Suppose $ker f = ker g$, show that $\exists \alpha \in \mathbb{C}$ such that $f(p)=\alpha g(p)$, $\forall p \in \mathcal{P}_7 (\mathbb{C})$

Question

Suppose $f$, $g$: $\mathcal{P}_7 (\mathbb{C})\to \mathbb{C}$ and ker $f$ = ker $g$, show that $\exists \alpha \in \mathbb{C}$ such that $f(p)=\alpha g(p)$, $\forall p \in \mathcal{P}_7 (\mathbb{C})$.

The hint is: first to show that, if $f(p) \ne 0$, then $\mathcal{P}_7 (\mathbb{C})$ = ker$f$ $\oplus$ span($p$).

I don't see how the hint fits in this problem.

Answer 1

If $\mathcal{P}_7(\mathbb{C}) = \ker f \oplus \operatorname{span} \{p\}$ then every $q \in \mathcal{P}_7(\mathbb{C})$ can be uniquely written as $q = r + \lambda p$ where $r \in \ker f$ and $\lambda \in \mathbb{C}$.

Since $r \in \ker f = \ker g$ we have $f(r) = g(r) = 0$ so $$f(q) = f(r+\lambda p) = f(r) + \lambda f(p) = \lambda f(p)$$ $$g(q) = g(r+\lambda p) = g(r) + \lambda g(p) = \lambda g(p)$$ Therefore since $f(p) \ne 0$ we have $$g(q) = \frac{g(p)}{f(p)}f(q)$$

If we set $\alpha = \frac{g(p)}{f(p)} \in \mathbb{C}$ we get $g = \alpha f$.