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 ?

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

1 Answer 1


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


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


$W=Im(\phi)$ 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


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.

  • 1
    $\begingroup$ Thanks a lot ! Didn't think I could link the dimension of the intersection to the sum of ranks. $\endgroup$
    – aleph0
    Jan 2, 2020 at 12:35

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