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:

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

Solution 1:

Hint: Sometimes diagonalization, but sometimes even easier.

$$A^2=I$$

Solution 2:

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

Solution 3:

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

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

Solution 4:

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