Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n=3$.

share|improve this question
4  
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. –  Brian M. Scott Nov 17 '12 at 7:04
3  
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. –  EuYu Nov 17 '12 at 7:06
    
Very nice, thank you all! –  drawar Nov 17 '12 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.