Sutter et al. [1] in their paper "Multivariate Trace Inequalities" give an intuitive proof of the following Golden-Thompson inequality:

For any hermitian matrices $A,B$:

$$ \text{tr}(\exp{(A+B)}) \le \text{tr} \exp{(A)}\exp{(B)}. $$

Lemmas involve the spectral pinching method which uses the eigendecomposition $A=\sum_{\lambda}\lambda P_\lambda$ where the $\lambda$ are eigenvalues and $P_\lambda$ corresponding projectors which are mutually orthogonal.

The spectral pinching map with respect to $A$ is defined as $$ \mathcal{P}_A: X \mapsto \sum_{\lambda} P_\lambda XP_\lambda $$ and then come these properties for any $X \geq 0$:

1) $\mathcal{P}_A[X]$ commutes with $A$

2) $\text{tr} \mathcal{P}_A[X]A = \text{tr} X A$

3) \begin{align} \mathcal{P}_A[X] &= \sum_{\lambda \in \text{spec}(A)} P_\lambda XP_\lambda\\ &= \frac{1}{|\text{spec}(A)|} \sum_{y=1}^{|\text{spec}(A)|}U_yXU_y^*\\ &\geq \frac{1}{|\text{spec}(A)|} X \end{align}

where $\text{spec}(A) = \{\lambda_1, \lambda_2, \dots, \lambda_{|\text{spec}(A)|}\}$ and $U_y = \sum_{z=1}^{|\text{spec}(A)|} \exp{\frac{i2\pi yz}{|\text{spec}(A)|}}P_{\lambda_z}$ satisfies $UU^T=I$

1) and 2) are straightforward to follow. But I cannot understand how the second equality and the first inequality hold true for 3).

[1] https://arxiv.org/abs/1604.03023


1 Answer 1


$\newcommand{\al}{\alpha} \newcommand{\la}{\lambda}$ Let $\la_1,\dots,\la_n$ be the distinct eigenvalues of $A$, so that $|\text{spec}(A)|=n$. Then \begin{multline} \sum_{y=1}^n U_yXU_y^*=\sum_{y=1}^n\sum_{u,v=1}^n e^{i2\pi yu/n}P_{\la_u}XP_{\la_v}e^{-i2\pi yv/n}\\ = \sum_{u,v=1}^n P_{\la_u}XP_{\la_v}\sum_{y=1}^n e^{i2\pi y(u-v)/n} =\sum_{u,v=1}^n P_{\la_u}XP_{\la_v}n1_{\{u=v\}} =n\sum_{u=1}^nP_{\la_u}XP_{\la_u}, \end{multline} so that the 2nd equality in 3) holds.

Now, as pointed out in the cited paper, $U_yXU_y^*\ge0$ for all $y$, whereas $U_n=I$. So, the inequality in 3) follows as well.


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