I'm stuck finding a counterexample of this exercise. Let the linear system $\dot{x} = Ax$ with $x \in \mathbb{R}^n$ and $\lambda$ an eigenvalue of $A$. I know if $w \in \mathbb{R}^n$ such that $A^{T}w = \lambda w$, then $w^T x = 0$ is invariant with respect of the flow of the differential equation.
Is it true that if $w^T x = 0$ is invariant, then $A^Tw = \lambda w$?
My answer is that is a false statement, but I cannot find a counterexample to prove my hypothesis. I guess, in $\mathbb{R}^2$, some matrix such that $det(A)=0$ would help me, but I do not know how to proceed.
Any help?
There is no such counterexample, the statement is true. Let $V(x)=w^T x$, $w\ne 0$. If the hyperplane $$S=\{x\in\mathbb R^n:\; V(x)=0 \}$$ is invariant, then the derivative along the trajectories $\dot V= w^T\dot x= w^T Ax$ is zero on $S$: $$ \forall x\in S\quad \dot V =0. $$ Thus we have $$ \forall x\in\mathbb R^n\quad w^T x =0 \Rightarrow w^T Ax=0. $$ If $w^T A\ne 0$, then $\{x\in\mathbb R^n:\; w^T Ax=0 \}$ is also a hyperplane. It contains the hyperplane $S$, and hence coincides with it. If $w^T x=0$ and $w^T Ax=0$ are the equations of the same hyperplane, then $w^T A=\lambda w^T$.
If $w^TA=0$, then also $w^T A=\lambda w^T$, where $\lambda=0$.