Eigenvalues and transpose Please help me with the following question.
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...
 A: 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. 
A: 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 $n$.
