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 you have distinct eigenvalues, you have a complete eigenbasis, which I will denote as $\{ v_1, v_2, v_3 \}$
Claim: $ v = v_1 + v_2 + v_3 $ works.
Complete this. Show that $av + bTv + cT^2 v = 0 \Leftrightarrow a, b, c = 0 $.
Hint: You have a quadratic equation that is satisfied by 3 distinct values.
