# Cyclic Vector and Diagonalizable operator

I've been working on this exercise:

Prove that a diagonalizable operator $$T$$ on an $$n$$-dimensional vector space has a cyclic vector iff it has $$n$$-distinct eigenvalues.

I tried to do the following:

Let $$v$$ be the cyclic vector of $$T$$. Then $$C=\{v, Tv, \dots, T^{n-1}v\}$$ is basis of $$V$$ and since $$T$$ is diagonalizable there is a basis of eigenvectors, say $$B=\{b_1, \dots, b_n\}$$.

Now, you can write $$v$$ in terms of basis $$B$$. And hence the other vectors of basis $$C$$, that is $$Tv, T^2v, \dots, T^{n-1}v$$.

In the end, I believe that the result will follow of the linear independence of $$\{v, Tv, \dots, T^{n-1}v\}$$, however I couldn't show how. It seems to me that contradiction could be used, assuming that there are distinct eigenvalues ​​and somehow have that the set $$\{v, Tv, \dots, T^{n-1}v\}$$ will not be linearly independent.

Is my idea correct? Any tips to continue? If there is a simple way to solve, I would be grateful for the help.

## 2 Answers

Write $$T^nv$$ in terms of $$v,Tv,\dots,T^{n-1}v$$, and then use the following two standard facts:

Fact 1: A linear operator on a finite dimensional vector space is diagonalizable if and only if its minimal polynomial is $$(x-\lambda_1) \cdots (x-\lambda_r)$$, where $$\lambda_1,\dots,\lambda_r$$ are all its distinct eigenvalues.

Fact 2: The minimal polynomial of the matrix $$\begin{pmatrix}0&0&\dots &0&-c_{0}\\1&0&\dots &0&-c_{1}\\0&1&\dots &0&-c_{2}\\\vdots &\vdots &\ddots &\vdots &\vdots \\0&0&\dots &1&-c_{{n-1}}\end{pmatrix}$$ is $$c_0+c_1x+\cdots+c_{n-1}x^{n-1}+x^n$$.

Hint: Let $$B$$ a diagonalising basis, take an arbitrary vector $$v$$ and compute the determinant of the matrix whose columns are the coordinates of $$v, Tv,\ldots,T^{n-1}v$$ in $$B$$.

Hint 2: