A problem on dual space Let $V$ be a finite dimensional vector space. Let $f$ and $g$ are two elements in the dual space $V'$. If $\ker f \subset \ker g$. I need to prove that there exists a scalar $\lambda$ such that $g=\lambda f$.
Now how to guess why this should happen intuitively and what can be a possible $\lambda$?
 A: We may suppose $f \neq 0$ (otherwise $f=g=0$); in particular, $f$ is surjective.


*

*Let $r \in \mathbb{R}$ and $v \in V$ such that $r=f(v)$, and define $\varphi(r)=g(v)$.

*Because $\mathrm{ker}(f) \subset \mathrm{ker}(g)$, $\varphi : \mathbb{R} \to \mathbb{R}$ is well-defined.

*Morevoer, $\varphi$ is linear so $\varphi : r \mapsto \lambda r$ for some $\lambda \in \mathbb{R}$.

*Clearly, $g= \varphi \circ f$ by construction, ie. $g= \lambda f$.


More generally, with an analogous argument, you can show that if $\bigcap\limits_{i=1}^n \mathrm{ker}(f_i) \subset \mathrm{g}$ then $g = \sum\limits_{i=1}^n \lambda_i \cdot f_i$ for some $\lambda_i \in \mathbb{R}$.
A: Intuitive observation:
Let $x_1, \cdots, x_n$ be a basis of $V$ and $dx^1, \cdots, dx^n$ a dual basis of $V'$ (that is, $dx^i(x_j) = \delta^i_j$). Put
$$
f = \sum_{i = 1}^n f_i dx^i, \qquad
g = \sum_{i = 1}^n g_i dx^i.
$$
For any $v = \sum_{j = 1}^n v^j x_j \in V$, 
\begin{align*}
f(v) &= \sum_{i, j}f_i v^j \delta^i_j = f_1 v^1 + \cdots f_n v^n,
\\
g(v) &= \sum_{i, j}g_i v^j \delta^i_j = g_1 v^1 + \cdots g_n v^n.
\end{align*}
So the condition $\ker f \subset \ker g$ becomes
$$
f_1 v^1 + \cdots + f_n v^n = 0 \implies g_1 v^1 + \cdots g_n v^n = 0
$$
for all $v = \sum_{j = 1}^n v^j x_j \in V$. This can be happen only if there exists $\lambda$ such that $g_i = \lambda f_i$, in other words $g = \lambda f$. If, in particular, $f_i \neq 0$ then that scalar is determined as
$$
\lambda = g_i/f_i = g(x_i)/f(x_i).
$$
After this observation, you'll become comfortable with the statement. In addition, the key point was the choice of $v \in V$ such that $f(v) \neq 0$ to write $\lambda = g(v)/f(v)$. This is what walcher's answer says (and this expression is free from the choice of basis!).
A: Hint: What is the dimension of $\text{ker}f$ for $f\in V'$? If $v\notin \text{ker}f$ what is $\displaystyle\frac {g(v)}{f(v)}?$
