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.

Question: If $A=\left[\begin{array}{ccc} 1 & 0 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\end{array}\right]$ then show that $A^n=A^{n-2}+A^2-I$, $n\geq3$. Hence find $A^{50}$.

Where do I begin? I solved upto $A^8=4A^2-I$ and $A^9=4A^2+A-I$ with similar pattern emerging but I'm stuck where to go from here. And since I'm starting from the equation I was supposed to arrive at, I don't think this is the solution. But I don't see anything I can do with $A$ to prove that equation except maybe calculate $A^2$...

share|improve this question
1  
To show the recurrence relation you can use induction. –  user38268 Feb 15 '12 at 10:56
    
You refer to a "similar pattern emerging" but you need to see what that pattern actually is. Check what the even powers $A^{2n}$ are for $n=2,3,4,\dots$ in terms of $A^2$ and $I$. Can you conjecture a general formula? If you figure it out correctly, proving it with induction should be easy: use the formula on the power $A^{2n+2}$ and substitute in for $A^{2n}$ (we assume the hypothesis holds for evens up to $2n$). –  anon Feb 15 '12 at 11:04

1 Answer 1

up vote 2 down vote accepted

$1.~$ show for $n=3$ that :

$$A^3 =A^1+A^2-I \Rightarrow A(A^2-I)=A^2-I$$

$2.~$ suppose that :

$$A^n=A^{n-2}+A^2-I$$

EDIT :

As Brian correctly observed step $3$ should be :

$3.~$

$$\begin{align*} &A^{n+1}=A(A^{n-2}+A^2-I)=A^{n-1}+A(A^2-I)=A^{n-1}+A^2-I \end{align*}$$

which is true according to first induction step , so :

$$A^n=A^{n-2}+A^2-I\text{ for all }n \geq 3$$

Therefore :

$$\begin{align*} A^4&=2A^2-I\\ A^6&=3A^2-2I\\ A^8&=4A^2-3I\\ &\;\vdots\\ A^{50}&=25A^2-24I \end{align*}$$

share|improve this answer
2  
The argument in step (3) is invalid as stated: showing that $P$ implies a true statement does not show that $P$ is true. You need to argue that if $A^n=A^{n-2}+A^2-I$, then $A^{n+1}=A(A^{n-2}+A^2-I)=A^{n-1}+A(A^2-I)=A^{n-1}+A^2-I$ by step (1). –  Brian M. Scott Feb 15 '12 at 12:01

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.