Prove that the elements of these two sequences are not null

Question

Let $x_{n+1}=x_n+2y_n$ and $y_{n+1}=y_n-x_n$, where $x_1=1$ and $y_1=-1$.

I tried proving by contradiction, I tried by induction, I got nothing.

This is a question I had on an exam, I didn't manage to solve it, and afterwards I spent one day thinking about it and still came up with nothing.

Please note that a full solution is not necessary, if you could just provide a hint, that would be awesome.

The origin of the question was to show that $\left( \begin{array}{ccc} 1 & 2 \\ -1 & 1 \\ \end{array} \right)^n$ has no zero elements for any positive integer $n$.

Answer 1

The pair $(x_n, y_n)^T$ is equal to $$ \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}^{n-1} \begin{pmatrix} 1 \\ -1 \end{pmatrix}. $$ If you want to prove that there is no $n$ s.t. $x_n=y_n=0$, then you just use that fact that a regular matrix never maps a nonzero vector to zero. If you want to prove that for any $n$ neither $x_n$ nor $y_n$ is zero, do the eigenvalue decomposition and compute $x_n, y_n$ exactly.

Answer 2

Let $A= \left( {\begin{array}{*{20}c} 1 & 2 \\ { - 1} & 1 \\ \end{array}} \right) $. Let try us to diagonalize this matirx, we have to find first the eigenvalue of $A$.

\begin{align} \left| {\lambda I - A} \right| = \left| {\begin{array}{*{20}c} {\lambda - 1} & { - 2} \\ 1 & {\lambda - 1} \\ \end{array}} \right| = \left( {\lambda - 1} \right)^2 + 2 = \lambda ^2 - 2\lambda + 3 \ne 0\,\,\left( {\text{for}\,\,\text{all}\,\,\text{real}\,\,\lambda } \right) \end{align} so that no "real" eigenvalues and thus the matrix $A$ is not diagonalizable; i.e., there is NO a non-singular matirx $P$ such that $PAP^{-1}=D$ (D:=diagonal matirx), or we write $A=P^{-1}DP$

Answer 3

Hint: look at their remainders $\pmod p$ for some small primes $p$.