Using diagonalization to find A^k

[1up20x6]

Homework Statement

$$A = \begin{pmatrix}<br /> 1 & 4\\<br /> 2 & -1<br /> \end{pmatrix}$$

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

Homework Equations

The Attempt at a Solution

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

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

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 $n$?