Using diagonalization to find A^k

[1up20x6]

Homework Statement

[tex]A = \begin{pmatrix}
1 & 4\\
2 & -1
\end{pmatrix}[/tex]

Find [itex]A^n[/itex] and [itex]A^{-n}[/itex] where n is a positive integer.

Homework Equations

The Attempt at a Solution

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

[tex]\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}[/tex]
[tex]\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}[/tex]
[tex]P = \begin{pmatrix}
λ_1 & λ_2
\end{pmatrix}
= \begin{pmatrix}
2 & -1\\
1 & 1
\end{pmatrix}[/tex]
[tex]P^{-1} = \begin{pmatrix}
\frac{1}{3} & \frac{1}{3}\\
\frac{-1}{3} & \frac{2}{3}
\end{pmatrix}[/tex]
[tex]D = \begin{pmatrix}
λ_1 & 0\\
0 & λ_2
\end{pmatrix}
= \begin{pmatrix}
3 & 0\\
0 & -3
\end{pmatrix}[/tex]
[tex]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}[/tex]
[tex]= \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}[/tex]
[tex]= \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}[/tex]

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 [itex]A^{-n}[/itex] 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
[tex]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}[/tex]

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

 

[1up20x6]

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