properties of, 3x3 matrix, determinant 1, real eigenvalues Let A be a 3x3 matrix with determinant 1. Suppose there exists x such that 
$$\lim_{n\to\infty} (A^n x) = v$$ and $x\neq v$, $v \neq 0$. Then I can prove that $A$ must have real eigenvalues. I want to show that there exists $y$ such that $$\lim_{n\to\infty}(A^{-n}y) = v,$$ $y \neq v$. From the condition on $x$ above, we know $Av = v$, so one eigenvalue of $A$ is 1. As I mentioned above, the others are real, and so the 3 eigenvalues of $A$ are $1, \lambda, 1/\lambda$. If $\lambda > 1$, then $1/\lambda < 1,$ so let $w$ be the eigenvector corresponding to $\lambda$. $$ \lim_{n\to\infty} (A^{-n}(v + w)) = lim_{n\to\infty} v + \lambda^{-n}w = v.$$ In other words, this choice of $y$ works. However, I cannot see how to find a $y$ when all the eigenvalues of $A$ are 1. Assuming there is always a $y$ the question I am trying to solve says that $y,x,v$ are linearly independent. How can I prove this?
 A: If $A$ has eigenvalues $\lambda, 1, \frac 1\lambda$ 
If $|\lambda| = 1$ there will be a contradiction with the critera $\lim_\limits{n\to \infty} A_n x = v$ with $x\ne v\ne 0$
If all of the eigenvalues exactly equal $1$ then A is the identity and 
$A^nx = x$
and if $|\lambda|  = 1$ and $\lambda \ne 1$ (either $\lambda  = -1$ or $\lambda$ is complex) then  
$\lim_\limits{n\to \infty} A^nx$ does not exist the if eigenvectors associated with $\lambda, \frac 1{\lambda}$ are components of $x$
and if they are not.
$\lim_\limits{n\to \infty} A^nx = x$
let $|\lambda| > 1$
Let $\bf{u,v,w}$ be the eigenvectors associated with eigenvalues $\lambda, 1, \frac 1\lambda$ respectively
$\lim_\limits{n\to\infty} A^n (a\mathbf v + b \mathbf w) = a\mathbf v$
$\lim_\limits{n\to\infty} A^{-n} (a\mathbf v + b \mathbf u) = a\mathbf v$
Update
In response to comments below.
Does 
$\lim_\limits{n\to\infty} A^n \mathbf v$ exist?
Suppose $A = \pmatrix {\lambda \\&1\\&&\frac 1\lambda}$
$\pmatrix {\lambda \\&1\\&&\frac 1\lambda}\pmatrix{0\\1\\0} = \pmatrix {0\\1\\0}\\
\pmatrix {\lambda \\&1\\&&\frac 1\lambda}^n\pmatrix{0\\1\\0} = \pmatrix {\lambda^n \\&1\\&&(\frac 1\lambda)^n}\pmatrix {0\\1\\0} = \pmatrix {0\\1\\0}\\
\lim_\limits{n\to\infty}  \pmatrix{\lambda^n \\&1\\&&(\frac 1\lambda)^n}\pmatrix{0\\1\\0} = \pmatrix {0\\1\\0}$
and
$\pmatrix{\lambda^n \\&1\\&&(\frac 1\lambda)^n}\pmatrix{0\\0\\1} = \pmatrix {0\\0\\(\frac 1\lambda)^n}\\
\lim_\limits{n\to\infty}  \pmatrix{\lambda^n \\&1\\&&(\frac 1\lambda)^n}\pmatrix{0\\0\\1} = \pmatrix {0\\0\\0}$
A: This partial answer is only to rule out the non-diagonalizable case.
Assume there is a repeated eigenvalue $\lambda_1$ with only one eigenvector $u$ and another vector $v$ to complete the basis. Also, the simple eigenvalue $\lambda_2$ corresponds to eigenvector $w$.
If $\lambda_1<1$, then $\lambda_2>1$, so:
$$\lim_{n\to\infty}A^n u=0$$
$$\lim_{n\to\infty}A^n v=\infty$$
$$\lim_{n\to\infty}A^n w=\infty$$
Hence there is no way to build a vector $x$ with $\lim_{n\to\infty}A^n x$ exists and is not null.
If $\lambda_1>1$, then $\lambda_2<1$, so:
$$\lim_{n\to\infty}A^n u=\infty$$
$$\lim_{n\to\infty}A^n v=\infty$$
$$\lim_{n\to\infty}A^n w=0$$
and the rationale is analogous.
If $\lambda_1=1$, then also $\lambda_2=1$, so:
$$\lim_{n\to\infty}A^n u=u$$
$$\lim_{n\to\infty}A^n v=\infty$$
$$\lim_{n\to\infty}A^n w=w$$
and there is no way to build a vector $x$ with $\lim_{n\to\infty}A^n x$ exists and is not $x$.
Finally, if $\lambda$ is a triple eigenvalue, it must be necessarily $1$. Let $u$ be the only eigenvector, so:
$$\lim_{n\to\infty}A^n u=u$$
$$\lim_{n\to\infty}A^n v=\infty$$
$$\lim_{n\to\infty}A^n w=\infty$$
Again there is no way that the conditions hold.
The key (shown by Doug M's answer) is that you need at least two eigenvectors, associated to different eigenvalues.
