Let $ \mathbf{M} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ and $\mathbf{X}_{n} = \begin{bmatrix}x_n \\ y_n \end{bmatrix}$. Solve $\displaystyle \mathbf{X}_{n+1} = \mathbf{MX}_n$ for $n \ge 0$ with the i.c. $x_0 =2, y_0 = 0$.
I'm not sure how to solve this. My idea is to write this as $\mathbf{X}_{n+k} =\mathbf{M}^k \mathbf{X}_n$ for $k \ge 0$. But it's not clear how I'd get the solution from that, after I've gone through the calculation?
On
Alt. hint (without matrix powers): the recurrence relations can be written as:
$$ \begin{align} \begin{cases} x_{n+1}=2 x_n + y_n \\ y_{n+1}=x_n + 2y_n \end{cases} \end{align} $$
Subtracting the two equalities gives $\,x_{n+1}-y_{n+1}=x_n-y_n\,$, which telescopes to:
$$x_n-y_n=x_{n-1}-y_{n-1}=\ldots=x_0-y_0=C$$
Thus $\,y_n=x_n-C\,$, and substituting back into the first relation gives the simple recurrence in $\,x_n\,$:
$$ x_{n+1}=2x_n+(x_n-C) \quad\iff\quad x_{n+1}=3x_n - C $$
If you diagonalize the matrix,
$$M=PDP^{-1}$$ and
$$M^n=PDP^{-1}PDP^{-1}\cdots PDP^{-1}=PD^nP^{-1}$$
so that the $n^{th }$ power of a diagonal matrix is the diagonal made of the $n^{th}$ powers.
Depending on the values of the Eigenvalues, the diagonal elements converge to zero or diverge to infinity (and $1$ remains and $-1$ alternates).
But in this exercise, all of this is useless as the initial conditions are null.