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Let $\mathbb{C}^n$ be given. Consider a matrix $A$ on it with one simple eigenvalue zero and all other eigenvalues having strictly negative real part.

Now, let $v$ be the eigenvector to eigenvalue $0$ and $V$ be a subspace of $\mathbb{C}^n$ such that $v \notin V$ and $AV \subset V.$

Does this imply that on $V$ all eigenvalues have strictly negative real part?

If anything is unclear, please let me know.

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Yes.

Since

$AV \subset V, \tag{1}$

$A$ may be considered a linear operator on $V$, and since $V \subset \Bbb C^n$ is indeed a subspace, it is a complex vector space in its own right; thus, $A$ has its own eigenstructure on $V$. If $w \in V$ is an eigenvector of $A$, we have

$Aw = \lambda w \tag{2}$

for some $\lambda \in \Bbb C^n$. But now we have

$w \in V \subset \Bbb C^n, \tag{3}$

thus $w \in \Bbb C^n$ is also an eigenvector of $A$ acting on $\Bbb C^n$. Since $\lambda \ne 0$, it follows from our hypothesis that $\Re(\lambda) < 0$.

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