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 $n=3$.

  • 4
    $\begingroup$ Looks to me as if you’re done. For (i) you could also note that $B-\lambda I$ sends $e_3$ to $e_2$, $e_2$ to $e_1$, and $e_1$ to $0$, so $(B-\lambda I)^2$ collapses everything to multiples of $e_1$, and $(B-\lambda I)^3$ kills off everything. $\endgroup$ Nov 17, 2012 at 7:04
  • 3
    $\begingroup$ A useful thing to remember for matrices like these (Called Jordan Blocks by the way) is that the level of their nilpotence is based on the location of the diagonal of $1$s; each power will move the diagonal up one level. $\endgroup$
    – EuYu
    Nov 17, 2012 at 7:06
  • $\begingroup$ Very nice, thank you all! $\endgroup$
    – drawar
    Nov 17, 2012 at 7:20

1 Answer 1


When dealing with Jordan blocks, it helps to keep in mind the shift operator $S$ which sends $e_j$ to $e_{j-1}$ when $j>1$, and kills $e_1$. Every Jordan block is the sum of a scalar operator with a shift. In your case, $B=\lambda I+S$.

i) The $k$th power of $S$ is the shift by $k$ indices: $S^k$ sends $e_j$ to $e_{j-k}$ when $j>k$, and kills $e_j$ otherwise. This makes it identically zero once $k$ reaches the dimension of the space, which is $3$.

ii) nothing to add here: a nullspace of any linear operator is a linear subspace

iii) is indeed just i) in another language, with same answer.


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