# The cyclic vector-space

First, we define the cyclic vectorspace how:

Let $T$ be a linear operator on a vector space $V$, and let $x$ be a nonzero vector in $V$. The subspace $$W=\text{span}\left\{ x,T(x),T(T(x)),...,T^{k}(x),...\right\}$$is called the $T$-cyclic subspace of $V$ generated by $x$.

And my problem is the next:

Let $T$ be a linear operator on an $n$-dimensional vector space $V$ such that $T$ has $n$ distinct eigenvalues. Prove that $V$ is a $T$-cyclic subspace of itself.

I know that, let $W$ a $T-$cyclic vectorspace generated by $v$ with $\text{dim}(W)=k$ the set $\left\{v,T(v),T^{2}(v),..,T^{k-1}(v)\right\}$ is a basis of the cyclic space generated by $v$.

If I consider the set of eigenvectors $\left\{u_1,u_2,...,u_n \right\}$, the vector $y=\displaystyle\sum_{j=1}^{n}u_j\in V$.

We claim that the set $\left\{y,T(y),T^{2}(y),..,T^{n-1}(y) \right\}$ is linearly independent.

Let $\lambda_1y+\lambda_2T(y)+\lambda_3T^{2}(y)+...+\lambda_nT^{n-1}(y)=0$, but, is my claim true? I have no idea how can I prove.

Since they correspond to distinct eigenvalues, the set of vectors $\left\{u_1,u_2,...,u_n \right\}$ is linearly independent, and thus we can use it as a basis for $V$. With respect to this basis, $y=[1,1,\ldots,1]^t$, $T(y)=[\lambda_1,\lambda_2,\ldots,\lambda_n]^t$, $T^2(y)=[\lambda_1^2,\lambda_2^2,\ldots,\lambda_n^2]^t$, $\ldots$, $T^{n-1}(y)=[\lambda_1^{n-1},\lambda_2^{n-1},\ldots,\lambda_n^{n-1}]^t$. And now Vandermonde's determinant proves that they are linearly independent.