Show that $T$ is nilpotent operator

Question

Let $T$ be a linear operator defined on a n dimensional vector space $V$. Suppose there exists a vector $v$ such that $\beta=\{v,Tv,…,T^{n-1}v\}$ is linearly independent, show that $T$ is nilpotent.

By given assumption $\beta$ is a basis for V. I tried computing matrix of $T$ with respect to $\beta$ but could not compute last column of matrix. Would it be the case that $T^nv=0$?

Answer 1

It's false. Consider the linear transformation which interchanges the two standard basis vectors.

See companion matrices.