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The matrix $\mathbf{M}$ is given by 
\[
\mathbf{M}=\begin{pmatrix}\cos(2\pi/m) & -\sin(2\pi/m)\\
\sin(2\pi/m) & \cos(2\pi/m)
\end{pmatrix},
\]
where $m$ is an integer greater than $1.$ Prove that 
\[
\mathbf{M}^{m-1}+\mathbf{M}^{m-2}+\cdots+\mathbf{M}^{2}+\mathbf{M}+\mathbf{I}=\mathbf{O},
\]
where $\mathbf{I}=\begin{pmatrix}1 & 0\\
0 & 1
\end{pmatrix}$ and $\mathbf{O}=\begin{pmatrix}0 & 0\\
0 & 0
\end{pmatrix}.$

The sequence $\mathbf{X}_{0},\mathbf{X}_{1},\mathbf{X}_{2},\ldots$
is defined by 
\[
\mathbf{X}_{k+1}=\mathbf{PX}_{k}+\mathbf{Q},
\]
where $\mathbf{P,Q}$ and $\mathbf{X}_{0}$ are given $2\times2$
matrices. Suggest a suitable expression for $\mathbf{X}_{k}$ in terms
of $\mathbf{P},$ $\mathbf{Q}$ and $\mathbf{X}_{0},$ and justify
it by induction.

The binary operation $*$ is defined as follows: 
\[
\mathbf{X}_{i}*\mathbf{X}_{j}\mbox{ is the result of substituting \ensuremath{\mathbf{X}_{j}}for \ensuremath{\mathbf{X}_{0}}in the expression for \ensuremath{\mathbf{X}_{i}}. }
\]
Show that if $\mathbf{P=M},$ the set $\{\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\ldots\}$
forms a finite group under the operation $*$.

Solution (Optional)

$\mathbf{M}^m = \mathbf{I}$, we also have $\mathrm{det}(\mathbf{M - I}) = \cos^2(2\pi/m) - 2\cos(2\pi/m) + 1 + \sin^2(2\pi/m) = 2(1-\cos(2\pi/m))$ therefore $\mathbf{M-I}$ is invertible.

Therefore since $\mathbf{(M-I)(M^{m-1} + M^{m-1} + \cdots + M^2 + M + I)= M^m-I = 0}$ we can cancel the $\mathbf{M-I}$ to obtain the desired result.

$\mathbf{X_0 = X_0}$
$\mathbf{X_1 = PX_0+Q}$
$\mathbf{X_2 = P(PX_0+Q)+Q = P^2X_0 + PQ + Q}$

Claim: $\mathbf{X_k = P^k X_0 + (P^{k-1} + P^{k-2} + \cdots + I)Q}$

Proof: (By induction on $k$). Base case $k = 0$ is true.

Assume it's true for some $k = l$, then consider $k = l+1$

$\mathbf{X_{l+1} = PX_l + Q = P( P^l X_0 + (P^{l-1} + P^{l-2} + \cdots + I)Q) + Q = P^{l+1}X_0 + (P^l + P^{l-1} + \cdots + P)Q + Q =  P^{l+1}X_0 + (P^l + P^{l-1} + \cdots + P + I)Q}$

Suppose $\mathbf{P} = \mathbf{M}$, then consider the set $\{\mathbf{X_1, X_2}, \ldots\}$ with the operation $*$ as defined.

$\mathbf{X_i * X_j} =  M^{i}(X_j) + (M^{i-1} + M^{i-2} + \cdots + M + I)Q = M^{i}(M^jX_0 + (M^{j-1} + M^{j-2} + \cdots + M + I)Q) + (M^{i-1} + M^{i-2} + \cdots + M + I)Q = M^{i+j}X_0 + (M^{i+j-1}+\cdots + M + I)Q = X_{i+j}$

Since $X_m = X_0$ we can check all the requirements of the group, but this is going to be isomorphic to the cyclic group with $m$ elements.

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