I was stuck for a long time to prove the following trace inequality
\begin{equation}\label{ine1}
{\text {tr}} ({\bf A}_1^{-1}) \leq {\text {tr}} ({\bf A}^{-1}), \tag{1}
\end{equation}
where ${\bf A} \in {\mathbb C}^{K \times K}$ is a positive definite matrix and ${\bf A}_1$ consists of the diagonal elements of ${\bf A}$, i.e., ${\bf A}_1 = {\text {diag}} ( {\text {diag}} ({\bf A}))$.
I have run a lot of simulation using Matlab and this inequality always holds.
(1) in the simple case with $K=2$ can be easily proven.
However, I can not prove it for the more general $K>2$ case.
Some of my efforts are as follows.
Denote ${\bf A}_1 = {\text {diag}} \{ a_1, \cdots, a_K \}$ and the eigenvalues of ${\bf A}$ by $ \{ \lambda_1, \cdots, \lambda_K \}$.
Then, it is obvious that
\begin{equation}\label{e1}
\sum_{k=1}^K a_k = \sum_{k=1}^K \lambda_k. \tag{2}
\end{equation}
In addition, using Hadamard's inequality, we have
\begin{equation}\label{ine2}
\prod_{k=1}^K a_k \geq \prod_{k=1}^K \lambda_k. \tag{3}
\end{equation}
Since (1) is equivalent to
\begin{equation}
\sum_{k=1}^K \frac{1}{a_k} \leq \sum_{k=1}^K \frac{1}{\lambda_k}, \tag{4}
\end{equation}
I tried to prove (4) using (2), (3), and the relationship between different means (Harmonic mean, etc.). But they didn't work.
Anyone providing the proof or relevant hints will be much appreciated. Thank you.
Best Answer
The attempt at using Hadamard's Inequality is a nice idea since it is one of the original results in majorization. In general, let $\mathbf w\in \mathbb R_{\gt 0}^n$ have the $n$ components on the diagonal of $A$ and $\mathbf x\in \mathbb R_{\gt 0}^n$ have the $n$ eigenvalues of $A$. Then
$\mathbf w \preceq \mathbf x$
where $\preceq$ denotes (strong) majorization
There are many ways to prove this result. E.g. it is implied by Maximize $\mathrm{tr}(Q^TCQ)$ subject to $Q^TQ=I$
now for $u\gt 0$ the mapping $u\mapsto u^{-1}$ is convex and for $\mathbf u\in \mathbb R_{\gt 0}^n$
$f(\mathbf u) = \sum_{k=1}^n u_k^{-1}$
is symmetric, hence Schur Convex. Combining majorization with Schur Convexity gives
$\text{trace}\Big(\big(A_1\big)^{-1}\Big)$
$=\text{trace}\Big(\big(A\circ I\big)^{-1}\Big)$
$=f\Big(\mathbf w\Big)$
$\leq f\Big(\mathbf x\Big)$
$=\text{trace}\Big(\big(A\big)^{-1}\Big)$
which is the inequality OP has sought