The question is as follows:
Suppose that $V$ is a vector space over $\mathbb{C}$ of dimension $3$. Fix a non-zero vector $v\in V$ and define $$U:=\{T\in\mathcal{L}(V):v\mbox{ is and eigenvector of }T\}.$$ Now fix an arbitrary basis $B$ of $V$. Show that there is a nonzero linear transformation $S\in U$ such that $$\mathcal{M}(S,B)=\begin{bmatrix}a&a&c\\b&a&c\\b&b&c\end{bmatrix}.$$
Note that the notation $\mathcal{L}(V)$ is used to denote the linear operators on $V$ and $\mathcal{M}(S,B)$ is the matrix of $S$ with respect to $B$. I'm not sure if this notation is conventional or not, and it's probably clear from context...
I am able to show that $U$ is a subspace of $\mathcal{L}(V)$ of dimension $7$, but beyond that I'm not sure how to approach this. I know a nonconstructive solution exists, but I keep finding myself actually trying to compute values of $a,b,c$ (which is a poor approach even without knowledge of a nonconstructive solution, since the basis is arbitrarily chosen). This is a homework question, so I am requesting hints, not a full solution.
Edit: Here is the solution presented in the comments below. My original solution is after this.
Note that $\dim V\ge\dim(U+W)=\dim U+\dim W-\dim(U\cap W)$. Let $W$ be the subspace of linear operators with matrices of the form of $M$. Then since $\dim V=9$, $\dim U=7$, and $\dim W=3$, it follows that $\dim(U\cap W)\ge1$. The desired result follows.
Observe that the dimension of the space of linear operators with matrices of the form of $M$ (with respect to $B$) is $3$. Since the dimension of the full space of $3\times3$ matrices is $9$, and $\mbox{dim }U=7$, by the pigeonhole principle, the desired result follows.