147
  • 6