Trace of elements in a finite complex matrix group is bounded

[jahlex]

Homework Statement

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

 

[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.

 

[pasmith]

Homework Statement

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

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?