Linear functional $f$ is a linear combination of $f_1,..., f_r$ if and only if $\cap N_i\subseteq N_f$

Question

I have seen this question, but it is one-way. I would like to know if someone has some idea for the other side.

I should mention that $N_f$ is null space of $f$ and $N_i$ is null space of $f_i$, for $i = 1, ..., r$, which are $r$ linear functional on a vector space $V$.

Edited! First, I thought the equivalent condition for $f$ being a linear combination of $f_1, ..., f_r$ is $N_f\subseteq \cap N_i$. Then, by the answer below, I got this is a wrong condition (my mistake). Afterwards, I realized the equivalent condition is $\cap N_i\subseteq N_f$.

I should add that now the proof of the reverse is obvious.

Answer 1

If $f_1 = f_2$ then $f= f_1-f_2$ is a linear combination of $f_1, f_2$, but $N_f$ is not a subset of $N_1 \cap N_2$.

So the other side is not true.