Let [itex]T : V \to V[/itex] be a linear operator on an n-dimensional vector space V. Assume that the characteristic polynomial of T splits over F (the field underlying V). Prove that the following are equivalent:(adsbygoogle = window.adsbygoogle || []).push({}); I can prove (1) implies (2). In my book, the section on Jordan canonical form comes before the section on minimal polynomials, and then the section on minimal polynomials says very little about Jordan canonical form, so the hint given to prove (2) implies (3) isn't helping me too much. Any hints and suggestions would be helpful, thanks.

- There exists a vector [itex]x \in V[/itex] such that [itex]\{x,T(x),T^2(x),\dots ,T^{n - 1}(x)\}[/itex] is linearly independent.
- The characteristic polynomial of T is equal to [itex](-1)^n[/itex] times the minimal polynomial of T.
- There exists a basis [itex]\beta[/itex] such that

[tex][T]_{\beta} = \left (\begin{array}{ccccc}0 & 0 & \dots & 0 & (-1)^{n - 1}a_0\\1 & 0 & \dots & 0 & (-1)^{n - 1}a_1\\0 & 1 & \dots & 0 & (-1)^{n - 1}a_2\\ \vdots & \vdots & \ddots & \vdots & \vdots \\0 & 0 & \dots & 1 & (-1)^{n - 1}a_{n - 1}\end{array}\right )[/tex]

where [itex](-1)^nt^n + a_{n - 1}t^{n - 1} + \dots + a_1t + a_0[/itex] is the characteristic polynomial of T.

(Hint: For proving that (2) implies (3), it is helpful to show that the Jordan canonical form of T is the same as the Jordan canonical form of the matrix given in part (3)).

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# Homework Help: How do I prove this? [Linear Algebra]

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