Given a matrix $A$, find $A^n$ Given the matrix $$A = \left[{9\atop20}{-4\atop-9}\right]$$ how do I find $A^7$ or $A^{54}$ or $A^{2008}$ (etc.) ?
I know I need the eigenvalues of A, but I'm not sure what to do afterwards.


*

*Is the answer a $2 \times 2$ matrix?

*What role do the eigenvalues play in solving the problem?

*Whats the set up for the solution?

*Can this process be used on larger square matrices? ($A_{3\times3}$, or $A_{4\times4}$, or $A_{n\times n}$)?



Eigenvalues:
$$\det(\lambda I_n-A)=0$$
$$\det\left( \lambda \left[{1\atop0}{0\atop1}\right] -\left[{9\atop20}{(-4)\atop(-9)}\right] \right)=0$$
$$\det\left( \left[{\lambda\atop0}{0\atop\lambda}\right] -\left[{9\atop20}{(-4)\atop(-9)}\right] \right)=0$$
$$\det\left( \left[{(\lambda-9)\atop(-20)}{4\atop(\lambda+9)}\right] \right)=0$$
$$(\lambda-9)(\lambda+9) -(20)(-4)=0$$
$$(\lambda^2-9\lambda+9\lambda-81) +80=0$$
$$\lambda^2-1=0$$
$$\lambda^2=1$$
$$\lambda=\pm1$$
 A: Diagonalize your matrix $A$
$$A = P^{-1}DP$$
$$A^{n} = P^{-1}D^{n}P$$
$D^n$ is easily solvable in the form $$D^{n} = \begin{bmatrix} \lambda1^{n} & 0 \\ 0 & \lambda2^{n} \end{bmatrix}$$
Multiply it $P^{-(1)}D^nP$ out and you have your answer.
In the case of the matrix that you gave $A$ to an even power $= I A^{odd}$ $^{power} = A$
A: Diagonalize a matrix is difficult (since finding the roots of a polynomial is really difficult in general. In particular if the degree of your polynomial is greater or equal to $5$, there is no general formula to get the roots of your polynomial). But you can always find a polynomial $P$ such that $P(A)=0$ (if $A$ is your matrix), for example the characteristic polynomial. Then to evaluate $A^n$ you simply write the division of $X^n$ by $P$
$$ X^n = P(X)Q(X)+R(X) $$
with the degree of $R$ < degree of $P$. And then
$$A^n = P(A)Q(A)+R(A)=R(A).$$ 
A: Hint: Sometimes diagonalization, but sometimes even easier.
$$A^2=I$$
A: Hint Ingeneral Diagonalize the matrix $A$
$D=QAQ^{-1}$ for some invertible $Q$ and $D=diag\{1,-1\}$
now $D^n=QA^nQ^{-1}$
so $A^7=Q^{-1}D^7Q$
Now can you find the matrix $Q$?
but as in your case we see $A^2=I$ so $A^{54}=I,A^7=A^6\times A=I\times A=A$ 
A: $$A^2=I$$
$$A^3=A^2A=IA=A$$
$$(A^{even}=I)$$
$$(A^{odd}=A)$$ 

Diagonalize the matrix $A$
$$A=PDP^{-1}$$
$$A^n=(PDP^{-1})^n = PD^nP^{-1}$$
$$D=\left[{\lambda_1\atop\lambda_2}\right]\left[{1\atop0}{0\atop1}\right]=\left[{\lambda_1\atop0}{0\atop\lambda_2}\right]=\left[{1\atop0}{0\atop-1}\right]$$
$$P=\left[v_1 v_2\right]$$
$$(A-\lambda_i I)v_i={0}$$
$$-----------------------$$
$$\left(\left[{9\atop20}{(-4)\atop(-9)}\right]-(1)\left[{1\atop0}{0\atop1}\right]\right)\left({v_1x_1\atop v_1x_2}\right)=0$$
$$\left[{8\atop20}{-4\atop-10}\right]\left({v_1x_1\atop v_1x_2}\right)\rightarrow {8v_1x_1-4v_1x_2=0\atop 20v_1x_1-10v_1x_2=0} \rightarrow {v_1x_1=1\atop v_1x_2=2}\rightarrow v_1=\left[{1\atop2}\right]$$
$$--$$
$$\left(A-(-1)I
\right)\left({v_2}\right)=0 \rightarrow v_2=\left[{1\atop5/2}\right]$$
$$--$$
$$P=\left[{1\atop 2}{1\atop (5/2)}\right]$$
$$P^{-1}={-1\over det(A)}\left[{\searrow\atop-}{-\atop\nwarrow}\right]=\left[{-10\atop8}{4\atop-4}\right]$$
$$-----------------------$$
$$A^{2008}= PD^{2008}P^{-1}$$
$$A^{2008}=\left[{1\atop 2}{1\atop (5/2)}\right] \left[{1\atop0}{0\atop-1}\right]^{2008}  \left[{-10\atop8}{4\atop-4}\right] = \left[{1\atop0}{0\atop1}\right]$$
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{A = \pars{\begin{array}{rr}9 & -4\\ 20 & -9\end{array}}}$

Also,
$$
A^{2} = \pars{\begin{array}{rr}9 & -4\\ 20 & -9\end{array}}
\pars{\begin{array}{rr}9 & -4\\ 20 & -9\end{array}}
=
\pars{\begin{array}{rr}1 & 0\\ 0 & 1\end{array}} = {\cal I}\,,\quad A^{3} = A\,,\quad A^{4} = {\cal I}\,,\quad\mbox{etc}\ldots
$$

$$\color{#0000ff}{\large%
A^{n}
=\left\lbrace%
\begin{array}{lcl}
\pars{\begin{array}{rr}9 & -4\\ 20 & -9\end{array}} & \mbox{if} & n\ \mbox{is odd}
\\[3mm]
\pars{\begin{array}{rr}1 & 0\\ 0 & 1\end{array}} & \mbox{if} & n\ \mbox{is even}
\end{array}\right.}
$$
