Exercise 4 from section 17 of Halmos' Finite-dimensional Vector Spaces states:
If $y(x) = \xi_1 + \xi_2 + \xi_3$ whenever $x = (\xi_1, \xi_2, \xi_3)$ is a vector in $\mathbb{C}^3$, then $y$ is a linear functional on $\mathbb{C}^3$; find a basis of the subspace consisting of all those vectors $x$ for which $[x, y] = 0$.
My idea is to consider instead the $x'' \in {\mathbb{C}^3}''$ (where ${\mathbb{C}^3}''$ is the space of all linear functionals on the linear functionals on $\mathbb{C}$, and $x''$ is the dual of the dual of $x$). Since we have $[x,y] = [y,x'']$, we are interested in the set $S$ of $x''$ where $[y,x'']=0$; but this is the annihilator of $\{y\}$, so (by a previous theorem) $S = \{y\}^0 = \textrm{span}\{a', b'\}$ where $\{a, b, y\}$ is any basis for ${\mathbb{C}^3}'$ and $a'$ and $b'$ are the dual of $a$ and $b$. For this problem we can simply pick say $a(x) = \xi_1$ and $b(x) = \xi_2$ whenever $x = (\xi_1, \xi_2, \xi_3)$. The issue is that it seems pretty difficult to convert $S$ (which is a subspace of ${\mathbb{C}^3}''$) back to the subspace of $\mathbb{C}^3$ of which $S$ is the dual of the dual; but it is the latter set that we are really after.