An element both in $V_\mu$ and in $V_\overline{\mu}$ is zero?

Question

I was doing an exercise with following notations:

• $V$ is a $\mathbb{R}$-vectorspace
• $\alpha \in$Hom$_\mathbb{R} (V,V)$ and $\alpha_\mathbb{C}\in$Hom$_\mathbb{C} (V_\mathbb{C},V_\mathbb{C})$ its complex extension
• $V_\mu$ and $V_\overline{\mu}$ are the generalized eigenspaces for eigenvalues $\mu,\overline{\mu}$ respectively

Further in the exercise $V_\mu \oplus V_\overline{\mu}$ is mentioned. So I was wondering, why is it a direct sum?

I have already proven that $V_\overline{\mu}=\overline{V_\mu}$, so any element in $V_\mu \cap V_\overline{\mu}$ needs to be in $V$ (since if $v=\overline{v}$, $v$ is real). How do I prove that $v$ is $0$? Or have I maybe made a mistake along the way?

Answer 1

Generalised eigenvectors corresponding to distinct values are linearly independent. A vector is linearly independent from itself only if it's the zero. Therefore, $v\in V_\mu \cap V_{\overline{\mu}}$ implies $v=0$ if $\mu\notin\mathbb R$.