I'm trying to make my way through two problems in Curtis's Linear Algebra, chapter 25. One of the two problems is this one, #5:
Prove that $V$ is cyclic relative to a linear transformation $T \in \mathcal{L}(V)$ if and only if the minimal polynomial of $T$ is equal to the characteristic polynomial.
Part of the issue is that I'm not really sure what it's asking. The statement that $V$ is cyclic seems to suggest that the entire vector space (say it has dimension $n$) is generated by some vector $v \in V$, along with $Tv, T^2v, \ldots T^{n-1}v$; Curtis's definition of a cyclic subspace (as opposed to a space, though I can't see why there should be a difference) seems to reinforce this:
A $T$-invariant subspace $V_1$ of $V$ is called cyclic relative to $T$ if $V_1 \neq 0$ and there exists a vector $v_1 \in V_1$ such that $V_1$ is generated by $\{ v_1, Tv_1, T^2v_1, \ldots T^kv_1 \}$ for some $k$.
However, I've seen several examples that seem to directly contradict this interpretation. For example, we recently had a take-home midterm that required we find elementary divisors, rational canonical form, etc., of a $6\times 6$ matrix; a classmate who received a perfect score on her test found that her matrix's characteristic polynomial was equal to its minimal polynomial — it was $(x+8)^2 (x+14)^2 (x^2+61x+80)$ — and yet the vector space could be written as $\left< v_6 \right> \oplus \left< v_4 \right> \oplus \left< -v_2+v_3+v_6 \right>$ — i.e., as the direct sum of three cyclic subspaces, not as one cyclic subspace.
So that suggests that "$V$ is cyclic" might mean the direct sum of cyclic subspaces. However, while working on the exercise that follows (#6), I came across this:
$$S= \left(\begin{array}{ccc} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 2 \end{array}\right)$$
Here, the characteristic polynomial is not equal to the minimal polynomial, which is $m(x) = (x-1)(x+1)(x+2)$. The eigenspace associated with $\lambda = 1$ has dimension $2$. And so it seems as though, if $V$ has basis $\{ v_1, v_2, v_3, v_4 \}$, we can write $V = \left< v_1 \right> \oplus \left< v_2 \right> \oplus \left< v_3 \right> \oplus \left< v_4 \right> $. This would seem to contradict my second interpretation of the original question.
Obviously I'm missing something. Could someone fill me in on what it might be? (I've tried to be thorough, but I'm also trying not to make this ridiculously long; if I've left out necessary information, I'll be happy to append it.)