I am trying to work out problems from Linear Algebra, by Hoffman and Kunze and came across this problem in the exercise of Section 3.4, I have a difficulty solving the (c) part of the problem.
Problem 12
If $V$ is an $n$ – dimensional vector space over the field $F$, and let $B = \{\alpha_1,\alpha_2,\alpha_3, \cdots \alpha_n\}$ be an ordered basis for $V$.
(a) There is a unique Linear Operator $T$ on $V$ such that :
$$
\begin{align}
T(\alpha_j) = \alpha_{j+1}, \quad j=1, \cdots ,n-1 \quad T(\alpha_n)=0
\end{align}
$$
What is the matrix $A$ of $T$ in the ordered basis $B$?
(b) Prove that $T^n=0$, but $T^{n-1} \neq 0$
(c) Let $S$ be any linear operator on $V$ such that $S^n=0$, but $S^{n-1} \neq 0$ . 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).
(d) Prove that if $M$ and $N$ are $n \times n$ matrices $F$ such that $M^n = N^n = 0$, but $M^{n-1} \neq 0 \neq N^{n-1}$, then $M,N$ are similar.
However the solution for the problem (c) can be obtained assuming (d) is true, in the following way:
Let $[S]_B$ represent the matrix corresponding to the linear operator $S$ under some basis $B$, and we want to show that this operator has the matrix $[S]_{B^{'}}=A$ under some basis $B^{'}$.
If the two nilpotent matrices of order $n$ are similar, then there must be an invertible matrix $P$ such that $[S]_{B^{'}} = P^{-1}[S]_BP$. The matrix $P$ can be used to prove the existence of another basis $B^{'}$ such that:
$$
\begin{align}
[\alpha]_{B} = P[\alpha]_{B^{'}}
\end{align}
$$
and therefore, there indeed exists a basis $B^{'}$ such that
$$
\begin{align}
[S]_{B^{'}} = P^{-1}[S]_BP
\end{align}
$$
I am also having a problem solving part (d), however, I was wondering if one could prove (c) without using (d) explicitly ? I will be thankful for any hints to solve the problem!
Best Answer
Since $S^{n-1}\neq 0$, there exists a vector $\alpha_0$ (necessarily non-zero) such that $S^{n-1}(\alpha)\neq 0$. Consider the vectors $\alpha,S(\alpha),S^2(\alpha),\dots S^{n-1}(\alpha)$ They are all non-zero (or else the last one would be too by linearity of $S$). Moreover, they are linearly independent, for suppose there are scalars $a_k\in F$ so that $\sum_{k=0}^{n-1}a_kS^k(\alpha)=0$, where $S^0=I$. Applying $S^{n-1}$ we get $\sum_{k=0}^{n-1}a_kS^{k+n-1}(\alpha)=0$, so $a_0S^{n-1}(\alpha)=0$, because all the terms with $k+n-1\geq n$ vanish, and we are left just with the first one. Since $S^{n-1}(\alpha)\neq 0$, we must have $a_0=0$. So now we have $\sum_{k=1}^{n-1}a_kS^k(\alpha)=0$. Applying $S^{n-2}$, we get $a_1=0$, and continuing this way we get $a_0=a_1=\cdots = a_{n-1}=0$. Thus, the vectors $\alpha,S(\alpha),\dots S^{n-1}(\alpha)$ are linearly independent. Since $V$ is $n$-dimensional, these $n$ vectors form a basis for $V$. Furthermore, the matrix representing $S$ with respect to this basis satisfies our requirements, because $S(\alpha_k)=\alpha_{k+1}$ for $k=0,1,\dots, n-2$, and $S(\alpha_{n-1})=S^n(\alpha)=0$, which by part (a) is what it takes for the matrix representation of an operator to have the desired form.