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Show that no positive power of the matrix $\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$ equals $I_2$.

I claim that given $A^{n}, a_{11} = 1$ and $a_{12} >0, \forall n \in \mathbb{N}$. This is the case for $n=1$ since $A^{1} = \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$ with $1=1$ and $1>0$.

Now assuming that $a_{11} = 1$ and $a_{12}>0$ for $A^{n}$ show that $A^{n+1} = A^{n}A = \left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right)\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right) = \left( \begin{array}{cc} a_{11} & a_{11} + a_{12} \\ a_{21} & a_{21} + a_{22} \end{array} \right)$.

According to the assumption $a_{11} = 1$ and $a_{12}>0 \Rightarrow 1+a_{12} = a_{11}+a_{12}>0$. Taken together, this shows that $A^{n} \neq I_{2} \forall n\in \mathbb{N}$ since $a_{12}\neq0 = i_{12}$.

First of all, was my approach legitimate and done correctly? I suspect that I did not solve this problem as intended by the author (if at all!), could anyone explain the expected solution please? Thank you!

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Your solution seems perfectly fine to me. In fact, see if the same approach actually gives you the precise value of $A^n$. – Alex B. Apr 10 '11 at 9:54
up vote 11 down vote accepted

Your solution seems OK to me. You can also find $A^n$ explicitly: let $E=\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$. Then $A=I+E$ and $E^2=0$. So $(I+nE)(I+E)=I+(n+1)E$ and so, by induction, $A^n=I+nE\ne I$ for $n\ge1$.

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Not to appear pedantic, but maybe it's worth noting that this fact is true only assuming that you're working over a field (or a ring) of characteristic $0$ (I think you're implicitly assuming that your matrices have real entries).

Working over fields (or rings) of positive characteristic $n$, a power of $A$ does indeed become equal to the identity matrix $I_2$, namely $A^n=I_2$ (work as in mac's answer and remember that $n=0$ in your coefficient ring).

As a matter of fact, any matrix $A$ over a finite field $\Bbb F$ with non-zero determinant has the property that $A^N=I_2$ for some $N>0$. This follows from general basic properties of finite groups.

But even over a finite field there are matrices whose powers never equal $I_2$, just take a matrix $A$ such that $A^2=A$.

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