# Is this $\mathbb{C}[t]$ module cyclic?

Let $T$ be the linear operator on $\mathbb{C}^2$ whose matrix is \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}

Is the corresponding $\mathbb{C}[t]$ module cyclic?

Since the characteristic polynomial is $(t-2)(t-1)$ which a relatively prime, I believe the module is isomorphic to $\mathbb{C}[t]/((t-2)(t-1))$ and thus is clearly cyclic.

However, this answer is dissatisfying. If I really want to show it's cyclic, I want to find some $v\in \mathbb{C}$ such that all of $\mathbb{C}^2$ can be written in the form $aTv+bv$ (I think I only need two terms since $\mathbb{C}^2$ is two dimension over $\mathbb{C}$). How should I find this $v$? What can I use from module theory to help?

Also, could you give a non-trivial example of a matrix for the operator $T$ in which the corresponding module would NOT be cyclic? It seems it should always be cyclic since $\mathbb{C}[t]$ is a PID.

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An element generates the module, iff it does not belong to a maximal non-trivial submodule.

Because your module is 2-dimensional (as a vector space), the non-trivial maximal submodules must be 1-dimensional, and hence need to be eigenspaces of $T$. So if you pick a vector $v$ that is not an eigenvector it will not belong to any proper submodule, and hence generates all of $\mathbb{C}^2$.

If your matrix were larger, say 3x3, and three distinct eigenvalues, then the non-trivial maximal submodules would be direct sums of two eigenspaces. If there are repeated eigenvalues, then it gets a bit more complicated.

To answer your last question. The simplest counterexample is $T$= zero matrix. For any vector $v\in\mathbb{C}^2$, the $\mathbb{C}[T]$-submodule generated by $v$ is simply $\mathbb{C}v$. Hence the module $\mathbb{C}^2$ is not cyclic in that case. The same holds, if we let $T$ be any scalar matrix. There are other examples, [edit:added] when the matrices are larger than 2x2 (see further down)[/edit], but they all involve repeated eigenvalues.

In your example case we can select $v=(0,1)^T$. Then $T\cdot v=(1,1)^T$. You see that these two vectors are linearly independent over $\mathbb{C}$, and therefore they form a vector space basis.

In 3x3 case you can let $$T=\pmatrix{1&1&0\cr 0&1&0\cr 0&0&1\cr}.$$ There are infinitely many maximal submodules $M$. We have $M_\infty$ spanned by $(1,0,0)^T$ and $(0,0,1)^T$ and $M_z$ spanned by $(1,0,0)^T$ and $(0,1,z)^T$, for any complex number $z$. The vector $(z_1,z_2,z_3)$ belongs to $M_\infty$, if $z_2=0$ and to $M_{z_3/z_2}$, if $z_2\neq0$. So any vector belongs to a non-trivial maximal submodule. Hence the module $\mathbb{C}^3$ is not cyclic.

The method of constructing non-cyclic modules in this way is to select $T$ in such a way that it has several Jordan blocks belonging to the same eigenvalue. Equivalently, the resulting module is cyclic, iff the characteristic and minimal polynomials of $T$ are equal. This follows from the theory of invariant factors. The minimal polynomial is a factor of the characteristic polynomial and has the same set of zeros, so for the two polynomials not to be equal it is necessary (but not sufficient) for $T$ to have repeated eigenvalues.

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I'm a little bit confused about your language: when you say the module is 2-dimensional, what do you mean since dimension doesn't apply to modules? For my second question, are you saying any case where there is a repeated eigenvalue (thus giving the characteristic polynomial the form $(t-\alpha)^2$ is the only one in which the corresponding module is not cyclic? – Steven-Owen Apr 29 '12 at 20:24
@jake, in this case the modules are also vector spaces over $\mathbb{C}$, and I talked about dimensions as a vector space. Sorry about not making that clear right away. – Jyrki Lahtonen Apr 29 '12 at 20:28
Characteristic polynomial of the form $(t-\alpha)^2$ does not guarantee that the module is not cyclic. For example, with $$T=\pmatrix{\alpha&1\cr0&\alpha\cr}$$ we do get a cyclic module generated by $v=(0,1)^T$. In this case the minimal polynomial is also $(t-\alpha)^2$. We need the minimal polynomial to be of a smaller degree than the characteristic polynomial in order to get a non-cyclic module. – Jyrki Lahtonen Apr 29 '12 at 20:32

If you write your matrix in rational canonical form, the first basis vector will do what you want. So, conjugate your matrix till it's in RCF (there's an algorithm for this) then conjugate the first basis vector backwards.

In this case it's easier to guess and check: the second basis vector is already a cyclic vector.

For your second question, in the 2 by 2 case there's not really a non-trivial example. In higher dimensions, take a matrix that's the block sum of the 2 by 2 identity matrix and some ugly matrix, then conjugate it by something disgusting until it's unrecognizable.

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If a finite dimensional $\mathbb C[t]$-module $M$ is cyclic, then in fact if you choose an element of $M$ at random, there are enormous chances that it will generate $M$.

Carry out this experiment: pick a random element of $\mathbb C^2$ (for example, by asking your mother and father for two numbers, and using them as the coodinates of the element) and see what submodule it generates. Unless your parents are trying hard to trick you —and good parents know better!— the submodule will be the whole of $\mathbb C^2$, showing that $\mathbb C^2$ is cyclic.

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