<s>[INST] Compute $A^{2013}$. 
QUESTION 6
Let $$A=\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}-1&0&0\\0&\frac12&-\frac{\sqrt3}2\\0&\frac{\sqrt3}2&\frac12\end{bmatrix}\cdot\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}^{-1}$$
Compute $A^{2013}$ (10 marks)


Solution First, observe that
$$A^{2013}=\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}-1&0&0\\0&\frac12&-\frac{\sqrt3}2\\0&\frac{\sqrt3}2&\frac12\end{bmatrix}^{2013}\cdot\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}^{-1}$$
Further, let $T:\mathbb R^3\to\mathbb R^3$ be the linear operator of the left multiplication with the matrix $$\begin{bmatrix}-1&0&0\\0&\frac12&-\frac{\sqrt3}2\\0&\frac{\sqrt3}2&\frac12\end{bmatrix}$$
Notice that $T$ acts as follows: it sends $x_1$ to $-x_1$ and rotates the plane $(x_2,x_3)$ by the angle $\pi/3$ in the counter-clockwise direction. It means that $T^6$ is the identity transformation and hence $T^{2010}$ is also the identity (because $2010$ is a multiple of $6$). Thus $T^{2013}=T^3$, which, according to its geometric meaning, does the following: sends $x_1$ to $-x_1$ and rotates the plane $(x_2,x_3)$ by the angle $\pi$. The rotation by the angle $\pi$ is minus identity and hence the whole transformation $T^3$ is minus identity. Thus, $$A^{2013}=\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}-1&0&0\\0&\frac12&-\frac{\sqrt3}2\\0&\frac{\sqrt3}2&\frac12\end{bmatrix}^{2013}\cdot\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}^{-1}\\=\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}\cdot\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}^{-1}=\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}$$

I don't really understand the part where the solution talks about the way $T$ acts (second half of the solution). Can anyone explain to me why is that so?
 [/INST] Notice that we can write $A$ on this form
$$A=P\operatorname{diag}(-1,R(\pi/3))P^{-1}$$
where
$$R(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$
is the matrix of rotation of angle $\theta$. Now using the bloc multiplication we get
$$A^{2013}=P\operatorname{diag}(-1,R^{2013}(\pi/3))P^{-1}=P\operatorname{diag}(-1,R(2013\times\pi/3))P^{-1}\\=P\operatorname{diag}(-1,R(\pi))P^{-1}=P\operatorname{diag}(-1,-1,-1)P^{-1}$$
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