## Eigenvalues and transpose

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Let $F:\mathbb{R}^{k}\to\mathbb{R}^{k}$ be a continuously differentiable mapping;

$F_{n}(x)$ be $n$-th iteration of $F(x)$, i.e. $F_{1}(x)=F(x)$, $F_{n}(x)=F(F_{n-1}(x))$;

$J_{n}(x)=(F_{n}(x))'$ be Jacobian matrix of $F_{n}(x)$;

$\lambda_{n}^{(1)}(x), \lambda_{n}^{(2)}(x), \ldots, \lambda_{n}^{(k)}(x)$ be eigenvalues of $J_{n}(x)$;

$\mu_{n}^{(1)}(x), \mu_{n}^{(2)}(x), \ldots, \mu_{n}^{(k)}(x)$ be eigenvalues of $(J_{n}(x))^{T}J_{n}(x)$.

How to prove that for all $i=\overline{1,k}$ there exists $j=\overline{1,k}$ such that $\lim\limits_{n\to\infty}\big|\lambda_{n}^{(i)}(x)\big|^{\frac{1}{n}}=\lim\limits_{n\to\infty}\big(\mu_{n}^{(j)}(x)\big)^{\frac{1}{2n}}$? (if this statement is true)

As I see, $\big|\lambda_{n}^{(i)}(x)\big|\neq\big(\mu_{n}^{(j)}(x)\big)^{\frac{1}{2}}$ in common case...

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## 2 Answers

In general, this may not be true.

EDIT (Atending OP´s objection) The important thing is that as stated, the problem is reduced to a linear algebra problem since it is possible to construct a diffeomorphism of $\mathbb{R}^n$ such that matrix $J_n(0)$ for the orbit of $0$ is the product of any sequence of invertible matrices.

To construct this, consider a translation of $\mathbb{R}^n$ (say $F(x)= x+b$) and in a neighborhood of $nb$ modify the diffeomorphism so that the derivative in that point is the desired matrix $A_n$.

In dimension $2$, a way of getting the desired counterexample is to consider the two times two upper triangular matrices $A_n$ with both eigenvalues $1$ and $K^n$ in the upper right corner (I am not being able to write matrices).

We get that the eigenvalues of $J_n$ will be always $1$, but the norm of $J_n$ grows exponentially, so, it is not true that the limits coincide.

I haven't thought on how to make a counterexample where the norms of $A_n$ are bounded but it should be not very difficult.

However, when some recurrence is added into the game, some results in the direction of what you are looking for are available. A key word for searching is Oseledets Theorem (or Multiplicative ergodic theorem, notice that in some places it is named Oseledec, or with some variations, othe key word for searching is: Lyapunov exponents). In particular, given an invariant probability measure, what you look for is satisfied for almost every point.

Playing with the proofs of this results, other more natural'' counterexamples can be constructed.

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But this is not a counterexample. You proposed $F(x)=Ax+b$ if $\|x\|\leq\delta$, $F(x)=x+b$ if $\|x\|>\delta$ where $\|b\|$ is sufficiently large.

Here $A$ is a matrix with eigenvalues $\lambda^{(1)}, \lambda^{(2)}, \ldots\lambda^{(k)}$;

$\mu^{(1)}, \mu^{(2)}, \ldots\mu^{(k)}$ are eigenvalues of $A^{T}A$;

$A$ is such that $\big|\lambda^{(i)}\big|\neq\big(\mu^{(i)}\big)^{\frac{1}{2}}$.

Then, as you write, $J_{n}(0)=A$ and we have

$\lim\limits_{n\to\infty}\big|\lambda_{n}^{(i)}(0)\big|^{\frac{1}{n}}=\lim\limits_{n\to\infty}\big(\mu_{n}^{(i)}(0)\big)^{\frac{1}{2n}}=1$

since $\lambda_{n}^{(i)}(0)=\lambda^{(i)}$ do not depend on $n$.

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Alexandra: generally this should be a comment on rpotrie's answer. Now, the person who posts the question generally can comment on all answers to her questions. In your case, since you provided two different log-ins, the software wasn't able to recognize you as the same person, and demands you to have 50 reputation before allowing the comment. For future reference, please log-in to MathOverflow using the same credentials every time. This way you can post comments such as above directly as a response comment to the answers given, and not as a separate answer.  – Willie Wong Oct 26 2010 at 13:45
An additional benefit for posting comments instead of new answers is that the answerer, in this case rpotrie, will be notified of your comment. New answers don't have this benefit and so rpotrie may not notice your objection until much later.  – Willie Wong Oct 26 2010 at 13:46
You are right, I wasn't carefull enough to construct the example. The idea is that with that construction (considering translations and including the matrix you want in the orbit of a chosen point, say $0$) you get that you can realize any product of matrices as the product of the derivatives along the orbit.  – rpotrie Oct 26 2010 at 16:24
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