Finding diagonalizable matrix's basis If $T: R^3 \rightarrow R^3$ is diagonalizable and has distinct eigenvalues, find $0 \neq v \in R^3$ such that
{$v,T(v), T^2(v)$} is a basis for $R^3$.  
I think that statement: {$v,T(v), T^2(v)$} is a basis for $R^3$, is obvious by theorem, but don't know what the question asks to find. It means that I really have to find exact value of $v$, $v=(1,0,0)$ something like that?

Since $T$ is diagonalizable, let $T=diag (t_1, t_2, t_3)$. Suppose $v=(v_1, v_2, v_3)\in R^3$ is a desired vector, then 
$$
\det 
\pmatrix{
v_1& v_2& v_3\\
t_1v_1& t_2v_2& t_3v_3\\
t_1^2v_1& t_2^2v_2& t_3^2v_3
}=v_1v_2v_3\text{Vandermonde matrix}
$$
so $v_1v_2v_3\ne 0$ and $t_1, t_2, t_3$ are distinct.
