Does the inclusion of the kernels of two linear forms imply that one of the linear forms is a multiple of the other?

Question

Does the following statement hold:

"Let $\mathbb{K}$ be a field and let $V$ be a $\mathbb{K}$- vector space, $\dim V\lt\infty$. Let $f,g\in V^*$, $f\not =0$. If $\ker (f)\subseteq \ker (g)$, then $g=\lambda f, \lambda \in \mathbb{K}$"?

If it holds, is it true for infinite dimensional vector spaces too?

I know that $\ker (f)=\ker (g)$ iff $g=\lambda f, \lambda \in \mathbb{K}$, but I don't know if I could use this fact o not.

Answer 1

You don't need the fact that $V$ is finite-dimensional. If $V$ is a vector space over $K$ and $f, g \in V^{*}$, we can prove that $\ker f \subseteq \ker g$ if and only if there exists some scalar $\lambda$ such that $g = \lambda f$.

$(\Leftarrow)$ This direction is easy.

$(\Rightarrow)$ This requires a little bit more work.

Notice that $K$ is a one-dimensional vector space, so $\text{im }f$, which is a subspace of $K$, must be finite-dimensional as well. If $\text{im }f = \{ 0 \}$, then $f$ is the zero linear map from $V$ to $K$, so $g$ must be the zero map as well and any $\lambda \in K$ will do.

Suppose $\text{im } f \neq \{ 0 \}$. Then $\text{im } f$ must be all of $K$. Let $u$ be the vector in $V$ such that $f(u) = 1$. Then $f(u)$ is a basis of $K$, so there must be some scalar $\lambda$ such that $g(u) = \lambda f(u)$.

Now let $v \in V$ be arbitrary. Then $f(v) = \alpha f(u)$ for some scalar $\alpha$. We see that $f(v - \alpha u) = 0$, so $v - \alpha u \in \ker f$. By our hypothesis, $v - \alpha u$ is in $\ker g$ as well, so $$ g(v - \alpha u) = 0 .$$ This implies that $g(v) = \alpha g(u) = \alpha (\lambda f(u)) = \lambda (\alpha f(u) ) = \lambda f(v).$ Since $v$ was arbitrary, we can conclude that $g = \lambda f$.

Addendum: This result depends on the fact that $K$ is a one-dimensional vector space. In a more general setting, the following two results hold:

1. Suppose $T_{1}, T_{2}$ are linear maps from $V$ to $W$ and $W$ is finite-dimensional. Then $\ker T_{1} \subseteq \ker T_{2}$ if and only if there exists a linear map $S$ from $W$ to $W$ such that $T_{2} = ST_{1}$.

2. Every linear map from a one-dimensional vector space to itself is multiplication by some scalar.