## Problem

Consider two *d* x *d* complex matrices, *R* and *S*, whose entries lie in the unit disk:

$\quad |R_{i,j}|<1 \quad$ and $\quad |S_{i,j}|<1 $.

Say that *R* is constructed by randomly choosing complex numbers from the unit disk, but *S* is constructed as

$\quad S_{i,j} = f(i/d,j/d)$

where $f(x,y)$ is a smooth function for $x,y \in [0,1]$, with $|f(x,y)|<1$. In other words, the entries of *S* are smooth functions of the indices (in the limit of large *d*), but those of *R* are not.

## Question

How do the trace norms

$\quad ||R||=Tr[\sqrt{R^\dagger R}] \quad$ and $\quad ||S||=Tr[\sqrt{S^\dagger S}]$

of these matrices behave as $d \to \infty$ ?

## Numerical Evidence

A few lines of Mathematica strongly suggest that

$\quad ||R|| \propto d^{3/2}$

but

$\quad ||S|| \propto d$

for large *d*. (The proportionality constants depend on the probability distribution used to pick numbers from the unit disk for *R* and the function $f(x,y)$ used to pick entries for *S*, respectively.)

What explains this behavior?

## Addendum

After Willie's excellent answer below, I thought I'd mention that it's really fast to see the scaling behavior once you discretize the function. Let $F$ be some matrix of discrete values for the function, and let $J_n$ be the $n \times n$ matrix with all elements equal to unity.

$||F \otimes J_n|| = \mathrm{Tr} \sqrt {(F^\dagger \otimes J_n)( F \otimes J_n)} = \mathrm{Tr} \sqrt {(F^\dagger F) \otimes (J_n J_n)} = ||F|| \cdot ||J_n|| = n ||F||$

Basically, the idea is that once the dimension of $F$ is large enough to capture the important detail in the function, increasing the dimension is really just increasing the dimension of $J$.