Domain of linear operator being the direct sum of kernel and another subspace. Let, $F: X \rightarrow Y$ be a linear operator. Let, $Q = F(X)$ be an $n$-dimensional vector space. How do I demonstrate that,
$$X = \ker(F) \oplus Z,$$
for some subspace $Z$ with dimension equal to $n$?

Let $\alpha$ be a basis for $\ker F$, which we can extend to a basis $\beta$ for $X$. Then
$$X = \rm{Span}(\alpha) \oplus \rm{Span}(\beta \setminus \alpha) = \ker(F) \oplus \rm{Span}(\beta \setminus \alpha).$$
Let $Z = \rm{Span}(\beta \setminus \alpha)$; we need only show that $\dim Z = n$. Well, we must have $\dim Z \geq n$ since
$$n = \dim F(X) = \dim \rm{Span}\{F(v)\}_{v \in \beta} = \dim \rm{Span}\{F(v)\}_{v \in \beta \setminus \alpha} \leq \dim Z$$
where the third equality follows from the fact that $\alpha \subseteq \ker F$. Can you take it from here? (Why must we have $\dim Z \leq n$?)
