Find the dimension of the annihilator of a nonzero vector 
Let $V=L(\mathbb{R}^5,\mathbb{R}^5)$ be the real vector space of linear operators on $\mathbb{R}^5$. Let $x$ be a nonzero vector in $\mathbb{R}^5$ and define
  $$W=\{T\in V\,|\,Tx=0\}.$$
  What is the dimension of $W$?

I have shown that $W$ is indeed a subspace of $V$, so the question is well-formed. Now, I am trying to find a basis for $W$ in order to get its dimension. I know that the dimension of $V$ is $25$, so the basis must have less than $25$ elements. However, I'm not having much successes. Any help would be appreciated.
 A: You can consider the mapping:
$$\varphi:L(\mathbb{R}^5,\mathbb{R}^5)\longrightarrow\mathbb{R}^5:T\longmapsto Tx.$$
Clearly, $\varphi$ is well-defined and linear. The space $W$ you're looking for is $W=\operatorname{Ker}\varphi$. Now, by the Rank–Nullity Theorem,
$$\dim W=\dim\operatorname{Ker}\varphi=\dim L(\mathbb{R}^5,\mathbb{R}^5)-\operatorname{rk}\varphi.$$
It is well-known that $\dim L(\mathbb{R}^5,\mathbb{R}^5)=5\times5=25$. Moreover, it is not difficult to show that $\varphi$ is onto, hence $\operatorname{rk}\varphi=5$. Hence
$$\dim W=25-5=20.$$

Another method is the following: Since $x$ is a non-zero vector, by the Incomplete Basis Theorem, there exists a basis $\mathscr{B}$ of $\mathbb{R}^5$ that contains $x$, say $\mathscr{B}=(x,u_2,u_3,u_4,u_5)$. Now, let $T\in L(\mathbb{R}^5,\mathbb{R}^5)$. Clearly, $T$ belongs to $W$ if an only if the first column of the matrix of $T$ in the basis $\mathscr{B}$ is nil. Hence, there are $4\times5=20$ coefficients to uniquely determine this matrix, hence $\dim W=20$.
A: Hint: the function $e_x:L(\Bbb R^5,\Bbb R^5)\longrightarrow\Bbb R^5$ defined by $e_x(T) = Tx$ is linear.
