Let T : $R^3 → R^3$ be the linear transformation defined by $T(x, y, z) = (0, 2x, 2y)$. Show that T is nilpotent of index 3. Find a vector v such that $\{v, T v, T^2v \}$ is a basis for $\Bbb R^3$.
I am struggling with this question i looked over my notes and i google it to understand how to solve it but no result
$T^3(x,y,z) = T^2(T(x,y,z)) = T^2(0,2x,2y) = T(T(0,2x,2y)) = T(0,0,4x) = (0,0,0) = 0\in\mathbb{R}^3$, for your first question.
Based on the above calculation, you just need to find $v=(x,y,z)$ such that $\{(x,y,z),(0,2x,2y),(0,0,4x)\}$ forms a basis for $\mathbb{R}^3$.