I was reading "A Course in Linear Algebra" by David Damiano and John Little. The third paragraph of chapter 6.2 confused me. Let me quote:
Let $\text{dim}(V) = n$, and let $N: V \rightarrow V$ be a nilpotent mapping. We know that $N^n(\mathbf{x})=\mathbf{0}$. For a given $\mathbf{x}$, however, it may very well be true that $N^k(\mathbf{x}) = \mathbf{0}$ for some $k<n$. Now, for each $\mathbf{x} \in V$, either $\mathbf{x} = \mathbf{0}$ or there is a unique integer $k$, $1 \le k \le n$, such that $N^k(\mathbf{x}) = \mathbf{0}$, but $N^{k-1}(\mathbf{x}) \ne 0$. It follows then, that if $\mathbf{x} \ne \mathbf{0}$, then the set $\{N^{k-1}(\mathbf{x}), N^{k-2}(\mathbf{x}), ... , N(\mathbf{x}), \mathbf{x}\}$ consists of distinct nonzero vectors.
There are two things that I don't understand from this paragraph. First, why did it say that either $\mathbf{x} = \mathbf{0}$ or there must exists an integer $k$, $1 \le k \le n$ that $N^k(\mathbf{x}) = \mathbf{0}$ but $N^{k-1}(\mathbf{x}) \ne \mathbf{0}$? Is it not possible to have $N^k(\mathbf{x}) = \mathbf{0}$ for all $1 \le k \le n$ for the non-zero $\mathbf{x}$?
Secondly, I cannot see how the elements in the set $\{N^{k-1}(\mathbf{x}), N^{k-2}(\mathbf{x}), ... , N(\mathbf{x}), \mathbf{x}\}$ must be distinct? Is it some properties of matrix multiplications that makes $N^{k}$ ... $N^1$ to be distinct matrices?
Fix a vector $x\ne 0$.
The author's $k$ is the least positive integer such that $N^k(x)=0$.
Since it's the least, it's automatic that it's unique.
If $k=1$, then $N^{k-1}(x) = x$, so $N^{k-1}(x)\ne 0$.
If $k>1$, then $N^{k-1}(x)\ne 0$, by minimality of $k$.
Thus, in either case, we have $N^{k-1}(x)\ne 0$.
Now suppose $N^i(x) = N^j(x)$, for some $i,j$, where $0 \le i < j < k$.
By minimality of $k$, we have $N^j(x) \ne 0$.
But from $N^{i}(x)=N^{j}(x)$, it follows that the infinite sequence $$N^{i}(x), N^{i+1}(x), N^{i+2}(x),...$$ repeats in blocks of length $j-i$.
To see this in action, \begin{align*} &N^i(x) = N^j(x)\\[4pt] \implies\;&N^{i+1}(x)= N(N^i(x)) = N(N^j(x))=N^{j+1}(x)\\[4pt] \implies\;&N^{i+2}(x)= N(N^{i+1}(x)) = N(N^{j+1}(x))=N^{j+2}(x)\\[4pt] \implies\;&N^{i+3}(x)= N(N^{i+2}(x)) = N(N^{j+2}(x))=N^{j+3}(x)\\[1pt] &\;\;\;\vdots\\[4pt] \end{align*} so in general, for all $h \ge i$, we have $$ N^h(x)=N^{h+(j-i)}(x) \qquad\qquad\qquad\qquad\qquad\;\, $$ In particular, we get $$ N^i(x)=N^j(x)=N^{2j-i}(x)=N^{3j-i}(x) = \cdots \; $$ contradiction, since $N^h(x)=0$, for all $h\ge n$.
It follows that the $k$ vectors $$x,N(x),N^2(x),...,N^{k-1}(x)$$ are distinct.