Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
5  
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

2 Answers 2

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 $n\ge1$.

share|improve this answer

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 $A^2=A$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.