Product of random matrices which commute almost surely

Question

In the paper "Matrix concentration for products" it is stated, that the following is easy to show. Let $X_1,\dots X_n$ be independent, bounded, square matrices, which commute almost surely. Define $Y_i=I+\frac{X_i}n$. Then $$ \log \mathbb{E}\|Y_n\dots Y_1\|\leq \frac 1 n \|\sum_{i=1}^n\mathbb{E}X_i\| +O\left(\sqrt{\frac{\log d}n}\right) $$ $\|\cdot\|$ is the spectral norm and $d$ is the dimension of our matrices.
I know the weaker inequality for matrices which do not necessarily commute almost surely with $\sum_{i=1}^n\|\mathbb{E}X_i\|$ instead of $\|\sum_{i=1}^n\mathbb{E}X_i\|$.
Is this really easy to show? How do I prove it?

Answer 1

It's true if the $X_i$'s are Hermitian. Then with probability 1 the matrices are simultaneously diagonalizable, so we may as well write everything in a basis where they are diagonal.

Then $$\mathbb E \|Y_n \dots, Y_n\| = \mathbb E \max_{j \in [d]} |(Y_n)_{jj} \dots (Y_1)_{jj}| = \mathbb E \max_{j \in [d]} \left|\prod_{i=1}^n \left(1 + \frac{(X_i)_{jj}}{n}\right) \right|.$$

If $|(X_i)_{jj}| \leq C$ almost surely, then each term is positive so it suffices to bound $\log \mathbb E \max_{j \in [d]} e^{\sum_{i=1}^n (X_i)_{jj}/n}$. Now, this quantity is less than

$$ \log \mathbb E \max_{j \in [d]} e^{\sum_{i=1}^n ((X_i)_{jj} - \mathbb E (X_i)_{jj})/n} + \max_{j \in [d]} \frac 1n \mathbb E \sum_{i=1}^n (X_i)_{jj}.$$

The second term is bounded by $\frac 1n \left\|\mathbb E \sum_{i=1}^n X_i \right\|$. The first term is $O(\sqrt{\log d/n})$ by standard scalar concentration bounds for bounded random variables.