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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?

mikasa
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    Check that generalised eigenvectors corresponding to distinct values are linearly independent. In present case, you have to assume $\mu\notin\mathbb R$. – AlvinL Oct 16 '22 at 14:48

1 Answers1

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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 $\mu\notin\mathbb R$.

AlvinL
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