question about span

Question

Let $\dot x = Ax$ and $S$ be an orthogonal basis of null space of $A$. Does it imply that $x$ converge to a point in $\operatorname{span}\{S\}$?

Answer 1

This is only try is all remaining eigenvalues has a negative real part and for the eigenvalues of zero the algebraic multiplicity is the same as the geometric multiplicity. All other eigenvalues have to have a negative real part to ensure that all other modes decay to zero, for example $$ A = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} $$ can blow up to infinity. And the algebraic multiplicity has to be equal to the geometric multiplicity to ensure that the eigenvector of the eigenvalue of zero is actually a stationary point, for example $$ A = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} $$ can also increase linearly in time, however the solution will converge to the span of the generalized eigenvectors of eigenvalue of zero (but not necessary to a point).

Answer 2

No. You don't specify whether you're asking about the limit of $x(t)$ as $t\to\infty$ or $t\to-\infty$; for a counterexample to both let $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$.