The following paragraph is in page 16 of "Introduction to Lie Algebras and Representation Theory - Humphreys"

$K$ is a subalgebra of $\mathfrak gl$($V$). Denote $W$={$w\in V$ : $x.w=\lambda(x)w$, for all $x \in K$}, $i.e.$ the set of common eigenvectors of $K$. Let $L=K+Fz$ (for any $z\in L\setminus K$) and [use the fact that $F$ is algebraically closed to find and eigenvector $v_0\in W$ of $z$.] Then $v_0$ is obviously a common eigenvector for $L$.

I can't understand the part [ ]. I mean, I don't know how to use the algebraically closedenss of $F$. I guess it's quite easy part but I'm stucked. Thanks for any help in advance.


1 Answer 1


If $F$ is an algebraically closed field and $V$ is a finite-dimensional vector spae, then every linear map $L\colon V\longrightarrow V$ has some eigenvalue, since the eigenvalues are the roots of the characteristic polynomial of $L$. If $F$ is not algebriacally cloed, then $L$ may not have eigenvalues.

  • $\begingroup$ So, just restrict z:V->V to z:W->W (this is possible since we proved that z leaves W invariant in the textbook) and then, by your answer, we get the result we wated. right? $\endgroup$
    – learner
    Commented Jan 23, 2021 at 9:37
  • $\begingroup$ Yes, that is correct. $\endgroup$ Commented Jan 23, 2021 at 9:40
  • $\begingroup$ Thanks a lot!!! It really helped me Thank you! $\endgroup$
    – learner
    Commented Jan 23, 2021 at 10:29
  • $\begingroup$ I'm glad I could help. $\endgroup$ Commented Jan 23, 2021 at 12:11

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