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 $W$).


2 Answers 2


Take $V=K$ and $W=K^2$; consider $f\colon V\to W$ given by $f(1)=(1,0)$ and $f_1\colon V\to W$ given by $f_1(1)=(0,1)$. Of course $\ker f_1\subseteq\ker f$, but $f$ is not linearly dependent on $f_1$.

Let's look at the general case. If $U=\bigcap_{i=1}^n\ker f_i$, then your maps induce linear maps $\bar{f}_i\colon V/U\to W$ $(i=1,\dots,n)$ and $\bar{f}\colon V/U\to W$. The problem is now whether any linear map $g\colon V/U\to W$ can be obtained as a linear combination of the given ones.

Thus we can assume $U=0$ and that the maps are linearly independent, so the problem becomes

Let $f_1,\dots,f_n\colon V\to W$ be linearly independent maps of $K$-vector spaces with $\bigcap_{i=1}^n\ker f_i=0$; is it true that any linear map $f\colon V\to W$ is a linear combination of the given maps?

What we can say is that there is an embedding

$$ V=\frac{V}{\bigcap_{i=1}^n\ker f_i}\to\bigoplus_{i=1}^n \frac{V}{\ker f_i}. $$

so $\dim V\le n\dim W$. Therefore, increasing the dimension of $V$ in the original problem doesn't help in the "normalized" situation. This shows also that the assertion is, in general, false: the span of the given maps has dimension $n$, while $\dim\mathrm{Hom}(V,W)=(\dim V)(\dim W)$. You should be able to show a counterexample, now.

Notice that, in the case when $W=K$, it suffices to show that the span of the given map has dimension $\dim V$, which follows from the above embedding.

  • $\begingroup$ Thanks for your counterexample. I included the condition $\dim V \ge \dim W$. $\endgroup$
    – Ralph
    May 9, 2013 at 18:14

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 ?


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