Product of Matrices I

Question

Given the matrix \begin{align} A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right) \end{align} consider the first few powers of $A^{n}$ for which \begin{align} A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right) \hspace{15mm} A^{2} = \left( \begin{matrix} 7 & 6 \\ 9 & 10 \end{matrix} \right) \hspace{15mm} A^{3} = \left( \begin{matrix} 25 & 26 \\ 39 & 38 \end{matrix} \right). \end{align} Notice that the first rows have the values $(1,2)$, $(7,6)$, $(25,26)$ of which the first and second elements are rise and fall in a cyclical pattern. The same applies to the bottom rows.

1. Is there an explanation as to why the numbers rise and fall in order of compared columns?
2. What is the general form of the $A^{n}$ ?

Answer 1

It is often possible to diagonalize a square matrix, that is find a diagonal matrix $D$ and another matrix $P$ such that $$A=PDP^{-1}.$$ This is a nice property because $$A^n=PDP^{-1}PDP^{-1}...PDP^{-1}=PD^nP^{-1}.$$ The $P$ and $P^{-1}$ matrices cancel except on the ends, so we only have to take a power of the diagonal matrix, which is easy. If $$D=\left(\begin{array}{cc} a & 0\\ 0 & b\\ \end{array}\right),$$ then $$D^n=\left(\begin{array}{cc} a^n & 0\\ 0 & b^n\\ \end{array}\right).$$ In fact, it is possible to diagonalize your matrix as $$A=\left(\begin{array}{cc} 2 & -1\\ 3 & 1\\ \end{array}\right) \left(\begin{array}{cc} 4 & 0\\ 0 & -1\\ \end{array}\right) \left(\begin{array}{cc} 1/5 & 1/5\\ -3/5 & 2/5\\ \end{array}\right).$$ Thus, $$A^n=\left(\begin{array}{cc} 2 & -1\\ 3 & 1\\ \end{array}\right) \left(\begin{array}{cc} 4^n & 0\\ 0 & (-1)^n\\ \end{array}\right) \left(\begin{array}{cc} 1/5 & 1/5\\ -3/5 & 2/5\\ \end{array}\right)=\left( \begin{array}{cc} \frac{3 (-1)^n}{5}+\frac{1}{5} 2^{2 n+1} & \frac{1}{5} (-2) (-1)^n+\frac{1}{5} 2^{2 n+1} \\ \frac{1}{5} (-3) (-1)^n+\frac{3\ 4^n}{5} & \frac{2 (-1)^n}{5}+\frac{3\ 4^n}{5} \\ \end{array} \right).$$ As for the top row business, you can see that each term in the top row has two terms. The second of each is the same. In $n$ is odd, the first terms are $-3/5$ and $2/5$ respectively. If $n$ is even, the two terms are $3/5 $ and $-2/5$. Can you see why this relationship causes the behavior you noticed?

Answer 2

In answer to part 2, the general form is $$M^n=\frac{1}{5}\begin{pmatrix} 2\times 4^n+3 \times(-1)^n & 2\times 4^n-2(-1)^n\\3\times 4^n+3(-1)^{n+1} & 3\times 4^n+2(-1)^{n}\\\end{pmatrix}$$

But you know this already because it has already been posted!

(it just took me longer to write out in MathJax)