I seem to be stuck at formally showing something that intuitively seems to be true. I have a linear non-autonomous system of the form
$$ \dot{x} = A(t)x $$
where $A(t):\mathbb{R}\to \mathbb{R}^{n\times n}$ is a uniformly bounded square upper block triangular matrix, such that its $m$ diagonal blocks $\{B_i\}_{i=0}^m$ are strictly negative definite (SND) for all $t\ge0$, and $\lim_{t\to\infty} B_i(t)$ is bounded from above by a SND matrix (i.e. there is an SND matrix $B_0$ such that $B_i(t)-B_0$ is also SND). By strictly negative definite I mean a matrix such that $y^TB_i y<0$ for all nonzero $y$.
Q: Are Lyapunov exponents of such a system all negative? If so, how to show this?
Now, I know that if the whole $A(t)$ is negative definite, the Lyapunov exponents (LE) of the system are all negative. I also know from theorem 5.1 that if $A(t)$ is triangular with negative elements on the diagonal, its LE's are also all negative (the time average of a negative function is negative). I can't figure out how to couple these two statements to arrive at a result for a block-triangular matrix. don't see