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.




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

$A$ 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 $\Re(\lambda) < 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.