# proof about commutative operators and T-cyclic vectors

Let $V$ be a finite dimensional vector space over $F$. Let $T:V \to V$ be a linear operator. Prove that if every linear operator $U$ which commutes with $T$ is a polynomial of $T$, than $T$ has a $T$-cyclic vector.

I don't really know where to start... can someone please point me in the right direction?

With some general theory this can be done quite painlessly. By the structure theorem for finitely generated modules over a principal ideal domain (with $F[X]$ as PID, and $V$ considered as $F[X]$-module with $X$ acting as $T$), $V$ decomposes as a finite direct sum of cyclic modules; the decomposition is not unique, but the number and types of the modules are unique (and so $V$ has a cyclic vector if and only if there is (at most) one cyclic module in such a decomposition). The cyclic factors are either isomorphic to $F[X]$ as module over itself (but this cannot happen for $V$ because it is finite dimensional over $F$), or of the form $F[X]/P$ for a non-constant monic polynomial $P\in F[X]$; moreover these polynomials $P_i$ in the successive cyclic modules (called the invariant factors of the $F[X]$ module $V$) divide each other: $P_i\mid P_{i+1}$ whenever both polynomials exist.

Now the question is to show that if all linear operators commuting with $T$ are polynomials in $T$ (in other words they are just scalar multiplications in the $F[X]$-module, where "scalar" means element of $F[X]$) then there cannot be two or more cyclic modules in the decomposition (the converse is also true, see this question). So assume a decomposition given by the theorem has at least two factors $F[X]/P_1$, $F[X]/P_2$, with $1\neq P_1\mid P_2$. Since the factors in the decomposition are submodules, every scalar multiplication stabilises them; it will suffice to find a $F[X]$-module morphism $V\to V$ that does not stabilise all factors (the module morphism property says that it is $F$-linear and commutes with $T$). But we have a non-zero morphism $\pi:F[X]/P_2\to F[X]/P_1$ of reducing modulo $P_1$, which is defined because $P_2$ is a multiple of $P_1$. To make this into a $F[X]$-module endomorphism $V$, it suffices to first project $V\to F[X]/P_2$ parallel to all other cyclic factors, then apply$~\pi$, and finally inject the factor $F[X]/P_1$ back into $V$. The resulting endomorphism is easily seen to not stabilise the cyclic factor $F[X]/P_2$, as desired. QED

To my understanding this is the so called cyclic vector theorem. There are many versions of the proof accessible. One quite comprehensive comes here:

http://planetmath.org/proofofcyclicvectortheorem

Would hardly make sense to copy down all to here.

Hope this helps.

• The reference is quite hard to read, though basically correct. The crucial definition of $g$ needs repair though: the left hand side should be $g(f^{k_i}(v_i))$, and the first clause in the right hand side deals with $i=1$. Also note that the cyclic decomposition theorem used has reverse divisibility relation among the annihilator polynomials, so that the first factor, on which $g$ vanishes, has the largest of the annihilator polynomials, namely the minmal polynomial of $T$. Jun 13, 2013 at 10:26

Here is another answer, less general that the one I gave five years ago as I am going to assume $F$ is algebraically closed (for instance $F=\Bbb C$), but also using less advanced theory. I will show that if every linear operator that commutes with $T$ is a polynomial in $T$, then $T$ has a cyclic vector.

Since $F$ is algebraically closed, $V$ decomposes into a direct sum of generalised eigenspaces, which are of the form $V_i=\ker((T-\lambda_i I)^{m_i})$, taken for the distinct eigenvalues $\lambda_i$ of$~T$ (the exponents $m_i$ just need to be big enough; taking the algebraic multiplicity of $\lambda_i$ or even just the fixed value $n=\dim(V)$ for them will do).

A first reduction is to show that if the restrictions to $T$ to each $V_i$ all admit a cyclic vector, then so does $T$; this will allow us to then reason for each of these restrictions separately. To prove this point, view $V=V_1\oplus V_2\oplus\cdots\oplus V_k$ as a Cartesian product, and taking a cyclic vector $v_i$ in each summand, form $v=(v_1,v_2,\ldots,v_k)$ which is going to be our cyclic vector. By hypothesis, the vectors $T^j(v_i)$ for $0\leq j<\dim(V_i)$ together span $V_i$. It suffices to show that for every $i,j$ there exists a polynomial $P_{i,j}$ such that $P_{i,j}[T](v)=(0,\ldots,0,T^j(v_i),0,\ldots,0)$ (with the nonzero component in position $i$), as then polynomials in $T$ applied to $v$ span everything. Now since the annihilator polynomial $(X-\lambda_i)^{m_i}$ of $V_i$ is relatively prime to the product $\prod_{l\neq i}(X-\lambda_l)^{m_l}$ of the annihilator polynomials of the other $V_l$, there exists by the Chinese remainder theorem a polynomial multiple $Q$ of that product that satisfies $Q\equiv 1\pmod{(X-\lambda_i)^{m_i}}$. Then $Q[T](v_l)=0$ for $l\neq i$ while $Q[T](v_i)=v_i$. But then taking $P_{i,j}=X^jQ$ does the job.

With that proven, focus on the restriction to a single $V_i$, or equivalently assume there is exactly only one eigenvalue$~\lambda$. We claim that our hypothesis of "commutation implies polynomial in $T$" forces the the Jordan decomposition to have only one block (equivalently, the eigenspace for$~\lambda$ must have dimension$~1$). For if there were more than one block (for$~\lambda$), take the first block to be at least as large as the second, and consider the linear operator that on the Jordan basis for this Jordan normal form vanishes on all basis vectors except those for the second block, which it sends to the corresponding basis vectors for the first block: this operator commutes with $T$, but is no polynomial in$~T$ (since those stabilise the individual blocks).

We are done: if $T$ has a single Jordan block, than it has a cyclic vector (the final basis vector of the Jordan basis).

• Professor, why is it ok to view $V$ as a Cartesian product of $V_1, \ldots, V_k$ given it is actually a direct sum of them? Jun 20, 2020 at 12:54
• @Zhanxiong: because the Cartesian product of finitely many vector spaces provides a model for their direct sum; is this news for you? I was not going into category theoretic finesses in this answer, just wanted to form a tuple of vectors in the components of the direct sum and consider their sum in the whole space. Jun 20, 2020 at 17:07
• I can intuitively understand this view. But I also felt like the direct sum of different cyclic subspaces will not be cyclic anymore...Is it possible to construct the vector $v$ so that it has the same dimension (length) as $v_1, \ldots, v_k$? I don't mind it will take a much longer writing. Jun 20, 2020 at 17:27
• My reasoning is based on that if $V_i$ is cyclic, then the matrix of $T|_{V_i}$ under the cyclic basis $B_i = \{v_i, T(v_i), \ldots, T^{\dim(V_i) - 1}(v_i)\}$ is a companion matrix. After combining $B_1, \ldots, B_k$, would the matrix under it still a companion matrix? Jun 20, 2020 at 17:33
• @Zhanxiong: I can see why you find this not very intuitive, but the same happens with groups: the direct product of cyclic groups can be cyclic. Indeed the Chines remainder theorem says that if a strong condition of relative primality is met, it will be cyclic. You won't (for obvious reasons) find a cyclic generator in any of the direct summands, but the sum of generators for each of the summands will be a generator globally. Jun 20, 2020 at 18:06

$$\newcommand{\complex}{\mathbb{C}}$$ $$\newcommand{\A}{T}$$ $$\newcommand{\B}{U}$$ $$\newcommand{\diag}{\mathrm{diag}}$$

When $$F = \complex$$, I will give a proof that is based on Jordan form theory.

Let $$\lambda_1, \lambda_2, \ldots, \lambda_t$$ be $$\A$$'s all distinct eigenvalues, and $$\A$$'s group of elementary divisors be \begin{align*} & (\lambda - \lambda_1)^{m_{11}}, (\lambda - \lambda_1)^{m_{12}}, \ldots, (\lambda - \lambda_1)^{m_{1k_1}}, \\ & (\lambda - \lambda_2)^{m_{21}}, (\lambda - \lambda_2)^{m_{22}}, \ldots, (\lambda - \lambda_2)^{m_{2k_2}}, \\ & \phantom{(\lambda - \lambda_1)^{m_{11}},} \ldots\ldots\ldots\ldots\ldots \\ & (\lambda - \lambda_t)^{m_{t1}}, (\lambda - \lambda_t)^{m_{t2}}, \ldots, (\lambda - \lambda_t)^{m_{tk_t}}, \end{align*} where $$m_{j1} \geq \cdots \geq m_{jk_j}, j = 1, 2, \ldots, t$$. Then there exists a basis $$\{\alpha_1, \ldots, \alpha_n\}$$ of $$V$$ under which $$\A$$'s matrix is $$J = \diag(J_{11}, J_{12}, \ldots, J_{1k_1}, \ldots, J_{t1}, J_{t2}, \ldots, J_{tk_t})$$, where $$J_{jl}$$ is the Jordan block associated with the elementary divisor $$(\lambda - \lambda_j)^{m_{jl}}$$, $$l = 1, 2, \ldots, k_j, j = 1, 2, \ldots, t$$.

If $$\A$$ is not a cyclic operator, then there exists at least one $$j \in \{1, 2, \ldots, t\}$$ such that $$k_j \geq 2$$. Otherwise, $$\A$$'s characteristic polynomial and minimal polynomial are identical, which would imply $$\A$$ is a cyclic operator. Without loss of generality, we can assume $$j = 1$$. Suppose $$C = \begin{pmatrix} 0_{m_{12} \times {(m_{11} - m_{12}})} & I_{(m_{12})}\end{pmatrix}$$ and define $$\begin{equation*} \tilde{J} = \diag\left(\begin{pmatrix}0 & 0 \\ C & 0 \end{pmatrix},\underbrace{0, \ldots, 0, \ldots, 0, \ldots, 0}_{k_1 + \cdots + k_t - 2 \text{ blocks}}\right). \end{equation*}$$

It is easy to verify that $$J$$ commutes with $$\tilde{J}$$. Define $$\B: V \to V$$ by $$\B(\alpha_1, \ldots, \alpha_n) = (\alpha_1, \ldots, \alpha_n)\tilde{J}$$, then $$\begin{equation*} \A\B(\alpha_1, \ldots, \alpha_n) = (\alpha_1, \ldots, \alpha_n)J\tilde{J} = (\alpha_1, \ldots, \alpha_n)\tilde{J}J = \B\A(\alpha_1, \ldots, \alpha_n), \end{equation*}$$ i.e., $$\B$$ commutes with $$\A$$. On the other hand, for any polynomial $$f$$, $$\begin{equation*} f(J) = \diag(f(J_{11}), f(J_{12}), \ldots, f(J_{1k_1}), \ldots, f(J_{t1}), \ldots, f(J_{tk_t})) \end{equation*}$$ does not equal to $$\tilde{J}$$ because the $$(2, 1)$$ block of $$f(J)$$ is zero while the $$(2, 1)$$ block of $$\tilde{J}$$ is $$C \neq 0$$. Therefore, $$\B$$ cannot be expressed as a polynomial in $$\A$$, contradiction. This completes the proof.

• $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix}$ has a cyclic vector and commutes with $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$ which are polynomials in the operator.

• $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$ does not have a cyclic vector. It commutes with the same projections, but they are not polynomials in this second operator.