Let $A\_n$ be a sequence of $d \times d$ matrices converging to a matrix $A$, all invertible and diagonalizable. We can define the Lyapunov spectrum of the corresponding dynamical system:
$$ \chi = \big\{ \lim\_{n \to \infty} \frac1n \log\|A^{(n)}v\| :\ v \in \Bbb R^d \big\}$$


where $A^{(n)} = A\_n A\_{n-1} A\_{n-2} \cdots A\_1$. Is it the case, in general, that $\chi$ equals the Lyapunov spectrum of the matrix $A$? If not, can we make this true by assuming that the sequence converges fast enough (for some notion of "fast enough")?



Lyapunov exponents of convergent sequence of matrices
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.


