So the answer is yes. A brief version is that it suffices to check this for the top exponent and then to use exterior powers to deduce the result for subsequent exponents.

Next, if $E$ is the sum of the generalized eigsnspaces of all the eigenvalues of maximal modulus, and $F$ is the sum of the remaining generalized eigenspaces, for any $\epsilon>0$, there is a power of $A$ such that $\lambda\|e\|\le \|A^ne\|\le e^{\epsilon n} \lambda \|e\|$ for all $e\in E$ and $\|A^nf\|\le \mu\|f\|$ where $\mu<\lambda$. Now you can build a “cone” of points whose $F$ component is at most $\epsilon$ their $E$ component. This cone is invariant under small perturbations of $A^n$. Now a bit of triangle inequality gives you the result you’re looking for.

So the answer is yes. A brief version is that it suffices to check this for the top exponent and then to use exterior powers to deduce the result for subsequent exponents.

(Slightly modified to make $\lambda$ the leading Lyapunov exponent)

Next, if $E$ is the sum of the generalized eigenspaces of all the eigenvalues of maximal modulus, and $F$ is the sum of the remaining generalized eigenspaces, for any $\epsilon>0$, there is a power of $A$ such that $e^{n(\lambda-\epsilon)}\|e\|\le \|A^ne\|\le e^{n(\lambda+\epsilon)}\|e\|$ for all $e\in E$ and $\|A^nf\|\le e^{n\mu}\|f\|$ where $\lambda$ is the logarithm of the absolute value of the dominant eigenvalue(s) of $A$ and $\mu<\lambda$. Now you can build a “cone” of points whose $F$ component is at most $\epsilon$ their $E$ component. This cone is invariant under small perturbations of $A^n$. Now a bit of triangle inequality gives you the result you’re looking for.

So the answer is yes. A brief version is that it suffices to check this for the top exponent and then to use exterior powers to deduce the result for subsequent exponents.

Next, if $E$ is the sum of the generalized eigsnspaces of all the eigenvalues of maximal modulus, and $F$ is the sum of the remaining generalized eigenspaces, for as my $\epsilon>0$, there is a power of $A$ such that $\lambda\|e\|\le \|A^ne\|\le e^\epsilon \lambda \|e\|$ for all $e\E$ and $\|A^nf\|\le \mu\|f\|$ where $\mu<\lambda$. Now you can build a “cone” of points whose $F$ conponent is at most $\epsilon$ their $E$ component. This cone is invariant under small perturbations of $A^n$. Now a bit of triangle inequality gives you the result you’re looking for.

So the answer is yes. A brief version is that it suffices to check this for the top exponent and then to use exterior powers to deduce the result for subsequent exponents.

Next, if $E$ is the sum of the generalized eigsnspaces of all the eigenvalues of maximal modulus, and $F$ is the sum of the remaining generalized eigenspaces, for any $\epsilon>0$, there is a power of $A$ such that $\lambda\|e\|\le \|A^ne\|\le e^{\epsilon n} \lambda \|e\|$ for all $e\in E$ and $\|A^nf\|\le \mu\|f\|$ where $\mu<\lambda$. Now you can build a “cone” of points whose $F$ component is at most $\epsilon$ their $E$ component. This cone is invariant under small perturbations of $A^n$. Now a bit of triangle inequality gives you the result you’re looking for.