Linear nilpotent functions Let $A \in M_{n}(K)$ be a nilpotent matrix, where $M_n(K)$ is the space of $n \times n$ matrices over the field $K$. Prove that the following statements are equivalent :
(i) $\mu _A = \chi _A (= \lambda^n).$ (The minimal and characteristic polynomials are equal.)
(ii) $A^{n-1}\neq 0. $
(iii) $\operatorname{rank} A =n{-}1.$ 
(iv) There exists a vector $x \in K^n$ such that $A^i x$, $i = 0, \dots, n-1$ is a $K$-basis of $K^n.$
I could prove the implications (i) $\iff$ (ii).
To prove that (ii) $\iff$ (iii), I have with me the following example:
Let $v_1, v_2, \dots, v_n$ be a basis of $K^n$. Define a $K$-linear map $f:K^n \to K^n$ where $f(v_i):=v_{i+1},\ i=1,...,n{-}1$ and $f(v_n)=0$. This is a nilpotent function with $f^{n-1}\neq 0$ and $\operatorname{rank} f=n{-}1$.
Now the question is: Can any nilpotent matrix with rank $n-1$ be expressed as the above function by change of basis? How can I prove that such a basis exists? If I can do this, the proof follows. Or else please suggest other ways to prove the equivalence of the statements.
 A: Let me prove (ii) $\Leftrightarrow$ (iv):
Assume $A^{n-1}\ne 0$. Then there is $x$ with $A^{n-1}x\ne0$. The vectors $x, Ax, \dots , A^{n-1}x$ are linearly independent: Assume there are coefficients $a_i$ such that
$$
a_0 x + a_1 Ax + \dots a_{n-1}A^{n-1} x =0.
$$
Let $i$ be the smallest index such that $a_i\ne0$. Multiplying the equation above by $A^{n-1-i}$ yields $a_i A^{n-1}x=0$, hence $a_i=0$. Thus, $a=0$ and the vectors are linearly independent (and a basis of $K^n$).
Assume that $x,Ax,\dots A^{n-1}x$ is a basis of $K^n$. Then $A^{n-1}x\ne0$, implying $A^{n-1}\ne0$.
The same argument also proves (ii) $\Rightarrow$ (iii): Applying $A$ to the basis vectors $x,Ax,\dots, A^{n-1}x$ shows that the image of $A$ is spanned by $Ax,\dots,A^{n-1}x$, which are linearly independent. Hence rank of $A$ is $n-1$.
Let me prove (iii) $\Rightarrow$ (ii): By the assumptions, we have $\dim\ker(A)=1$. Then $\dim(\ker A^{k+1})-\dim(\ker A^{k})\le1$: Assume $A^{k+1}x=0$ but $A^kx\ne0$. Then $A^kx\in \ker(A)$.
Now consider the chain of inclusions
$$
\ker(A)\subset \ker(A^2)\subset \dots\subset \ker(A^n) =K^n.
$$
The dimension of these subspaces increase at most by one from one inclusion to the other. This implies $\dim(\ker A^k)=k$, and $A^{n-1}\ne0$.
