Trace-logarithm inequality $\operatorname{tr}\log (A) \leq \operatorname{tr}(A-I)$ for matrices $A$ with strictly positive eigenvalues

inequalitylinear algebralogarithmsmatricestrace

  • I recently came across Klein's inequality, which states that for any Hermitian matrices $A, B$ of the same size and any differentiable concave function $f :(0,\infty) \to \mathbb R$, we have

$$\operatorname{Tr}\left[f(A)-f(B)-(A-B) f^{\prime}(B)\right] \leq 0$$

where $f(A)$ is the induced map
defined on the eigenvalues and corresponding projectors $P$ as $f(A) \equiv \sum_{j} f\left(\lambda_{j}\right) P_{j},$ given the spectral decomposition $A=\sum_{j} \lambda_{j} P_{j}$. When $f= \log$ and $B=I$ Klein's identity seems to give

$$\operatorname{Tr}\left[\log(A)-(A-I)\right] \leq 0$$

  • This resembles the usual identity that we have in $1D$ namely, $\log x \leq x-1$.

  • Similarly, for any matrix $A$ satisfying $\|A-I\|<1$ in operator norm we have, by the power series expression of $\log$ around $I$:

$$\log (A)=\sum_{k=1}^{\infty}(-1)^{k+1} \frac{(A-I)^{k}}{k}=(A-I)-\frac{(A-I)^{2}}{2}+\frac{(A-I)^{3}}{3} \cdots$$

Thus we can see in this case as well that $\operatorname{Tr}\log (A) \leq \operatorname{Tr}(A-I)$.

  • Recall that a complex matrix has a logarithm if and only if it is invertible. When the matrix has no negative real eigenvalues, then there is a unique logarithm that has eigenvalues all lying in the strip $\{z \in \mathbf{C} \mid-\pi<\operatorname{lm} z<\pi\}.$ This logarithm is known as the principal logarithm.

My question is: to what extent can we generalise this result? For example, does it hold that for any matrix $A$ with strictly positive eigenvalues we have $\operatorname{Tr}\log (A) \leq \operatorname{Tr}(A-I)$? If it helps we can add the requirement that $A$ is a product of two positive definite matrices.

Best Answer

The inequality holds when $A$ has a strictly positive real spectrum.

Proof:

First suppose that $A$ is diagonalisable.

  1. Let $A=VDV^{-1}$ where $D$ is diagonal and let $\lambda_i>0$ for $i=1, \ldots, d$ be the eigenvalues of $A$. Then \begin{equation} \begin{split} \operatorname{Tr}(\log A) = \operatorname{Tr}(\log (V D V^{-1})) = \operatorname{Tr}( V\log ( D )V^{-1}) = \operatorname{Tr}(\log ( D )) = \sum_{i=1}^d \log (\lambda _i). \end{split} \end{equation}
  2. Use the inequality $\log x \leq x -1:$ \begin{equation} \begin{split} \sum_{i=1}^d \log \lambda _i \leq \sum_{i=1}^d ( \lambda _i-1) = \sum_{i=1}^d \lambda _i-d = \operatorname{Tr}(A-I) \end{split} \end{equation}

Then extend the result to non-diagonalisable matrices by a density argument (since $\log$ is a continuous function).

Related Question