Q:

Is this Proof Correct? (Order of a Jordan Block without Leading Diagonal)

I am wanting to prove that the smallest positive integer $p$ such that
$(J_r(\lambda) - \lambda I_r)^p = 0_{r\times r}$ is $r$, where $J_r(\lambda)$ is the Jordan Block of size $r\times r$ with eigenvalue $\lambda$.
I think I have a proof by induction but it seems so silly/easy that it can't be right?!

Induction Start: Consider
  $(J_2(\lambda) - \lambda I_2) =\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$.
  Sure enough, the
  smallest positive integer $p$ such that 
  $(J_2(\lambda) - \lambda I_2)^p = 0_{2\times 2}$ is $2$.
Induction Step: Suppose that the smallest integer $k > 2$ such that
  $(J_r(\lambda) - \lambda I_r)^k = 0_{r\times r}$ is $r$ (for $r > 2$). 
Then 
  $(J_r(\lambda) - \lambda I_r)^{k+1} = $
  $(J_r(\lambda) - \lambda I_r)^k (J_r(\lambda) - \lambda I_r) = $
  $0_{r\times r} (J_r(\lambda) -\lambda I_r) = 0_{r\times r}$. 
  Then the smallest positive integer $k+1$ such that
  $(J_r(\lambda) - \lambda I_r)^{k+1} = 0_{r \times r}$ is $r$.

Is there something wrong with this? Or is it good to go?

A:

No, this is incorrect. First of all, your induction structure would require you to prove something about powers of $(J_{r+1}(\lambda) - \lambda I_{r+1})$, which you haven't even written down. Second, when you write
$$
(J_r(\lambda) - \lambda I_r)^{k+1} = (J_2(\lambda) - \lambda I_2)^k (J_r(\lambda) - \lambda I_r),
$$
this is quite mistaken: even besides changing $r$ to $2$ without justification, the right-hand side is the product of a $2\times 2$ matrix with an $r\times r$ matrix, which is undefined in general. Indeed, for this last reason (the statement you want to prove for many $r$ involves matrices that don't play well with one another), induction seems like it would be very hard to carry out.
Suggestion: experiment with calculating the actual powers when $r=3$ and $r=4$. Can you spot a pattern?

