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?


1 Answer 1


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 $x$.


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