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.
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2$\begingroup$ 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. $\endgroup$– TheSilverDoeJun 2, 2022 at 21:36
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$\begingroup$ @thesilverdoe Well ,it was easier than I thought .I concluded that f has eigenvalues. Thank you a lot . $\endgroup$– GGGJun 2, 2022 at 21:45
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$\begingroup$ 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? $\endgroup$– Another UserJun 2, 2022 at 21:58
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$\begingroup$ @anotheruser you are absolutely right... $\endgroup$– GGGJun 2, 2022 at 22:09
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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.
answered Jun 2, 2022 at 22:14
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$\begingroup$ 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 ? $\endgroup$– GGGJun 3, 2022 at 12:29
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$\begingroup$ Where did you mention $\Bbb C$ in your question? Isn't this a question about vector spaces over fields in general? $\endgroup$ Jun 3, 2022 at 14:18
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$\begingroup$ 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. $\endgroup$– GGGJun 3, 2022 at 15:06
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$\begingroup$ 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 $f(x)=x^2+1$. $\endgroup$– GGGJun 3, 2022 at 15:20