Dual space kernel equality

Question

I don't know how to show this one: given elements $f,g ∈ V^∗$ of the dual space. I should now show that if $g = λf$ ,with $λ ∈ K\setminus\{0\}$ and $f$ is not $0$ , it follows that $\ker(f) = \ker(g)$.

Answer 1

You have to show the equality of two sets.

$\ker(f)=\ker(g)$

So we have to show $\subseteq$ and $\supseteq$.

To show $\ker(f)\subseteq\ker(g)$ we take a vector $v\in\ker(f)$.

Then $f(v)=0$ but $g=\lambda f$ so $f=\lambda^{-1}g$. Then $(\lambda^{-1}g)(v)=\lambda^{-1}g(v)=0\Leftrightarrow g(v)=0$. So $v\in\ker(g)$.

You can show $\supseteq$ exactly the same way. I leave that for you.