I'm a bit panicky for my exam soon, would appreciate any help with this:
I'm not sure how to start it or think about it, but I'll leave my thoughts (because I literally am unsure how to start them). Sorry if it's a long question, would appreciate any hints/guide at all.
Question
let $B \in M_n(\mathbb{C})$ satisfy
$$\mathrm{nullity}(B^{m-1})<n \qquad \text{and} \qquad \mathrm{nullity}(B^m) = n$$
for some positive integer $m$. Fix $v\in M_{n,1}(\mathbb{C})-\ker(B^{m-1})$.
i) Prove that $0$ is the only eigenvalue of $B$.
ii) by considering the Jordan Form of $B$, or otherwise, prove that $B+I$ has determinant $1$.
(iii) Prove that $B^k v \in \ker(B^{m-k})-\ker(B^{m-k-1})$ for $k=1,\ldots,m-1$.
My attempts:
(i)Why is this true? I thought this would only be true if we are given that $$\mathrm{nullity}(B^n) = n.$$
(ii) This one I got up to (assuming (i)) is correct:
Let $B = PJP^{-1}$. Then $B+I = PJP^{-1} + I = PJP^{-1} + PIP^{-1} = P(J+I)P^{-1}$.
Which means $\det(B+I) = \det(J+I)$.
Edit: I just had a eureka moment; it'll equal $1$ because $\det(J)$ is the diagonal entries multiplied, and so $\det(J+I)$ is $(0+1)^{n} = 1$?
(iii) This one I have no clue how to start
Let's start with the good news: your Eureka-moment insight on ii) is correct.
About i): Note that $\dim \ker B^m = 0$ means that $B^m$ is the all zero matrix. The easiest way to think about i) is its contra-positive: IF there is an eigenvalue unequal to zero, THEN the number $m$ with property $B^m = 0$ cannot exist. I could tell you more, but maybe this hint is enough?
about iii): it is really two statements: 1. $B^kv \in \ker B^{m - k}$ and 2. $B^kv \not\in \ker B^{m - k-1}$. Just prove them one at the time.
What helps is to realize what these staments actually mean: for 1. this is $B^{m-k}B^kv = 0$ and for 2. you can write something similar.
Now note that that the definition of $v$ also falls apart in two statements of this type.
I hope this helps.