# Factorization of a linear map by multiple other maps

Let $$E$$ be a vector space of finite dimension $$n$$. Let $$p\ge 1$$, $$g, f_1, \ldots, f_p\in\mathcal{L}(E)$$ be such that $$\bigcap_{i=1}^p\mathrm{ker}f_i\subseteq \mathrm{ker} g$$ Show that their exists $$h_1, \ldots, h_p\in \mathcal{L}(E)$$ such that $$g = \sum_{i=1}^{p} h_i\circ f_i$$

I can show the result for $$p=1$$, by studying $$\varphi : h\in\mathcal{L}(E)\mapsto h\circ f$$. It is easy to see that $$\ker\varphi = \{h\mid \mathrm{Im} f\subseteq\ker h\}$$ so that $$\dim\ker\varphi = (n-\mathrm{rk} f)\cdot n = n\cdot \dim\ker f$$. So that by rank theorem, $$\mathrm{rank}\varphi = n(n - \dim\ker f) = n\cdot\mathrm{rank}(f)$$ .

To conclure, notice that $$\mathrm{Im}(f) \subseteq \{h\mid \ker f\subseteq \ker h\}$$ but the last vector-space is of dimension $$n\cdot\mathrm{rank}(f)$$ so we have equality. Therefore, $$g\in \mathrm{Im}\varphi$$

But how can one generalize that to $$p$$ linear maps ? The proof for $$p=1$$ can also be done by choosing a nice basis for example (which then also works in infinite dimension). To come back to the general case, I thought of looking at the following linear transformation : $$\phi :x\mapsto (f_1(x), \ldots, f_p(x))$$ So that $$\ker\phi = \displaystyle\bigcap_{i=1}^n \ker f_i$$ but now, $$\phi\in\mathcal{L}(E, E^p)$$. I thought of using $$g':x\mapsto (g(x), \ldots, g(x))$$ but the previous proof does not work. Any ideas ?

• For $p=1$, it seems you overcomplicated it a bit. For general $p$, I would try induction.. Jan 2, 2020 at 11:06

We set $$W:=\{g: \ker(f)\subseteq \ker(g)\}$$.

It is clear that $$W$$ is a subspace of $$L(E)$$. Let $$A:=\{v_{\dim(\ker(f))},\dots , v_n\}$$ a completion of a base of $$\ker(f)$$ to a base of $$E$$. Then in this case the linear map

$$F: W\to \times_{i=1}^{n-\dim(\ker(f))}E$$

that maps $$g$$ to $$(g(v_{\dim(\ker(f))}), \dots , g(v_n))$$

is an isomorphism. Thus

$$\dim(W)=n\cdot (n-\dim(\ker(f)) =n\cdot rk(f)$$

by nullity rank theorem.

In the other hand, we can observe that

$$Im(\phi)\subseteq W$$

but the dimension of $$Im(\phi)$$ is equal to

$$\dim(Im(\phi))=n^2-\dim(\ker(\phi))$$

$$=n^2-n\cdot(n-rk(f))=n\cdot rk(f)$$

So

$$W=Im(\phi)$$ that is what we wanted to prove.

If you want generalize it, you must consider

$$\phi: L(E)^p\to L(E)$$

that maps

$$(h_1,\dots , h_p)$$ to $$\sum_{i=1}^nh_i\circ f_i$$

In this case you have

$$Im(\phi)\subseteq W$$

where $$W$$ has dimension

$$\dim(W)=n\cdot (n-\dim(\cap_{i=1}^p\ker(f_i)))$$

$$= n\cdot (\sum_{i=1}^nrk(f_i))$$

because you can apply the nullity rank theorem to the map

$$G: E\to E^p$$

which maps $$x\in E$$ to $$(f_1(x),\dots ,f_p(x))$$.

On the other hand

$$\dim(Im(\phi))=p\cdot n^2-\sum_{i=1}^pn\cdot (n-rk(f_i))$$

$$=n\cdot (\sum_{i=1}^nrk(f_i))$$

and this means

$$W=Im(\phi)$$

that is what we wanted to prove.

This result can be generalized also if you consider $$L(E, V)$$ where $$E,V$$ are finite dimensional spaces.

• Thanks a lot ! Didn't think I could link the dimension of the intersection to the sum of ranks. Jan 2, 2020 at 12:35