Using diagonalization to find A^k

[1up20x6]

Homework Statement

$$A = \begin{pmatrix} 1 & 4\\ 2 & -1 \end{pmatrix}$$

Find $A^n$ and $A^{-n}$ where n is a positive integer.

Homework Equations

The Attempt at a Solution

$$(xI - A) = \begin{pmatrix} x-1 & -4\\ -2 & x+1 \end{pmatrix}$$
$$det(xI - A) = (x-3)(x+3)$$
$$λ_1 = 3\quad λ_2 = -3$$

$$\begin{pmatrix} 2 & -4\\ -2 & 4\end{pmatrix} \begin{pmatrix} a\\ b \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}\quad \begin{pmatrix} a\\ b \end{pmatrix} = \begin{pmatrix} 2\\ 1 \end{pmatrix}$$
$$\begin{pmatrix} -4 & -4\\ -2 & -2\end{pmatrix} \begin{pmatrix} c\\ d \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}\quad \begin{pmatrix} c\\ d \end{pmatrix} = \begin{pmatrix} -1\\ 1 \end{pmatrix}$$
$$P = \begin{pmatrix} λ_1 & λ_2 \end{pmatrix} = \begin{pmatrix} 2 & -1\\ 1 & 1 \end{pmatrix}$$
$$P^{-1} = \begin{pmatrix} \frac{1}{3} & \frac{1}{3}\\ \frac{-1}{3} & \frac{2}{3} \end{pmatrix}$$
$$D = \begin{pmatrix} λ_1 & 0\\ 0 & λ_2 \end{pmatrix} = \begin{pmatrix} 3 & 0\\ 0 & -3 \end{pmatrix}$$
$$PD^nP^{-1} = A^n = \begin{pmatrix} 2 & -1\\ 1 & 1 \end{pmatrix} \begin{pmatrix} 3^n & 0\\ 0 & -3^n \end{pmatrix} \begin{pmatrix} \frac{1}{3} & \frac{1}{3}\\ \frac{-1}{3} & \frac{2}{3} \end{pmatrix}$$
$$= \begin{pmatrix} 2(3^n) & -(-3)^n\\ 3^n & (-3)^n \end{pmatrix} \begin{pmatrix} \frac{1}{3} & \frac{1}{3}\\ \frac{-1}{3} & \frac{2}{3} \end{pmatrix}$$
$$= \begin{pmatrix} \frac{2}{3}(3^n) + \frac{1}{3}((-3)^n) & \frac{2}{3}(3^n) - \frac{2}{3}((-3)^n)\\ \frac{1}{3}(3^n) - \frac{1}{3}((-3)^n) & \frac{1}{3}(3^n) + \frac{2}{3}((-3)^n) \end{pmatrix}$$

I think that everything I've done so far is correctly, but I can't find any way to simplify this equation any further, and I don't think that I could find $A^{-n}$ with an equation this complicated, so I must be missing something.

 

[lurflurf]

There are different ways of writing that out, you might not have pick the most simple. Another way is something like

A^n=(3^n)(1/2)((1+(-1)^n)I+(1/3)(1-(-1)^n)A)

which you can see is lot like yours, note all that (-1)^n stuff is just to unify the even and odd terms

even A^n=(3^n)I

odd A^n=(3^n)(1/3)A

I=A^0 the 2x2 identity matrix

 

[HallsofIvy]

Frankly, the first thing I would have done would be to note that
$$A^2= \begin{pmatrix}1 & 4 \\ 2 & -1\end{pmatrix}^2= \begin{pmatrix}9 & 0 \\ 0 & 9\end{pmatrix}= 9\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$$

From that it follows immediately that if n is even, $A^n= 3^nI$ and if n is odd, $A^n= 3^{n-1}A$

 

[1up20x6]

Thanks for your responses. Are those equations self-evident or is there a proof that they apply for all values of $n$?