Consider the linear time-varying system given by
$\dot x(t)=A(t)x(t)$
Denote by $\lambda_\max(t)$ the maximum eigenvalue of $A(t)+A^T(t)$. Suppose that there exist constants $\alpha\gt0$ and $\gamma$ such that $\lambda_\max(t)$ satisfies
$\int_\tau^t\lambda_\max(s)ds\le-\alpha(t-\tau)+\gamma$ where $\forall t\ge\tau$
Prove that $x_e=0$ is uniformly exponentially stable.
I was given a hint: Examine $d\over dt$$\Vert x\Vert_2^2$
But still, have no clue how to solve this problem. Also why we consider maximum eigenvalue for $A(t)+A^T(t)$? Please help me!
Following the hint we examine the derivative of the square of the Euclidean norm. I use $\langle \cdot, \cdot \rangle$ to denote the standard inner product on $\mathbb{R}^n$. We then have for all $t$ $$\frac{d}{dt}\|x(t)\|^2 = \frac{d}{dt} \langle x(t), x(t) \rangle = \langle A(t)x(t), x(t) \rangle + \langle x(t), A(t)x(t) \rangle$$ $$=\langle (A(t)+A(t)^T)x(t), x(t) \rangle \leq \lambda_{max}(t)\|x(t)\|^2 $$ Integrating from $\tau$ to $t$ we get $$ \|x(t)\|^2 \leq \|x(\tau)\|^2 + \int_{\tau}^t \lambda_{max}(s)\|x(s)\|^2 ds$$ Gronwall's lemma yields $$ \|x(t)\|^2 \leq \exp\left( \int_{\tau}^t \lambda_{max}(s) ds\right) \|x(\tau)\|^2 $$ With the given assumption we obtain $$ \|x(t)\|^2 \leq e^\gamma e^{-a(t-\tau)} \|x(\tau)\|^2 $$ As $a>0$ and $\gamma$ are independent of $t,\tau$ this shows uniform exponential stability.