# Given a particular matrix $A \in \operatorname{GL}_n(F)$, what is an easy way to determine $A^n$ $\forall n \in \Bbb Z$?

$$\newcommand{\GL}{\operatorname{GL}} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\m}{\left( \begin{matrix} #1 \end{matrix} \right)} \newcommand{\ds}{\displaystyle}$$ Original Problem: Consider the matrix $$A \in \GL_3(\R)$$ given by, for $$x > 0$$,

$$A = \m{-1 & x & -x \\ 1 & -1 & 0 \\ 1 & 0 & -1 }$$

What is the general form of $$A^n$$ for any $$n \in \Z$$?

My Work So Far: By doing some particular cases by hand and doing some pattern matching, I was able to observe that

$$A^n= (-1)^n \m{ 1 & -nx & nx \\ -n &\ds \frac{n(n-1)}{2} x + 1 &\ds -\frac{n(n-1)}{2} x \\ -n & \ds \frac{n(n-1)}{2} x &\ds 1 - \frac{n(n-1)}{2} x }$$

Proving this through a sort of "bidirectional induction" (i.e. show it with $$A$$ as the base case for $$n \in \Z_{>0}$$, then with $$A^{-1}$$ as the base case for $$\Z_{<0}$$, and then $$A^0 = I_3$$ trivially) seems more than doable, if a little tedious.

However, I don't think that this is the way it was intended to be solved. This comes as a homework problem and was tied to some others I was given. Namely:

• Find the characteristic polynomial $$\mu_A$$ of $$A$$
• Deduce $$A$$ is not diagonalizable
• Find vectors $$U_i \in \R^3$$ such that $$AU_1 = -U_1 \qquad AU_2 = U_1 - U_2 \qquad AU_3 = U_1 + U_2 - U_3$$
• Show $$A$$ is similar to the matrix $$B = \m{ -1 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & -1}$$ (I did so by showing $$\mu_A = \mu_B$$.)

I've done all of these, so I don't really need help there. It would not be unusual for these to tie into the problem somehow, based on how this professor's homework tends to go, but I'm not sure where the connection lies.

So I'm left wondering: based on this information, what is some more elegant way to calculate $$A^n$$ or prove its general form as given, if any? Be it for this specific matrix tied to the information given, or a more general case of $$A \in \GL_n(F)$$.

• You have $\mu_A(A)=A^3+a_1A^2+a_2A+a_0I=0$, so $A^3=-(a_1A^2+a_2A+a_0I)$ and higher powers can be expressed as a linear combination of $I,A,A^2$. Does this help? Also, you have $A^n=PB^nP^{-1}$. – Shivering Soldier Nov 8 '20 at 5:54
• Ah, I see. I haven't tried that myself out yet (focusing on other homework at the moment since I have plenty of time), but I'll be sure to try that out later, it looks promising. Thanks for the insight! – Eevee Trainer Nov 8 '20 at 7:30

First find the Jordan matrix of $$A$$. The eigenvalues of $$A$$ are found from the characteristic equation $$0=\det\begin{pmatrix}-1-\lambda & x & -x \\ 1 & -1-\lambda & 0 \\ 1 & 0 & -1-\lambda \end{pmatrix}=-(1+\lambda)^3$$ to be $$\lambda=-1$$ thrice. There is only one associated eigenvector $$(0,1,1)$$.
A generalized eigenvector is found from $$(A+I)^2v=0$$, $$\begin{pmatrix}0&0&0\\0&x&-x\\0&x&-x\end{pmatrix}v=0\Longleftarrow (A+I)v=\begin{pmatrix}0\\1\\1\end{pmatrix}\Longleftarrow v=\begin{pmatrix}1\\0\\0\end{pmatrix}$$ The third, a solution of $$(A+I)^3v=0$$, satisfies $$(A+I)v=\begin{pmatrix}1\\0\\0\end{pmatrix}$$, namely $$\begin{pmatrix}0\\1/x\\0\end{pmatrix}$$. So letting $$P=\begin{pmatrix}0&1&0\\1&0&1/x\\1&0&0\end{pmatrix}$$ gives $$A=PJP^{-1}=P\begin{pmatrix}-1&1&0\\0&-1&1\\0&0&-1\end{pmatrix}P^{-1}$$
Hence $$A^n=PJ^nP^{-1}$$ where $$J^n=(-1)^n\begin{pmatrix}1&(-1)^nn&\binom{n}{2}\\0&1&(-1)^nn\\0&0&1\end{pmatrix}$$ Multiplying out $$PJ^nP^{-1}$$ should give your answer.