It is a given problem of the book LINEAR ALGEBRA by Hoffman & Kunze (page 96)
Let $V$ be an $n$- dimensional vector space over the field $F$, and let $B=\{\alpha_1,\alpha_2,...,\alpha_n\}$ be an ordered basis for $V$.
(a) $\exists$ a unique linear operator $T$ on $V$ such that $T\alpha_j =\alpha_k,k=j+1, j=1,...,n-1, T\alpha_n =0$. What is the matrix $A$ of $T$ in the ordered basis $B$ ?
(b) Prove that $T^n=0$ but $T^m \ne 0,m=n-1$
(c) Let $S$ be any linear operator on $V$ such that $S^n=0$ but $S^m \ne 0 ,m=n-1$ .Prove that there is an ordered basis $B^*$ for $V$ such that the matrix of $S$ in the ordered basis $B^*$ is the matrix $A$ of part (a).
my try : (a) and (b) is easy to compute. my problem is to prove part (c). for simplicity I have assumed $n=2$. We know that the matrix with respect to the standered ordered basis of a given linear map treats as the same linear map.So I name that matrix $B$ and $B= \{\begin {bmatrix} a&b\\ c&d\\ \end{bmatrix}\}$
Since $S^2=0 , S\ne 0$ the rank of $S$ is one.So I assume $c=\alpha a, d=\alpha b$. Let $X=$[1 0]$\in R^2$ then $B^2X=0$ gives $\alpha= -a/b$.
Thus we get $S(x,y)=xS(1,0)+yS(0,1)=x(a ,\alpha a)+y(b,\alpha b)=(ax+by)(1,-a/b)$. I take $B^*=$$\{\beta_1=(0,1);\beta_2=(b,-a)\}$ then $S\beta_1=\beta_2$ and $S\beta_2=0$. And hence we get the matrix of $S$ with respect to $B^*$ ,which is $\{\begin {bmatrix} 0&0\\ 1&0\\ \end{bmatrix}\}=A$.
So we are done, assuming $n=2$ but how to do the general case. It seems to me very difficult in $n$ dimensional case if I generalise my try.Please someone give me a hint to solve part (c).Thanks for reading.
Note that $\{0\}\varsubsetneq\ker S\varsubsetneq\ker S^2\varsubsetneq\cdots\varsubsetneq\ker S^{n-1}\varsubsetneq\ker S^n=V$. So, take an element of $\ker S\setminus\{0\}$, then an element of $\ker S^2\setminus\ker S$ and so on.