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?

  • $\begingroup$ 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

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

$\lim_\limits{n\to\infty} A^n (a\mathbf v + b \mathbf w) = a\mathbf v$

$\lim_\limits{n\to\infty} A^{-n} (a\mathbf v + b \mathbf u) = a\mathbf v$


In response to comments below.

Does $\lim_\limits{n\to\infty} A^n \mathbf v$ exist?

Suppose $A = \pmatrix {\lambda \\&1\\&&\frac 1\lambda}$

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


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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.