# Intuitive proof of Golden-Thompson inequality

Sutter et al.  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).

$$\newcommand{\al}{\alpha} \newcommand{\la}{\lambda}$$ 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.