# Kernel of a row vector

Consider a row vector

$$a = \begin{pmatrix} x & y\end{pmatrix}$$.

If I am not wrong, the kernel of $$a$$ is defined as

$$\mathrm{ker}\,a = \{v\in \mathbb R^2 : \, a\,v = 0\}$$.

I think the following is correct :

$$\mathrm{ker}\,a = \{ \begin{pmatrix} -y\,z/x & z\end{pmatrix}^\top\}$$, $$z\in \mathbb R$$, since $$\begin{pmatrix} x & y\end{pmatrix}\begin{pmatrix} -y\,z/x & z\end{pmatrix}^\top=0$$ .

Would it also be correct to write

$$\mathrm{ker}\,a = \{\begin{pmatrix} -y & x\end{pmatrix}\}$$,

since $$\begin{pmatrix} x & y\end{pmatrix}\begin{pmatrix} -y & x\end{pmatrix}^\top = 0$$?

Or is the last expression odd?

The kernel of $$a$$ contains all, not just one, vectors $$v$$ such that $$av=0$$. Let $$v=[v_1,v_2]^\top$$. Then $$v\in\text{ker}(a)$$ if and only if $$xv_1+yv_2=0$$. By considering the rank of $$a$$, we have:
1. If $$a\neq 0$$, then $$\text{ker}(a)=\text{span}([-y,x]^\top)$$. This is because $$\text{rank}(a)=1$$, so the nullity of $$a$$ is $$2-1=1$$. Therefore, any nonzero solution $$v=v_0$$ to $$av=0$$ generates the kernel of $$a$$. Here, $$[-y,x]^\top$$ is one such solution.
2. If $$a=0$$, then $$\text{ker}(a)=\mathbb{R}^2$$. This is trivial.
Your first attempt ignores the case where $$x=0$$. If that's the case, you cannot divide by $$x$$.