Suppose $\phi \in L(V,F)$ and $\phi$ is not the zero map. Prove that $\dim V / \ker(\phi) = 1$.

Question

I'm working on a quotient space problem from Axler (3.E #15):

Suppose $\phi \in L(V,F)$ and $\phi$ is not the zero map. Prove that $\dim V / \ker(\phi) = 1$. Note that $F$ is the ground field for $V$.

In this case, we cannot assume that $V$ is finite-dimensional and thus cannot use the rank-nullity theorem. Additionally, theorem 3.89 (Dimension of a Quotient Space) is restricted to finite-dimensional vector spaces, so I'm unsure where to begin.

Any suggestions on where to get started?

Answer 1

Note that $\phi$ has to be surjective. (The rank of $\phi$ is either $0$ or $1 = \dim F$. But the hypothesis $\phi \neq 0$ rules out the first possibility.) Hence by the first isomorphism theorem (this is just the content of 3.91 (d) in the book) we have $V/\ker \phi \cong F$. The claim follows.

Answer 2

Choose $v_0$ such that $\phi(v_0) \neq 0$.

Choose $v \in V$, then $\phi(v-{ \phi(v) \over \phi(v_0)} v_0) = 0 $ and so $v-{ \phi(v) \over \phi(v_0)} v_0 \in \ker \phi$.

Hence $[v_0]= \{v_0\} + \ker \phi $ is a basis for $V / \ker \phi$.