Proving that the elements of $A^n$ matrix are non-zero

Question

Let $ A = \begin{pmatrix} 1 & 2 \\-1 & 1 \\ \end{pmatrix}$. Prove that for every positive integer $n$ there exist integers $x_{n},y_{n}$ such that $A^n= \begin{pmatrix} x_{n} & -2y_{n} \\ y_{n} & x_{n} \\ \end{pmatrix}$. Prove that for all $n$ the numbers $x_{n}$ and $y_{n}$ are non-zero.

I proved the existence of numbers $x_{n}$ and $y_{n}$, with the recurrence relations $x_{n+1}=x_{n}+2y_{n}$ and $y_{n+1}=y_{n}-x_{n}$.

For proving that these numbers are non-zero, I assumed that if $x_{n}=0$, then $detA^n=3^n=2y_{n}^2$. Since $3^n$ is always an odd integer, and $2y_{n}$ is always an even integer, we reach a contradiction. However, I'm stuck at the case $y_{n}=0$. I've looked at other approaches to this problem, but no other has worked so far.

Answer 1

If you find the eigenvalues and eigenvectors of the matrix, you can diagonalize the matrix and therefore obtain an explicit expression for $A^n$. Do you know this method?

Answer 2

What about eliminating the $y$ terms from your pait of recurrence relations to obtain a two-stage recurrence for $x$ terms. This can be solved...do you know this method?

Answer 3

the eigenvalues of $A$ are $1 \pm i\sqrt2.$ that means $A$ is not diagonalizable over reals. that rules out $y_n = 0$ because the eigenvalues of $A^n$ are $(1\pm i\sqrt 2)^n \neq 1.$

we only need to see why $x_n \neq 0.$ suppose $x_n = 0,$ then the eigenvalues of $A^n$ are $x_n$ times the eigenvalues of $\pmatrix{0&-2\\1&0}.$ that is $x_n\sqrt 2i \neq (1\pm i\sqrt 2)^n.$

Answer 4

We have $A^2 \equiv -A\pmod{3}$, so $A^n \equiv (-1)^{n+1} A\pmod{3} $

Therefore the entries of $A^n$ are never divisible by $3$, and in particular they are nonzero.