# Find a vector $v$ such that $V=\mathbb{C}[T]\cdot v.$

Let $V=\mathbb{C}^2$ and let $\alpha:V\to V$ be the $\mathbb{C}$-linear map given by \begin{equation*} \begin{pmatrix} a\\ b \end{pmatrix} \longmapsto \begin{pmatrix} 2 & 3\\ -3 & 8 \end{pmatrix} \cdot \begin{pmatrix} a\\ b \end{pmatrix}. \end{equation*} Consider $V$ as a $\mathbb{C}[x]$-vector space with scalar multiplication $$\left(\sum_{i=0}^nc_ix^i\right)\cdot v\doteq\sum_{i=0}^nc_i\alpha^i(v),$$ where $\alpha^i$ denoted the $i$-times composition of $\alpha$. Find a vector $v\in V$ such that $V=\mathbb{C}[x]\cdot v.$

I'm not sure where to start. I suppose the problem is asking for a generator for $V$ as a $\mathbb{C}[x]$-vector space, but where I get confused is that $V=\mathbb{C}^2$ has rank $2$. How do I find a single generator? Any help/ hints are greatly appreciated.

• The polynomial ring $\;\Bbb C[x]\;$ is not a division ring, much less a field. I think the intention here is to give $\;V\;$ as module over $\;\Bbb C[x]\;$, not a vector space, which is a term usually reserved when working over fields or at least division rings. . – DonAntonio Apr 16 '16 at 22:39
• I believe the right-hand side term in the definition of your scalar multiplication should $\sum_{i=0}^nc_i\alpha^i(v)$. – Arnaud D. Apr 17 '16 at 9:27
• @ArnaudD. Yes, it should. My mistake. – Jfemdl Apr 18 '16 at 0:18

One way to do this is to take a vector $v$ such that $v$ and $\alpha(v)$ are linearly independent. Indeed, in this case they form a basis of $V$, so that for any $w\in V$ there exist unique $c_0,c_1\in \mathbb{C}$ such that $w=c_0v+c_1\alpha(v)$. Thus $$w=c_0v+c_1\alpha(v)=(c_0+c_1x)\cdot v\in \mathbb{C}[x]\cdot v,$$which shows that $V=\mathbb{C}[x]\cdot v$.
Now finding a vector $v$ such that $v$ and $\alpha(v)$ are linearly independent is quite easy; it suffices to take a vector that is not an eigenvector, so almost any vector will do the job.