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

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 $g = \alpha f$.
