Finite Order Matrix and its Trace

Problems and solutions in Linear Algebra

Problem 28

Let $A$ be an $n\times n$ matrix and suppose that $A^r=I_n$ for some positive integer $r$. Then show that

(a) $|\tr(A)|\leq n$.

(b) If $|\tr(A)|=n$, then $A=\zeta I_n$ for an $r$-th root of unity $\zeta$.

(c) $\tr(A)=n$ if and only if $A=I_n$.

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Proof.

(a) $|\tr(A)|\leq n$.

We fist show that the eigenvalues of $A$ are $r$-th roots of unity.
Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{x}$ be an eigenvector corresponding to $\lambda$. That is, $A\mathbf{x}=\lambda \mathbf{x}$.

Multiplying this equality by $A$ on the left, we see that
\[A^2 \mathbf{x}=\lambda A\mathbf{x}=\lambda^2 \mathbf{x}.\] Again multiplying this equality by $A$ on the left we get $A^3\mathbf{x}=\lambda^3\mathbf{x}$.

Inductively, we obtain $A^r \mathbf{x}=\lambda^r \mathbf{x}$.
Since $A^r=I_n$, we get $\mathbf{x}=\lambda^r \mathbf{x}$, Thus $(\lambda^r-1)\mathbf{x}=\mathbf{0}$ and since $\mathbf{x}$ is nonzero vector (as it is an eigenvector), we have $\lambda^r=1$. Therefore eigenvalues are $r$-th roots of unity.


Next, we consider the Jordan canonical form of $A$. There exists an invertible matrix $P$ such that $P^{-1}AP$ is an upper triangular matrix $T=(t_{ij})$ whose diagonal entries are eigenvalues of $A$.
Since the trace of $A$ is equal to the trace of $P^{-1}AP$, we see that $\tr(A)$ is a sum of $n$ $r$-th roots of unity. Thus
\begin{align*}
|\tr(A)|=|\sum_{i=1}^{n}t_{ii}| \leq \sum_{i=1}^{n}|t_{ii}|=n. \tag{*}
\end{align*}

(b) If $|\tr(A)|=n$, then $A=\zeta I_n$ for an $r$-th root of unity $\zeta$

 We first refine the proof of part (a).
Note that the matrix $A$ is a solution of the equation $x^r-1$,

So the minimal polynomial of $A$ divides $x^r-1$, thus the minimal polynomial has no repeated roots. Thereby the Jordan canonical form $T$ is actually a diagonal matrix.

Now we start the proof of (b). The inequality in (*) becomes the equality if and only if all the roots of unity $t_{ii}$ are the same root of unity $\zeta$.
Then  the Jordan canonical form is $T=\zeta I_n$, hence $A=\zeta I_n$.

(c) $\tr(A)=n$ if and only if $A=I_n$

By part (b), we have $A=\zeta I_n$ for some $r$-th root of unity $\zeta$ but then $n=\tr(A)=\zeta n$ implies $\zeta=1$. Thus $A=I_n$ as required.


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  1. 08/01/2016

    […] a similar problem, also see Finite order matrix and its trace. This problem is kind of the converse of the current […]

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