Condition for symmetric part of $A$ for $\|x(t)\|$ monotonically decreasing ($\dot{x} = Ax(t)$)

Question

Recall the following discussion:
When is the symmetric part of a matrix positive definite?

Suppose $\dot{x} = Ax$ is asymptotically stable.

I am no quite understand why the following is the condition for $\|x(t)\|$ monotonically decreasing?

$A^T+A \prec 0$, i.e., the symmetric part of $A$ is negative definite

This result is from S. Boyd's lecture p.19:

https://stanford.edu/class/ee363/lectures/lq-lyap.pdf

Can anyone give me a hint to prove this?

score 1 · Accepted Answer · answered Jan 17 '18 at 17:17

Consider the positive definite function $$ V(x)= \|x\|^2=x^Tx=x^T I x. $$ Its derivative along the trajectories of the system $$\tag{1} \dot x=Ax $$ is equal to $$ \dot V(x)=\frac{dV(x(t))}{dt}=x^T (A^T I+IA) x=x^T (A^T +A) x. $$ Since $A^T+A$ is negative definite, the quadratic form $x^T (A^T +A) x$ is negative for any $x\ne 0$, thus, $V(x)$ decreases along every nonzero trajectory of (1).

Condition for symmetric part of $A$ for $\|x(t)\|$ monotonically decreasing ($\dot{x} = Ax(t)$)

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