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Let $V$ be a vector space over $\mathbb{C}$ and let $n=\mbox{dim}_{\mathbb{C}}(V)$ be finite. Let $T\in\hom_{\mathbb{C}}(V,V)$ and suppose that for every $c\in\mathbb{C}$, $\{v\in V:Tv=cv\}$ has dimension no greater than $1$.

Then show that $\exists w\in V$ such that $\{w,Tw,\dots, T^{n-1}w\}$ is linearly independent.

My Attempt: The hypothesis suggest that the characteristic function of $T$ splits into distinct linear factors. Since the minimal polynomial divides the char poly and the char poly divides a power of the minimal polynomial, they have the same roots. But this means that the char poly is the minimal polynomial and it is of degree $n$. But this means that there is only one invariant factor (when applying the Fundamental Theorem of modules over PIDs and viewing $V$ as a $\mathbb{C}[x]$ module. So in particular $V$ is cyclic). Explicitly the image of $\bar{1}$ is a generator. Is this all correct?

CitizenSnips
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    I like Futurama, too. – The Count Jul 17 '19 at 02:01
  • You’re correct. I would just emphasize the importance of the linear factors in the char poly being distinct. – JCM Jul 17 '19 at 02:03
  • $\mathbb{C}$ is an algebraically closed field so any polynomial over it splits completely. It has nothing to do with your hypothesis. Also the fact that the minimal polynomial and characteristic polynomial have the same roots holds over any field. Also just because they have the same roots does not mean they are the same. The characteristic polynomial could have a double root. – Parthiv Basu Jul 17 '19 at 02:15
  • I'll edit it. I meant that the hypothesis implies that the factors are distinct. If they are distinct eigenvalues then there is no double root and they have to be the same. Thanks! – CitizenSnips Jul 17 '19 at 02:22
  • The characteristic function need not split into distinct linear factors. For instance, $$ T = \pmatrix{0&1\0&0} $$ satisfies the hypothesis, but has characteristic function $p(x) = x^2$. It is true, however, that the minimal and characteristic polynomials must coincide. – Ben Grossmann Jul 17 '19 at 02:37
  • This question is related, but I don't see how to arrive at an answer to your question from the result – Ben Grossmann Jul 17 '19 at 02:42
  • One approach is to show that the hypothesis implies that every block of the Jordan form has maximal size, and from there conclude that the minimal and characteristic polynomials are the same – Ben Grossmann Jul 17 '19 at 02:44
  • I see what your saying: this is referring to geometric multiplicity, not algebraic multiplicity. I can fix my proof by looking at the JCF and then showing each block corresponds to it's own eigenvalue. Then I determine the minimal polynomial (using some facts) and from the JCF conclude that the eigenvalues are the same. I'll fix this soon! Thanks! – CitizenSnips Jul 17 '19 at 06:28

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