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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?

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  • $\begingroup$ If $A$ is diagonalizable and all its eigenvalues are $1$, then it is the identity, so there is no hope that $y\ne v$. $\endgroup$ – Miguel Aug 15 '17 at 17:50
  • $\begingroup$ I agree, but $A$ may not be diagonalizable. So I was thinking that in this case I have to assume $A$ has Jordan blocks. $\endgroup$ – Krishnan Mody Aug 15 '17 at 17:53
  • $\begingroup$ Maybe your proof that $A$ must have real eigenvalues would be enlightening. Did you mean at least one real eigenvalue or all eigenvalues must be real? $\endgroup$ – Miguel Aug 15 '17 at 17:56
  • $\begingroup$ You can have $\lambda_2=\lambda_3=-1$, but then $\lim_{n\to\infty}A^n$ does not exist. $\endgroup$ – Miguel Aug 15 '17 at 18:00
  • $\begingroup$ you want to show that exists some pair $u\neq v$ such that $A^{-n}u=v$ for any $n\in\Bbb N_{>0}$, or just for some $n$? Moreover: the matrix $A\in\Bbb R^{3\times 3}$ or $A\in\Bbb C^{3\times 3}$? $\endgroup$ – Masacroso Aug 15 '17 at 18:01
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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

$\pmatrix{\lambda^n \\&1\\&&(\frac 1\lambda)^n}\pmatrix{0\\0\\1} = \pmatrix {0\\0\\(\frac 1\lambda)^n}\\ \lim_\limits{n\to\infty} \pmatrix{\lambda^n \\&1\\&&(\frac 1\lambda)^n}\pmatrix{0\\0\\1} = \pmatrix {0\\0\\0}$

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  • $\begingroup$ Then $\lim_{n\to\infty}A^n {\bf u}$ does not exist. $\endgroup$ – Miguel Aug 15 '17 at 18:41
  • $\begingroup$ @Miguel I agree. But I never mention $\lim_\limits{n\to\infty} A^n\mathbf u$. $\lim_\limits{n\to\infty} A^n\mathbf v$ and $\lim_\limits{n\to\infty} A^n\mathbf w$ both exist. $\endgroup$ – Doug M Aug 15 '17 at 18:44
  • $\begingroup$ In fact, $\lim_{n\to\infty}A^n $ does not exist. What is then $\lim_{n\to\infty}A^n x$ ? $\endgroup$ – Miguel Aug 15 '17 at 18:45
  • $\begingroup$ @Miguel I have added some detail for you. $\endgroup$ – Doug M Aug 15 '17 at 18:56
  • $\begingroup$ But then there is a biiiiig notation problem. The OP should have emphasized that $\lim_{n\to\infty}A^n x$ means $\lim_{n\to\infty} (A^n x)$. $\endgroup$ – Miguel Aug 15 '17 at 18:57
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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.

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  • $\begingroup$ Thank you. Would you know how to prove that x,v,y are linearly independent? $\endgroup$ – Krishnan Mody Aug 15 '17 at 20:56
  • $\begingroup$ @KrishnanMody These vectors are constructed in the other answer. You can see they are independent. $\endgroup$ – Miguel Aug 15 '17 at 20:59
  • $\begingroup$ Oh right. Thanks! $\endgroup$ – Krishnan Mody Aug 15 '17 at 21:00

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