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Can anyone provide some hint on how to proceed with the proof of the following:

The characteristic polynomial and minimal polynomial of a linear operator $T$ coincide if and only if for a certain vector $y$ in $V$ ,vectors $y,Ty,\ldots \ldots ,T^{n-1}y$ are linearly independent i.e. the set $\{y,Ty,\ldots,T^{n-1}y\}$ forms the basis of $V$

JAME
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1 Answers1

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Note that the minimal polynomial equals the characteristic polynomial iff it has degree $n$ (since both are monic). If there exists such a $y$, then in particular, $I, T, T^2, \ldots, T^{n-1}$ are linearly independent, so no polynomial of degree less than $n$ vanishes at $T$, so the minimal polynomial must have degree exactly $n$.

Conversely, if the minimal polynomial has degree $n$, then $I, T, T^2, \ldots, T^{n-1}$ are independent. To find such a $y$, I think you can write the matrix in Rational Canonical Form and let $y$ be one of the basis vectors used to conjugate $T$ to its RCF.

Nishant
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  • can you please elaborate the second line of your answer which impiles linear independence ..I can't get it.. – JAME Apr 05 '15 at 15:28
  • If there is a $y$ such that $y, Ty, T^2y\ldots, T^{n-1}y$ are independent, then the operators $I, T, T^2, \ldots, T^{n-1}$ must be independent, for if $\sum c_iT^i=0$, then in particular, $\sum c_iT^iy=0$, contradicting the independence of the images of $y$. – Nishant Apr 05 '15 at 15:35
  • Can you prove the second part without using Rational Canonical Form? I have not done it. – Balasubramannyan S Nov 11 '20 at 02:47