Given a matrix $A$, find $A^n$

Question

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:

$$\det(\lambda I_n-A)=0$$ $$\det\left( \lambda \left[{1\atop0}{0\atop1}\right] -\left[{9\atop20}{(-4)\atop(-9)}\right] \right)=0$$ $$\det\left( \left[{\lambda\atop0}{0\atop\lambda}\right] -\left[{9\atop20}{(-4)\atop(-9)}\right] \right)=0$$ $$\det\left( \left[{(\lambda-9)\atop(-20)}{4\atop(\lambda+9)}\right] \right)=0$$ $$(\lambda-9)(\lambda+9) -(20)(-4)=0$$ $$(\lambda^2-9\lambda+9\lambda-81) +80=0$$ $$\lambda^2-1=0$$ $$\lambda^2=1$$ $$\lambda=\pm1$$

Answer 1

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 $A^{54}=I,A^7=A^6\times A=I\times A=A$

Answer 2

Hint: Sometimes diagonalization, but sometimes even easier.

$$A^2=I$$

Answer 3

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 $= I A^{odd}$ $^{power} = A$

Answer 4

$$A^2=I$$ $$A^3=A^2A=IA=A$$ $$(A^{even}=I)$$ $$(A^{odd}=A)$$

---

Diagonalize the matrix $A$ $$A=PDP^{-1}$$ $$A^n=(PDP^{-1})^n = PD^nP^{-1}$$ $$D=\left[{\lambda_1\atop\lambda_2}\right]\left[{1\atop0}{0\atop1}\right]=\left[{\lambda_1\atop0}{0\atop\lambda_2}\right]=\left[{1\atop0}{0\atop-1}\right]$$ $$P=\left[v_1 v_2\right]$$ $$(A-\lambda_i I)v_i={0}$$

$$-----------------------$$

$$\left(\left[{9\atop20}{(-4)\atop(-9)}\right]-(1)\left[{1\atop0}{0\atop1}\right]\right)\left({v_1x_1\atop v_1x_2}\right)=0$$

$$\left[{8\atop20}{-4\atop-10}\right]\left({v_1x_1\atop v_1x_2}\right)\rightarrow {8v_1x_1-4v_1x_2=0\atop 20v_1x_1-10v_1x_2=0} \rightarrow {v_1x_1=1\atop v_1x_2=2}\rightarrow v_1=\left[{1\atop2}\right]$$

$$--$$

$$\left(A-(-1)I \right)\left({v_2}\right)=0 \rightarrow v_2=\left[{1\atop5/2}\right]$$

$$--$$

$$P=\left[{1\atop 2}{1\atop (5/2)}\right]$$ $$P^{-1}={-1\over det(A)}\left[{\searrow\atop-}{-\atop\nwarrow}\right]=\left[{-10\atop8}{4\atop-4}\right]$$

$$-----------------------$$

$$A^{2008}= PD^{2008}P^{-1}$$

$$A^{2008}=\left[{1\atop 2}{1\atop (5/2)}\right] \left[{1\atop0}{0\atop-1}\right]^{2008} \left[{-10\atop8}{4\atop-4}\right] = \left[{1\atop0}{0\atop1}\right]$$

Answer 5

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

$$A^n = P(A)Q(A)+R(A)=R(A).$$

Answer 6

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{A = \pars{\begin{array}{rr}9 & -4\\ 20 & -9\end{array}}}$

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

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