Using diagonalization to find A^k

[1up20x6]

Homework Statement

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

\begin{pmatrix}<br /> 2 &amp; -4\\<br /> -2 &amp; 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 &amp; -4\\<br /> -2 &amp; -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 &amp; λ_2<br /> \end{pmatrix}<br /> = \begin{pmatrix}<br /> 2 &amp; -1\\<br /> 1 &amp; 1<br /> \end{pmatrix}
P^{-1} = \begin{pmatrix}<br /> \frac{1}{3} &amp; \frac{1}{3}\\<br /> \frac{-1}{3} &amp; \frac{2}{3}<br /> \end{pmatrix}
D = \begin{pmatrix}<br /> λ_1 &amp; 0\\<br /> 0 &amp; λ_2<br /> \end{pmatrix}<br /> = \begin{pmatrix}<br /> 3 &amp; 0\\<br /> 0 &amp; -3<br /> \end{pmatrix}
PD^nP^{-1} = A^n<br /> = \begin{pmatrix}<br /> 2 &amp; -1\\<br /> 1 &amp; 1<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> 3^n &amp; 0\\<br /> 0 &amp; -3^n<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> \frac{1}{3} &amp; \frac{1}{3}\\<br /> \frac{-1}{3} &amp; \frac{2}{3}<br /> \end{pmatrix}
= \begin{pmatrix}<br /> 2(3^n) &amp; -(-3)^n\\<br /> 3^n &amp; (-3)^n<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> \frac{1}{3} &amp; \frac{1}{3}\\<br /> \frac{-1}{3} &amp; \frac{2}{3}<br /> \end{pmatrix}
= \begin{pmatrix}<br /> \frac{2}{3}(3^n) + \frac{1}{3}((-3)^n) &amp; \frac{2}{3}(3^n) - \frac{2}{3}((-3)^n)\\<br /> \frac{1}{3}(3^n) - \frac{1}{3}((-3)^n) &amp; \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 &amp; 4 \\ 2 &amp; -1\end{pmatrix}^2= \begin{pmatrix}9 &amp; 0 \\ 0 &amp; 9\end{pmatrix}= 9\begin{pmatrix}1 &amp; 0 \\ 0 &amp; 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?