Spectrum of Lyapunov exponents of a linear system Question: How to show that the eigenvalues of matrices $\mathbf{A}$ and
$
\mathbf{L} = \log \lim_{t \to \infty} \left((e^{\mathbf{A}t}e^{\mathbf{A^T}t})^{\frac{1}{2t}}\right)
$
have equal real parts?

Motivation:
Consider a system
$$
\dot{\mathbf{x}} = \mathbf{Ax}
$$
where $\mathbf{A}$ is some constant matrx and $\mathbf{M}(t) := e^{\mathbf{A}t}$ is the evolution operator. The Lyapunov exponents of this system are then given by the eigenvalues of
$$
\mathbf{L} :=
\log \lim_{t \to \infty} \left((\mathbf{M}(t)\mathbf{M^T}(t))^{\frac{1}{2t}}\right) = 
\log \lim_{t \to \infty} \left((e^{\mathbf{A}t}e^{\mathbf{A^T}t})^{\frac{1}{2t}}\right)
$$
Now, on the other hand, the rate of expansion for a system above is constant and is also given by the real parts of $\mathbf{A}$'s eigenvalues. Therefore, eigenvalues of $\mathbf{L}$ are equal to real parts of eigenvalues of $\mathbf{A}$. I have checked it numerically and it also makes sense intuitively, but I don't see why, technically, the relation holds.
 A: I assume that your matrix is square.
$$ \log \lim_{t \rightarrow \infty} (M(t) M(t)^T)^{ \frac{1}{2t}} = \lim_{t \rightarrow \infty} \log (M(t) M(t)^T)^{ \frac{1}{2t}} =  \lim_{t \rightarrow \infty} \frac{1}{2t} \log (M(t) M(t)^T) $$
since the logarithm is continuous. Now with $M(t) = e^{At}$, you get
$$   \lim_{t \rightarrow \infty} \frac{1}{2t} \log (M(t) M(t)^T) = \lim_{t \rightarrow \infty} \frac{1}{2t} \log (e^{At}e^{A^Tt}) =  \lim_{t \rightarrow \infty} \frac{1}{2t}(A + A^T)t$$
$$= \frac{1}{2}(A + A^T)$$
Here you need to show that $A$ ant $A^T$ commute.
Now suppose $\lambda = a+ ib$ is an eigenvalue of $A$ with eigenvector $v$. Then 
$$(A + A^T)v = (\lambda + \bar{\lambda} )v = 2a v $$
A: The theorem of Lyapunov ensures that the eigenvalues $\lambda_i$ of the matrix $\textbf{A}\in \Bbb R^{n \times n}$ satisfy $Re(\lambda_i)<0$ if and only if, for any given symmetric positive definite matrix $\textbf{P}$, there exists a unique positive definite symmetric matrix $Q$ satisfying the Lyapunov equation:
$$Q\textbf{A}+\textbf{A}Q=-\textbf{P}$$
In other words your question points directly to the stability theorem itself. If I understood your question correctly, I believe the answer points to the Lyapunov theorem itself.
