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 2


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:

Vandermonde matrix.


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