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Let $T:V→V$ be a linear map over a finite dimensional vector space V with $dim$ $V=n$.

Prove that if every linear operator which commutes with $T$ is a polynomial in $T$, then T has a cyclic vector $v$ in $V$.

Let $S$ = the set of all polynomials in $T$. Then $S=C(T)$ as every polynomial in $T$ commutes with $T$ and we are told the converse is also true.

Suppose we do not have a cyclic vector $v$ i.e the set $A$ = {$v,Tv,...,T^{n−1}v$} is not a basis for V. Then $A$ is either linearly dependent, or it does not span $V$.

Suppose A is linearly dependent but it spans V. Then we can get rid of one element in A and have a spanning set of size $< n$ which is impossible.

Suppose A is linearly independent and it does not span $V$. This is impossible as $dim$ $V =n$.

Suppose $A$ is linearly dependent and does not span V. Now in order to arrive at a contradiction, I need to somehow construct a linear operator which commutes with $A$ and is not a polynomial in $T$.

Any help would be great.

Edit: The question really has not been answered in the other link. The answer only refers to a different website in which the text is incredibly difficult to read. Having another answer on stackexchange in clear text would make it much more accessible for the average reader.

John
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  • See here: https://planetmath.org/proofofcyclicvectortheorem – SMM Sep 12 '18 at 18:54
  • I've seen that link before but the notation was too difficult for me to follow, although I can recognize things like the cyclic decomposition theorem, which I have done Edit : Also, a comment from another link on stackexchange said that the construction of the function $g$ on this website was incorrect – John Sep 12 '18 at 19:00
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    There was more than just the accepted answer at the linked question. Figuring that maybe you found my answer there a bit "savant", I've added a somewhat simplified answer. But it remains a long and subtle argument. – Marc van Leeuwen Sep 14 '18 at 15:13

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