Jordan Form of a Unipotent Operator Induced on a Quotient Space

Question

Let $K$ be an algebraically closed field, let $V$ be a finite dimensional $K$-vector space, and let $u:V \rightarrow V$ be a unipotent operator. For each $j \in \mathbb{Z}^+$, let $c_j$ be the number of Jordan blocks of dimension $j$ in the Jordan form of $u$. Let $F_1$ be a 1-dimensional, $u$-invariant subspace. We can induce a unipotent operator $u'$ on $V/F_1$. The statement I am trying to prove is the following: There is some $i \in \mathbb{Z}^+$ such that the number of Jordan blocks of $u'$ of dimension $i$ is $c_i - 1$, the number of Jordan blocks of $u'$ of dimension $i - 1$ is $c_{i-1} + 1$, and the number of Jordan blocks of $u'$ of dimension $j \neq i$ is $c_j$.

The way I would think to go about this is to show that for any eigenvector $v$ of $u$ (in particular, a generator of $F_1$), we can find a Jordan basis that contains $v$. If this is in fact true, then the statement I am trying to prove above follows. However I am not sure that it is true. Any comments on proving this or an alternate way to prove the above statement would be helpful.

Answer 1

It is indeed the case that for any eigenvector $v$ of $u$, there exists a Jordan basis containing $v$. One way to prove this is using the following result from Hoffmann and Kunze's Linear Algebra (p. 233 in the second edition)

Theorem 3: (Cyclic Decomposition Theorem) Let $T$ be a linear operator on a finite-dimensional vector space $V$ and let $W_0$ be a proper $T$-admissible subspace of $V$. There exist non-zero vectors $\alpha_1,\dots,\alpha_r$ in $V$ with respective $T$-annihilators $p_1,\dots,p_r$ such that

1. $V = W_0 \oplus Z(\alpha_1;T)\oplus \cdots \oplus Z(\alpha_r; T)$;
2. $p_k$ divides $p_{k-1}$, $k = 2,\dots,r$.

Furthermore, the integer $r$ and polynomials $p_1,\dots,p_r$ are uniquely determined by (i), (ii), and the fact that no $\alpha_k$ is $0$.

$Z(v;T)$ denotes the cyclic subspace generated by $v$, i.e. the span of all vectors of the form $T^k v$ for $k = 0,1,2,\dots$. $T$-admissible means the following:

A subspace $W$ of $V$ is $T$-admissible if it is invariant under $T$ and for any vector $v \in V$ and polynomial $f$, if $f(T)v \in W$, then there exists a vector $w \in W$ such that $f(T)v = f(T)w$.

One possible proof using the above theorem is as follows. First of all, I will consider the operator $T = u - \operatorname{id}_V$ rather than $u$ itself, since $T$ is nilpotent and $u$ and $T$ have the same Jordan bases.

Begin by constructing the following linearly independent set $S$:

• Begin with $S = \{v_1\}$ with $v_1 = v$.
• Beginning with $i = 1$, either there exists a vector $x$ such that $T(x) = v_i$ or no such vector exists. If no such vector exists, then the process terminates and we have our set. Otherwise, add the vector $v_{i+1}$ to the set $S$ and repeat this step using $i + 1$ instead of $i$.

The result will be a Jordan chain $S = \{v_1,\dots,v_j\}$. Let $W$ denote the span of this set $S$. Argue that $S$ is linearly independent and that $W$ is a $T$-admissible subspace. Now, there are two cases: if $W = V$, then $S$ is a Jordan basis of $V$ containing $v$, so we are done. Otherwise, the above version of the cyclic decomposition theorem implies that there exist vectors $\alpha_1,\dots,\alpha_r$ such that $$ V = W_0 \oplus Z(\alpha_1;T)\oplus \cdots \oplus Z(\alpha_r; T). $$ For each $k = 1,\dots,r$, the (ordered) set $\{\alpha_k, T\alpha_k,\dots, T^{\deg(p_r)-1}\alpha_k\}$ is a Jordan chain whose span is the subspace $Z(\alpha_k;T)$. Putting $S$ together with these Jordan chains produces a Jordan basis of $T$ that contains the vector $v$, which was what we wanted.