TITLE: Intersection of kernels and linear dependence of linear maps
QUESTION [11 upvotes]: Let $f_1,...,f_n,f: V \to W$ be linear maps of $K$-vector spaces. If $W=K$ it's known that 

$f$ is linear dependent from $f_1,...,f_n$ iff $\;\;\bigcap_{i=1}^n \ker(f_i) \subseteq \ker(f)$. 

Question: Is this statement true for general $W$ ? 
Remark: The direction $(\Rightarrow)$ is obviously true and  if it helps $W$ can be assumed to be finite dimensional. 
Edit: You can also assume $\dim V \ge \dim W$ (and if necessary the dimension of $V$ can be assumed to be much larger than that of $W$).

REPLY [3 votes]: Here is a simple geometrical proof using Hahn-Banach :
Your condition says : for each $x$ where every $f_i(x)=0$, $x$ verifies $f(x)=0$, so that
$$a:= (1,0,..0) \notin R:=Range[x \mapsto (f(x), f1(x),...fn(x))~]$$
So we can separate strictly $\{a\}$, which is compact, and $R$, which is closed, by a hyperplane $\phi$ (provided the $f_i$s are continous, as per Baire-Banach):
$$  \lambda := \phi a  < \alpha < \phi R$$
But R is a vector space, so the last inequality implies $\phi R = 0$, and the first implies $\lambda < 0$
Since an hyperplane is a condition on a linear form :
$$\phi [f, f1, ..., fn] = \lambda f + \sum \lambda_i f_i = 0$$
We have have exactly that $f$ is linearly dependant on the $f_i$s.
Now, what is the graphical intuition behind this ?