Answer
False
Work Step by Step
If $v_0$ is an eigenvector of A corresponding to $\lambda$ and
$v_1$ is a vector satisfying $(A − \lambda I)v_1 = v_0$ we obtain:
$Ax=A(e^{\lambda t}v_1)\\
=(Av_1)e^{\lambda t}\\
=(v_0+\lambda v_1)e^{\lambda t}\\
=e^{\lambda t}v_0+x'(t)$
Then $x(t) = e^{\lambda t}v_1$ is not a solution to the vector differential equation because $x′ \ne Ax$.
