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I'm learning linear algebra and need help with the following problem:

Let $A = \begin{pmatrix}-2 & -3 & -3\\-1 & 0 & -1\\5 & 5 & 6\end{pmatrix} \in M_{3x3}(\mathbb{R}).$ Find $A^n$ for all $n \in \mathbb{N}$.

My first thought was to compute $A^2$, $A^3$, $A^4$ and see if a pattern emerge. I used Mathematica to compute the power of $A$ to save me some time. The computations gave

$A^2 = \begin{pmatrix}-8 & -9 & -9\\-3 & -2 & -3\\15 & 15 & 16\end{pmatrix}, \quad A^3 = \begin{pmatrix}-20 & -21 & -21\\-7 & -6 & -7\\35 & 35 & 36\end{pmatrix}, \quad A^4 = \begin{pmatrix}-44 & -45 & -45\\-15 & -14 & -15\\75 & 75 & 76\end{pmatrix}$.

The above results are clearly showing a pattern relating the first four powers of $A$, for instance $(a_{31}) = (a_{32}) = (a_{33}) + 1$, but I'm unable to write the general form of a matrix $A^{k}$.

$(Q1)$ How do I find/write the general form of a matrix $A^{k}$?

$(Q2)$ I only compute the first four powers of $A$. How do I know that this pattern won't fail for some integer $k$?

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    Have you tried diagonalising $A$? – 2017-02-01
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    @Arthur I did not considered diagonalising $A$. Thanks, I can see how this is useful. If A is diagonalisable, then I can write the general form for $A^k$ as $A^k = PD^{k}P^{-1}$. Is that the correct idea? – 2017-02-01

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The matrix is similar to the diagonal matrix $D=diag(1,1,2)$. Hence we have $$ A^n=(SDS^{-1})^n=SD^nS^{-1} $$ with $$ S=\begin{pmatrix} -5 & -4 & 3\cr -1 & -1 & 1\cr 6 & 5 & -5\end{pmatrix}. $$

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The method descirbed in this answer still works even when $A$ is not diagonalizable. The upside of this method is that you are not required to calculate eigenvectors or inverses of matrices.

The characteristic polynomial is $P_A(X)=(2-X)(X-1)^2$. The Cayley-Hamilton theorem says that $P_A(A)=(2Id-A)(A-Id)^2=0$. Consider the polynomial $X^n$. We can write $$X^n=Q(X)P_A(X)+r(X)$$ with $\deg(r(X))<3$. Hence we may write $r(X)=aX^2+bX+c$.

We can plug in the eigenvalue $2$ into the previous equation to get $2^n=r(2)=4a+2b+c$. (Here we used that $P_A(2)=0$).

Similarly $1=r(1)=a+b+c$. Last but not least we take the derivative w.r.t. $X$. This yields $$nX^{n-1}=Q'(X)P_A(X)+Q(X)P_A'(X)+2aX+bX.$$ Plug in $1$ to get $n=2a+b$. You can solve for $a,b$ and $c$. Then you know $r(X)$. Plug in $A$ in the original equation to get $$A^n=r(A)=aA^2+bA+cId.$$

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Your matrix has the following form for any power $k$ (showing this should be routine): $$\begin{array}{ccc}-3a+1&-3a&-3a\\-a&-a+1&-a\\5a&5a&5a+1\end{array}$$ where $a=1$ for $A^{1}$. Calculating $A^{k}$ for first few $k$ shows that the $[2,1]$ element is $-(2^k-1)$.