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.)

  • 1
    $\begingroup$ That a space can be written as a direct sum of some subspaces does not necessarily mean it can't be written as a cyclic subspace. I think this explains your confusion. $\endgroup$
    – Timbuc
    Nov 23, 2014 at 18:40
  • $\begingroup$ @Timbuc: Thanks for the hint. Are you saying, for example, that $V$ might be written as a direct sum of, say, two cyclic subspaces, but that one of those subspaces might be contained in the other? $\endgroup$
    – dmk
    Nov 23, 2014 at 18:50
  • $\begingroup$ @dmk: consider for instance the diagonal matrix of size $2$, with values $1$ and $2$. Can you prove that the whole space is cyclic for this matrix? $\endgroup$ Nov 23, 2014 at 18:53
  • $\begingroup$ @JeremyDaniel: I almost said no. The eigenvectors are $v_1$ and $v_2$, and we can't get from one to the other. Then I had a look at $v_1 + v_2$; its image is linearly independent. And so if by "cyclic" we mean the whole space is generated by one vector, then I think this proves it is :). The problem is, I'm not sure why I chose that linear combination besides the fact that I've seen similar choices made before. As for why one does that, no idea. In particular, the connection between what I just did and the eigenvectors eludes me. Can you recommend a source, online or otherwise? $\endgroup$
    – dmk
    Nov 23, 2014 at 19:11
  • $\begingroup$ I don't have a precise reference for this. Your first definition of cyclic is the good one; can you prove the proposition with that? It will help clarify things. $\endgroup$ Nov 23, 2014 at 19:24

1 Answer 1


Your confusion comes from the assumption that if a space is a direct sum of several cyclic subspaces, then it cannot at the same time be cyclic. This is wrong: as long as the cyclic subspaces have relatively prime minimal polynomials, then their direct sum will still be cyclic. To see this, consider a vector $v$ whose projections onto each of the cyclic summands is a cyclic generator for that summand. Applying a polynomial $P[T]$ in $T$ to $v$ will only result in $0$ if all the projections of$~v$ are simultaneously annihilated by $P[T]$, and this will only happen when $P$ is a common multiple of all those minimal polynomials. But by relative primality their least common multiple is their product, whose degree is the dimension of the space; it follows that $v$ is a cyclic generator for the whole space. This is quite similar to what happens with cyclic subgroups in an Abelian group (with the minimal polynomials playing the role of the orders of the subgroups).

An answer to the actual question can be found here.

Here is a direct argument. If $v$ is a generator for a cyclic space of dimension$~n$, no nonzero polynomial$~P$ of degree less than$~n$ will satisfy $P[T](v)=0$, so the characteristic polynomial of$~T$ will be its minimal polynomial.

Conversely suppose the characteristic and minimal polynomials of$~T$ are equal; we need to show that the space is cyclic. Decomposing this polynomial $P=\prod_iP_i^{m_i}$ as a product of powers of distinct irreducible monic polynomials$~P_i$, it will suffice by the above argument to show that each of the ($T$-stable) subspaces $V_i=\ker(P_i^{m_i}[T])$ is cyclic (since one has $V=\bigoplus_iV_i$; a vector in$~V$ all of whose components on the$~V_i$ are cyclic vectors for those subspaces will be cyclic for$~V$). From the fact that $P$ is the characteristic polynomial of$~T$, it follows that $\dim(V_i)=\deg(P_i^{m_i})$, while from the fact that $P$ is minimal polynomial it follows that $P_i^{m_i}$ is the minimal polynomial of the restriction of$~T$ to$~V_i$. In other words, the restriction of$~T$ to$~V_i$ has equal characteristic and minimal polynomials, namely $P_i^{m_i}$.

We must show the existence of a cyclic generator of$~V_i$, i.e., vector $v_i\in V_i$ such that the minimal monic polynomial$~Q$ such that $Q[T](v_i)=0$ has $\deg(Q)=\dim(V_i)$. As the minimal polynomial of $T|_{V_i}$ is $P_i^{m_i}$, this polynomial $Q$ divides $P_i^{m_i}$ for any vector$~v\in V_i$: only powers of $P_i$ can occur. It therefore suffices to find a vector$~v_i\in V_i$ for which $P_i^{m_i-1}[T](v_i)\neq0$. Since $\ker(P_i^{m_i-1}[T])\neq V_i$ by mentioned minimality, such a vector exists (indeed almost all vectors of $V_i$ are cyclic generators of$~V_i$, and almost all vectors of $V$ are cyclic generators of$~V$).

In $V_i$ you may find that a more pleasant basis than $\{\, T^k(v_i)\mid 0\leq k< dm_i\,\}$, where $d=\deg P_i$ the following one: $\{\, (X^kP_i^l)[T](v_i)\mid 0\leq k<d, 0\leq l<m_i\,\}$, and similarly you can find a more pleasant basic of$~V$ than the purely cyclic one of images of $T^k(\sum_iv_i)$ to be the "Cartesian product" of the basis of the individual $V_i$. But the cyclic basis is one (i.e., linearly independent) because $v$ has no monicannihilating polynomial of degree less than $~\dim V=\sum_im_i\deg P_i$.


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