# Prove that for any positive integer $n$, $A^n ≠ I$.

Let $A$ be a $2\times 2$ matrix with $tr(A) > 2$. Prove that for any positive integer $n$, $A^n ≠ I$.

I feel like I should approach this with respect to eigenvalues, i.e. the sum of the eigenvalues of $A$ is greater than $2$. However, I don't know where to head from here. Any help or guidance to a direction would be greatly appreciated!

First note that if $\lambda$ is an eigenvalue of $A$, then $\lambda^n$ is an eigenvalue of $A^n$. As you pointed out, the trace being greater than $2$ implies that the sum of the eigenvalues is greater than $2$, so at least one eigenvalue $\lambda$ of $A$ satisfies $|\lambda|>1$. Thus $A^n$ has an eigenvalue $\lambda^n$, and $|\lambda^n|=|\lambda|^n>1$. This implies that $A^n\neq I$.
• Ah of course. Should be fixed if I replace $\lambda>1$ with $|\lambda|>1$, correct? May 20, 2015 at 3:36
• No problem! At first I forgot that the matrix need not have any real eigenvalues. Thus if $\lambda$ is complex, the expression $\lambda>1$ makes no sense. What we really care about is just that it has a strictly larger magnitude than $1$, so that $\lambda^n$ has a strictly larger magnitude than $1$. May 20, 2015 at 3:54
Recall the characteristic equation $$A^2-tr(A)A+I_2\det A=O_2.$$ Assume by contradiction that, for some $n\ge1$, $A^n=I$. Then $\det A=\pm1$, $A$ is invertible with inverse $A^{n-1}$, and $n\neq1,2$ because $tr(A)>2$. So $n\ge3$.
Multiply the characteristic equation by $A^{n-2}$ to get $I-tr(A)A^{n-1}+\det A\,A^{n-2}=O_2$, $A^{n-1}(tr(A)I-\det A\,A)=I$, that is, $tr(A)I-\det A\,A=A$ from which you can get several contradictions.