$\ker S$ is not contained in $\ker T$ implies $\dim \Im T \ge 1$

Question

Let $T,S:V\rightarrow W$.where $V$ is a finite vector space above $F$ and $W$ is one-dimensional vector-space above $F$ ($\dim W = 1$).

It is given that $\ker S$ isn't contained in $\ker T$. Why is implying that $T \ne 0$ and hence, $\Im T = W$ ($ \dim \Im T= 1 $)?

Answer 1

Recall the rank-nullity theorem:

$$\dim V=\dim\ker T+\dim\operatorname{im} T$$ Clearly if $T=0$ then $\ker T=V$ and then $\ker S\subset \ker T$ which contradicts the hypothesis. Hence $T\ne0$ and then $\dim \ker T=\dim V-1$ and $\dim \operatorname{im} T=1$ which means that $\operatorname{im} T=W$.