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?

  • 2
    $\begingroup$ I like Futurama, too. $\endgroup$ – The Count Jul 17 at 2:01
  • $\begingroup$ You’re correct. I would just emphasize the importance of the linear factors in the char poly being distinct. $\endgroup$ – Juan Chi Jul 17 at 2:03
  • $\begingroup$ $\mathbb{C}$ 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. $\endgroup$ – Parthiv Basu Jul 17 at 2:15
  • $\begingroup$ 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! $\endgroup$ – CitizenSnips Jul 17 at 2:22
  • $\begingroup$ 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. $\endgroup$ – Omnomnomnom Jul 17 at 2:37

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