# Trace of elements in a finite complex matrix group is bounded

1. Dec 12, 2013

### jahlex

1. The problem statement, all variables and given/known data

Let $G$ be a finite complex matrix group: $G \subset M_{n\times n}$. Show that, for $g \in G, |\text{tr}(g)| \le n$ and $|\text{tr}(g)| = n$ only for $g = e^{i\theta}I$.

2. The attempt at a solution

Since $G$ is finite, then every element $g \in G$ has a finite order: $g^r = I$ for some whole number $r$. By the formula for traces, $\text{tr}(g) = \displaystyle\sum_{i=1}^n \lambda_i$ and $\text{tr}(g^r) = \displaystyle\sum_{i=1}^n \lambda_i^r = n$ where $\lambda_i$ are eigenvalues of $g$. So how do I show that $|\displaystyle\sum_{i=1}^n \lambda_i| \le \displaystyle\sum_{i=1}^n \lambda_i^r$ for complex $\lambda_i$ ?

The problem comes from Exercise A2.11 on page 612 of Nielsen and Chuang's Quantum Computation and Quantum Information 10th Anniversary Edition. The textbook can easily be found, for example, here www.johnboccio.com/research/quantum/notes/QC10th.pdf [Broken]

Last edited by a moderator: May 6, 2017
2. Dec 12, 2013

### brmath

Suppose $\lambda_1 = \lambda_2 = 1/2$. Then it is not true that |1/2 + 1/2| < |1/4 + 1/4|. So in general that inequality just isn't true.

There must be something about G being finite that rules out the above case.

3. Dec 13, 2013

### pasmith

If you can show that $g \in G \subset GL(\mathbb{C},n)$ is diagonalizable, then the fact that $g^r = I$ requires that the eigenvalues of $g$ lie on the unit circle. You then have $$\left| \sum_{i = 1}^n \lambda_i \right| \leq \sum_{i= 1}^n |\lambda_i| = \sum_{i = 1}^n 1 = n$$
where the first inequality is a basic result.

To show that $g$ is diagonalizable, consider the Jordan normal form of $g$. Why does the requirement that $g$ have finite order mean that its normal form cannot contain non-diagonal Jordan blocks?

Last edited by a moderator: May 6, 2017