Suppose $\varphi \in \mathcal{L}(V, \mathbf{F})$. Suppose $u \in V$ is not in $\operatorname{null}(\varphi)$. Prove that $$ V = \operatorname{null}(\varphi) \oplus \{au : a \in \mathbf{F}\}. $$
$V$ is a vector space over the field $\mathbf{F}$, but note that it is not necessarily finite-dimensional.
I will start my attempt and indicate where I have done something I think I may not be allowed to.
Let $(v_1, \ldots, v_n)$ be a basis for $\operatorname{null}(\varphi)$. We can extend this to a basis of $V$, because $\operatorname{null}(\varphi) \subseteq V$ $$ (v_1, \ldots, v_n, v_{n+1}, v_{n+2}, \ldots). $$
This is where I think I might have done something wrong: I have chosen a potentially infinitely long basis. Is that allowed? I shall continue anyway.
Let $W = \operatorname{null}(\varphi) + \{au : a \in \mathbf{F}\}$. Then we can write $w \in W$ as so: $$ w = \alpha_1 v_1 + \cdots + \alpha_n v_n + au $$ where $\alpha_1, \ldots, \alpha_n, a \in \mathbf{F}$. We can rewrite $$ au = \beta_1 v_1 + \cdots + \beta_n v_n + \beta_{n + 1} v_{n+1} + \beta_{n + 2} v_{n+2} + \cdots $$ where $\beta_1, \beta_2, \ldots \in \mathbf{F}$ and not all $\beta_i = 0$, because $u \notin \operatorname{null}(\varphi)$ (is it required to state this final part about not all $\beta_i$ being 0?).
Hence $$ w = (\alpha_1 + \beta_1)v_1 + \cdots + (\alpha_n + \beta_n) v_n + \beta_{n + 1} v_{n+1} + \beta_{n + 2} v_{n+2} + \cdots. $$
Clearly then $w$ is a linear combination of the basis vectors for $V$, hence $V = W$.
Finally, to show that $V = \operatorname{null}(\varphi) \oplus \{au : a \in \mathbf{F}\}$ is indeed a direct sum, we must show that when $w = 0$, the coefficients for all $v_i$, are zero, and that this is the only solution.
Because all $v_i$ form a basis, they are linearly independent. Hence all coefficients of $v_i$ are indeed 0.
Is using a potentially infinitely long basis incorrect?
Best Answer
When you choose a basis $v_{1}, \ldots, v_{n}$ for $\ker \varphi$, you're assuming that it is is finite-dimensional. Since there is no assumption on $V$ being finite-dimensional or any other information about the dimension of $\ker \varphi$, we cannot assure that $\ker \varphi$ is finite-dimensional.
This problem has two parts:
We need to show that for every $v \in V$, there exist $w \in \ker \varphi$ and $\alpha \in F$ such that $v = w + \alpha u$. So you start by considering an arbitrary vector $v \in V$, and then try to find $w$ and $\alpha$ such that $v = w + \alpha u$. When you're trying to prove an existence result, it is sometimes useful to think in reverse: suppose you already have $v = w + \alpha u$ with the desired properties and now try to find out what $w$ and $\alpha$ should be. Let's see how that looks.
Suppose we have have $w \in \ker \varphi$ and $\alpha \in F$ such that $v = w + \alpha u$. Can we get some information about $w$ and $\alpha$ from this?
Well, applying $\varphi$ to both sides, we see that $$ \varphi(v) = \varphi(w) + \alpha\varphi(u) $$
and $w$ is in $\ker \varphi$, so
$$ \varphi(v) = \alpha \varphi(u). $$
We know that $\varphi (u) \neq 0$, so it's time to use that fact:
$$ \alpha = \frac{\varphi(v) }{ \varphi(u) } .$$
So it seems like we've found our candidate for $\alpha$. Up to this point, it must the case that
$$ v = w + \frac{\varphi(v) }{ \varphi(u) } u .$$
Can you now see what $w$ should be?
Of course, what I described above shouldn't be in the final proof. It is only the scratch work. In the final proof you would explicitly say what your candidates for $w$ and $\alpha$ are, and show that they satisfy the required properties.
For the second part, we need to show that the sum is direct. In this case, I think it would be easier to use the fact that the sum of two subspaces $U$ and $W$ is direct if and only if $U \cap W = \{ 0 \} $.