1. ## Quick basis question

Suppose V is a vector space of dimension n, that T is a linear transformation on V and that there exists $v \in V$ such that $\{v, Tv, ..., T^{n-1}(v)\}$ is a basis of V.

Then clearly $T^{n}(v)$ can be written as a linear combination of the basis elements, but can we find this linear combination explicitly? I suspect not...

2. Originally Posted by Amanda1990
Suppose V is a vector space of dimension n, that T is a linear transformation on V and that there exists $v \in V$ such that $\{v, Tv, ..., T^{n-1}(v)\}$ is a basis of V.
So that T has rank n and is an invertible matrix.

Then clearly $T^{n}(v)$ can be written as a linear combination of the basis elements, but can we find this linear combination explicitly? I suspect not...
This would depend on the "characteristic equation" of T, [tex]|T- \lambda I|= 0[/itex] Since every linear transformation satisfies its own characteristic equation, that gives an $n^{th}$ degree polynomial for T and that can be solved for $T^n$. That's not quite what you are asking but it's the best I can think of.

3. Originally Posted by Amanda1990
Suppose V is a vector space of dimension n, that T is a linear transformation on V and that there exists $v \in V$ such that $\{v, Tv, ..., T^{n-1}(v)\}$ is a basis of V.

Then clearly $T^{n}(v)$ can be written as a linear combination of the basis elements, but can we find this linear combination explicitly? I suspect not...
Here's a simple example, with n=3. Let T be the linear transformation of $\mathbb{R}^3$ given by the matrix $\begin{bmatrix}0&0&\alpha\\ 1&0&\beta\\ 0&1&\gamma\end{bmatrix}$. Let $e_1,\;e_2,\;e_3$ be the vectors $\begin{bmatrix}1\\0\\0\end{bmatrix}$, $\begin{bmatrix}0\\1\\0\end{bmatrix}$, $\begin{bmatrix}0\\0\\1\end{bmatrix}$ in the standard basis. If $v = e_1$ then $Tv=e_2$ and $T^2v=e_3$. So $\{v, Tv,T^2(v)\}$ is a basis. But $T^3v = Te_3 = \begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}$, which is an arbitrary vector. So you are right to think that there is no explicit way to determine it.