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How can one infer from the Lie's theorem (in terms of existence of a common eigenvector) that a complex irreducible representation of a solvable lie algebra has dimension 1?

What I know is that one can write $\sigma(x)(v) \in \mathbb{C}.v$ if $\sigma$ is the representation so that $\mathbb{C}.v$ is invariant. How to complete the argument from here?

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    $\begingroup$ Take an irreducible representation $V$. Using Lie's theorem, you have found a $1$-dimensional subspace which is in fact a subrepresentation of $V$. Since $V$ is irreducible, then it does not have proper non-zero subrepresentations so... $\endgroup$ – Mariano Suárez-Álvarez Apr 26 '15 at 0:28
  • $\begingroup$ @MarianoSuárez-Alvarez So then are we saying that either the subspace $\mathbb{C}.v$ has to be 0 or that the representation itself should be one-dimensional? $\endgroup$ – Iguana Apr 26 '15 at 0:42
  • $\begingroup$ $v$ cannot be the zero vector (if we allowed $v$ to be zero in Lie's theorem, the theorem would be a triviality!) $\endgroup$ – Mariano Suárez-Álvarez Apr 26 '15 at 0:48

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