Linear Algebra (Matrixes with powers)

Question

Let $$ M \colon= \left[ \begin{matrix} 2 & -1 \\ 2 & 5 \end{matrix} \right] $$

Find formulas for the entries of $M^n$, where $n$ is a positive integer.

$$ M^n = \text{?} $$

Answer 1

Look for the Eigenvalues and eigenvectors:

$$\det(M-\lambda I)=0$$

Then the eigenvectors:

$$M\mathbf{x}=\lambda\mathbf{x}$$

Put the eigenvectors as columns in a $2\times 2$ matrix $P$, and compute $P^{\textrm{T}}MP$. Note that $P^{\textrm{T}}P=PP^{\textrm{T}}=I$

Answer 2

I think simple diagonilization makes sense here for $M$ has two different eigenvalues, $\{3,4\}$. Two corresponding eigenvectors are $[1,-1]^\top$ and $[1,-2]^\top$. Setting \begin{align} P=\begin{pmatrix} 1 & 1\\ -1 & -2 \end{pmatrix} \end{align} can be expressed $M$ as $M= P\Delta P^{-1}$ where $\Delta=\operatorname{diag}(3,4)$, that is \begin{align} \begin{pmatrix} 2 & -1\\ 2 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 1\\ -1 & -2 \end{pmatrix}\begin{pmatrix} 3 & 0\\ 0 & 4 \end{pmatrix}\begin{pmatrix} 2 & 1\\ -1 & -1 \end{pmatrix} \end{align}

Thus \begin{align} M^n = \begin{pmatrix} 1 & 1\\ -1 & -2 \end{pmatrix}\begin{pmatrix} 3^n & 0\\ 0 & 4^n \end{pmatrix}\begin{pmatrix} 2 & 1\\ -1 & -1 \end{pmatrix} = \begin{pmatrix} 2\cdot3^n - 4^n & 3^n - 4^n\\ 2(4^n-3^n) & -3^n - 2\cdot4^n \end{pmatrix} \end{align}

This is of course the same as what was mentioned in one of the comments above (egorovik)