Let $B = \left( {\begin{array}{*{20}{c}} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \\ \end{array}} \right)$ where $\lambda \ne0$.
(i) Find the smallest positive integer $k$ such that ${(B - \lambda I)^k} = 0$.
(ii) Explain why for every $n \geq 1$, ${V_n} = \left\{ {v \in {R^3}|{{(B - \lambda I)}^n}v = 0} \right\}$, $V_{n}$ is a subspace of $R^3$.
(iii) Find the smallest positive integer $n$ such that $V_{n}=R^3$.
Here are my thoughts so far, I think I'm halfway to the final answer, just need some help and double-check on this, thanks!
For part (i), I could do nothing but to try plugging $k=1$ up to $k=3$ and got $k=3$ as the answer. For part (ii), since $V_{n}$ is the nullspace of $(B-\lambda I)^n$,$V_{n}$ is a subspace of $R^3$. For part (iii), $V_{n}=R^3$ means the nullity of $(B-\lambda I)^n$ equals 3, thus its rank is 0. This happens iff $(B-\lambda I)^n$ is the zero matrix, hence it follows from (i) that $n=3$.
