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 $n$?
 A: Choose a basis $y_{1}, \ldots, y_{n}$ of $F(X)$ and consider the vectors $x_{1}, \ldots, x_{n}$ in $X$ such that
$$ F(x_{1}) = y_{1} $$
$$ \vdots $$
$$ F(x_{n}) = y_{n} .$$
Define $Z = \text{span}(x_{1}, \ldots, x_{n})$. It can be shown that the $x_{i}$'s are linearly independent, so $\dim Z = n$. Now let $x \in X$ be arbitrary. Then
$$ F(x) = \alpha_{1}y_{1} + \ldots + \alpha_{n}y_{n} $$
for some scalars $\alpha_{1}, \ldots, \alpha_{n}$.
It is clear that
$$ x = \left( x - \sum_{i=1}^{n}\alpha_{i}x_{i} \right) + \sum_{i=1}^{n}\alpha_{i}x_{i},$$
and from this we deduce that $X = \ker F + Z$. It is not difficult to show that the sum is direct.
A: 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 $\dim Z \leq n$?)
