# Eigenvalues of invariant subspaces

Let $$V$$ be a vector space over $$F$$ and $$f:V\mapsto V$$ a linear map. If $$dimV=n \geq 2$$ and the only invariant subspaces of $$V$$ are $$V$$ itself and {$$0_V$$} ,then investigate if $$f$$ has eigenvalues.

I'm sorry if this is a trivial question, but my book has a little to no info about invariant subspaces or even the definition of what invariant really means and what follows that.

Any little i know about it comes from wikipedia so if anyone has some site or thread(question on this site) to propose it would be great.

• Hint : You just need the definition of "invariant subspace". Assume that $f$ has an eigenvalue, and find a (very natural) invariant subspace related to this eigenvalue. Then use the hypothesis and conclude. Jun 2, 2022 at 21:36
• @thesilverdoe Well ,it was easier than I thought .I concluded that f has eigenvalues. Thank you a lot .
– GGG
Jun 2, 2022 at 21:45
• Apparently, you assumed that $\lambda$ is an eigenvalue of $f$, and you deduced from it that $\lambda$ is an eigenvalue of $f$. What's the point? Jun 2, 2022 at 21:58
• @anotheruser you are absolutely right...
– GGG
Jun 2, 2022 at 22:09

If $$f$$ has some eigenvalue $$\lambda$$, let $$v$$ in $$V\setminus\{0\}$$ be such that $$f(v)=\lambda v$$. Then, for each $$\mu\in F$$,$$f(\mu v)=\mu f(v)=\,u\lambda v=\lambda(\mu v).$$So, $$Fv$$ is a one dimensional invariant subspace of $$V$$. But it is being assumed that $$V$$ has invariant subspaces other than $$V$$ or $$\{0_V\}$$, and $$Fv$$ is none of them.
• I don't get how a linear map of a vector spave that maps to itself can have no eigenvalues. Don't all square matrices have eigenvalues in $\mathbb{C}$ by fundumental theorem of algebra ?
• Where did you mention $\Bbb C$ in your question? Isn't this a question about vector spaces over fields in general? Jun 3, 2022 at 14:18
• Yes it is in general. But $\mathbb{C}$ is pretty huge and i can't think of any field F that is not a subset of $\mathbb{C}$ other than $\mathbb{Z}_p$ ,p be a prime.
• Nevermind i was thinking the wrong way , you are 100% correct .Thank you for your time !For example if $\begin{pmatrix} 0 & 1 \\ -1& 0 \\ \end{pmatrix} \in \mathbb{R}^{3 \times 3}$ has no eigenvalues, since its characteristic polynomial is $f(x)=x^2+1$.