Is the multiple of an eigenvector also an eigenvector in this case?

Question

Be $\mathbf{A}_{\mathbf{v}} \in \mathbb{R}^{n \times n}$ with $\mathbf{A}_{\mathbf{v}}=\mathbf{v} \mathbf{v}^{\top}$. For any vector $v \in \mathbb{N}^n$ with $||v|| > 1$ and $n > 2$, $-v$ is an eigenvector of $A$.

What I know is that $v$ is always an eigenvector of $A_v$. Because $v$ is an eigenvector off $A_v$ his multiple should be also an eigenvector right? So $-v$ has to be an eigenvector of $A_v$.

Answer 1

Yes the nonzero multiple is always an eigen vector, since if $Av=\lambda v$, then $A(-v)=-Av=-\lambda v=\lambda (-v)$.