Let $A$ be a matrix. If $A^{n-1} \neq \bf{0}$ but $A^{n} = \bf{0}$, what does it say about the matrix $A$? 


Since the question doesn't say find all matrices $A$, I will only produce one example for both:
a. $A = \begin{bmatrix}0 & 1\\0&0\end{bmatrix}$
b. $A = \begin{bmatrix}0 & 0 &0\\1&0 & 0\\0&1&0\end{bmatrix}$ from Find a matrix so that $A^2$ not equal to 0 but $A^3$ is [Strang P78 2.4.23]
Just based off of these two, I would conjecture that


*

*Matrices $A$ satisfying $A^{n-1} \neq \bf{0}$ but $A^{n} = 0$ must be such that they are of dimension $n$
However, there is a matrix $A \in \mathbb{R}^{3\times3}$ that satisfies $A\neq \bf{0}$ but $A^2= \bf{0}$, namely
$$\begin{bmatrix}0&0&0\\1&0&0\\0&0&0\end{bmatrix}$$ 
The next thing I could think of was:


*

*Matrices $A$ satisfying $A^{n-1} \neq \bf{0}$ but $A^{n} = 0$ must be such that they have a dimension of at least $n$
But I don't know if that's it or not...

Question(s):


*

*What was I meant to conjecture?

 A: The author concurs with your conjecture. He was hoping that you'd realize that a matrix with $A^n=O$ but $A^{n-1}\ne O$ must be of size at least $n\times n$. And of course you were expected to find examples, not find all. (Perhaps for $2\times 2$ matrices you should try to find all?)
What you might also notice is that you can have any number of $1$'s on the sub- (or super-)diagonal ... in order to get specific examples. You have $3\times 3$ examples with one $1$ and with two $1$'s.
But you would get credit for a correct answer in the author's view. :)
A: We can generalize your answers to (a), (b) by saying an $n\times n$ matrix $A_{ij}=\delta_{i,\,j-1}$ satisfies $A^n=O\ne A^{n-1}$. Indeed, defining a basis $(e_i)_j=\delta_{ij}$, $(Ae_i)_j=A_{jk}\delta_{ik}=A_{ji}=(e_{i-1})_j$, so repeatedly applying $A$ to the basis elements moves backwards through it until we turn $e_1$ into the zero vector.
But you've also conjectured no smaller matrices satisfy $A^n=O=A^{n-1}$. One way to see this is to note that a $k\times k$ choice of $A$ with $k<n$ has a degree-$k$ characteristic polynomial that would, if $A^n=O$, need to divide $\lambda^n$ by the Cayley-Hamilton theorem, so it must be of the form $\lambda^k$. But then $A^k=O$ by said theorem, whence $A^{n-1}=O$, a contradiction.
