# Intersection of kernels and linear dependence of functionals

I am trying to prove the following. I have seen it alluded to in other places of the internet (this site included) but without proof.

Let $$L,L_1\ldots L_n$$ be linear functionals on a vector space $$X$$. If $$\bigcap_{i=1}^n ker(L_i) \subset ker(L)$$ then there exists $$t_i$$ for $$i=1\ldots n \in \mathbb{R}$$ such that $$L = \sum_{i=1}^n t_i L_i$$.

In other words, if the intersection of kernels of linear functionals is contained by the kernel of another linear functional then they are linearly dependent.

Related:

Intersection of kernels and linear dependence of linear maps

Linear dependence of linear functionals

• "Walter Rudin: Functional Analysis, Lemma 3.9." (from a deleted answer, for future reference of others) Feb 22, 2017 at 20:19
• @Chill2Macht Exactly what I was looking for. Thanks!
– PtF
Dec 4, 2021 at 12:26
• So every non-injective linear functional is a scalar multiple of any injective one? Feb 20 at 0:15

Let $K$ denote the scalar field. Consider $F\colon X \to K^n$ given by

$$F(x) = \begin{pmatrix}L_1(x)\\ L_2(x)\\ \vdots \\ L_n(x)\end{pmatrix}.$$

Let $R = \operatorname{im} F \subset K^n$. We have an induced isomorphism $$\tilde{F}\colon X/\ker F \xrightarrow{\sim} R.$$

Since $\bigcap\limits_{k=1}^n \ker L_k = \ker F \subset \ker L$, we have an induced linear form $\tilde{L} \colon X/\ker F \to K$, and can pull that back to $R$ as $\hat{L} := \tilde{L} \circ \tilde{F}^{-1}$. We can extend $\hat{L}$ to all of $K^n$ (extend a basis of $R$ to a basis of $K^n$, and choose arbitrary values, e.g. $0$, on the basis vectors not in $R$). Thus there is a linear form $\lambda \colon K^n \to K$ with

$$\lambda \circ F = \lambda\lvert_R \circ F = \hat{L}\circ F = \tilde{L}\circ \tilde{F}^{-1}\circ F = \tilde{L} \circ \pi = L,$$

where $\pi \colon X \to X/\ker F$ is the canonical projection.

But every linear form $K^n\to K$ can be written as a linear combination of the component projections, so there are $c_1,\dotsc, c_n$ with

$$\lambda\begin{pmatrix}u_1\\u_2 \\ \vdots \\ u_n \end{pmatrix} = \sum_{k=1}^n c_k\cdot u_k,$$

and that means

$$L(x) = \lambda(F(x)) = \sum_{k=1}^n c_k\cdot L_k(x)$$

for all $x\in X$, or

$$L = \sum_{k=1}^n c_k\cdot L_k.$$

• Hello Daniel, do you by any chance know of a proof that uses Hahn-Banach? I'd greatly appreciate it. Jan 12, 2019 at 22:46
• @GuillermoMosse Lemma 3.2 of Brezis's Functional Analysis book. Jan 5 at 3:32

I have a proof in the case where $X$ is reflexive.

Suppose that $L$ is not a linear combination of the $L_i$'s. Let $C = \{\sum_{i=1}^n t_i L_i : t_i \in \mathbb{R}\} \subseteq X^*$. Then $C$ is a closed convex subset and $C \cap \{L\} = \emptyset$ by assumption. So by geometric Hahn-Banach, there exists $\xi \in X^{**}$ such that $\xi(C) \subseteq (-\infty,\alpha)$ and $\xi(L) > \alpha$. Since $C$ is a subspace, we actually have $\xi(C) = \{0\}$ and $\alpha > 0$. Assuming that $X$ is reflexive, then $\xi$ corresponds to evaluation at some $x \in X$. This shows that $L_i(x) = 0$ for all $i = 1, \ldots, n$ but $L(x) > 0$. This contradicts $\bigcap_{i=1}^n \ker L_i \subseteq \ker L$.

I'm not sure whether this extends to the case where $X$ is not reflexive, however.

• Well, a priori, there is no topology given but one can endow $X^*$ with the weak*-topology so that the continuous dual of $X^*$ is $X$ (more precisely, every continuous linear functional is an evaluation). However, to prove this general fact $(X^*,\sigma(X^*,X))^*=X$ you need the lemma in question. Note, that Daniel's proof also uses a kind of Hahn-Banach theorem (though a very simple one in finite dimensions). Mar 13, 2014 at 7:54
• @Jochen: Do you mind telling me which part of Daniel's proof used Hahn-Banach? Nov 28, 2020 at 12:54
• ...we can extend $\hat L$ to all of $K^n$... Nov 29, 2020 at 14:45

[This is from Prop 1.1.1 of Kadison-Ringrose Vol 1]

Th: Let $${ V }$$ be a $${ K-}$$vector space, and $${ \rho _1, \ldots, \rho _n \in V ^{\ast} }.$$
Then $${ \text{span}( \rho _1, \ldots, \rho _n ) }$$ $${ = \lbrace \rho \in V ^{\ast} : \ker(\rho) \supseteq \bigcap _1 ^n \ker(\rho _j) \rbrace. }$$
Pf: The inclusion $${ \subseteq }$$ is clear. For the $${ \supseteq }$$ part we can proceed by induction.
[n=1 case] Say $${ \rho \in V ^{\ast} }$$ with $${ \ker(\rho) }$$ $${ \supseteq \ker(\rho _1) }.$$ We should prove $${ \rho }$$ $${ \in \text{span}(\rho _1) }.$$
If $${ \rho = 0 }$$ its true anyways, so say $${ \rho \neq 0 }.$$ Now $${ V \neq \ker(\rho) \supseteq \ker(\rho _1) ,}$$ so even $${ \rho _1 \neq 0 }.$$ (Especially $${ \rho, \rho _1 }$$ are surjective, so both $${ V/{\ker(\rho)} },$$ $${ V/{\ker(\rho _1)} }$$ are isomorphic to $${ K }$$).

Recall that given a linear map $${ \mathscr{V} \overset{T}{\to} \mathscr{W} }$$ and a subspace $${ { \color{green}{\mathscr{V _0}} } \subseteq { \color{purple}{\ker(T)} } ,}$$ we get a linear map $${ \mathscr{V}/{\mathscr{V _0}} \overset{\tilde{T}}{\to} \mathscr{W} }$$ sending $${ (v + \mathscr{V _0}) \mapsto T(v) }.$$
(Once one shows $${ \tilde{T} }$$ is well-defined, linearity is clear. Say $${ v _ 1 + \mathscr{V _0} = v _2 + \mathscr{V _0} }.$$ Now $${ (v _1 - v _2) \in \mathscr{V _0} \subseteq \ker(T) },$$ so $${ T(v _1 - v_2) = 0 }$$ i.e. $${ T(v _1) = T(v _2) },$$ as needed).

As above, since $${ { \color{green}{\ker(\rho _1)} } \subseteq { \color{purple}{\ker(\rho)} } }$$ we get a functional $${ V/{\ker(\rho _1)} \overset{\tilde{\rho}}{\to} K }$$ sending $${ (v + \ker(\rho _1)) \mapsto \rho(v) }.$$
But $${ V/{\ker(\rho _1)} }$$ is $${ 1 }$$ dimensional, and we already have a usual nonzero functional $${ V/{\ker(\rho _1)} \overset{\rho _1 ^{\ast}}{\to} K }$$ sending $${ (v+\ker(\rho _1)) \mapsto \rho _1 (v). }$$ So $${ \tilde{\rho} }$$ must be a multiple of $${ \rho _1 ^{\ast} }.$$
There is a $${ \lambda \in K }$$ such that $${ \tilde{\rho} = \lambda \rho _1 ^{\ast} }.$$ Now $${ \tilde{\rho} (v + \ker(\rho _1)) }$$ $${ = \lambda \rho _1 ^{\ast} (v + \ker(\rho _1)) }$$ for all $${ v \in V },$$ that is $${ \rho (v) = \lambda \rho _1 (v) }$$ for all $${ v \in V }.$$ So $${ \rho \in \text{span}(\rho _1) ,}$$ as needed.
[Induction step] Say the theorem statement holds when $${ n = N }.$$ We will show it holds for $${ n = N+1 }$$ too.
Let $${ \rho _1, \ldots, \rho _{N+1} \in V ^{\ast} },$$ and $${ \rho \in V ^{\ast} }$$ with $${ \ker(\rho) \supseteq \bigcap _1 ^{N+1} \ker(\rho _j) }.$$ We should prove $${ \rho \in \text{span}(\rho _1, \ldots, \rho _{N+1}) }.$$
Consider the restrictions $${ \varphi := \rho \big{|} _{\ker(\rho _{N+1})} }$$ and $${\varphi _1 := \rho _1 \big{|} _{\ker(\rho _{N+1})} , }$$ $${ \ldots, \varphi _N := \rho _N \big{|} _{\ker(\rho _{N+1})} }.$$
Note $${ \ker(\varphi) }$$ $${ \supseteq \bigcap _1 ^N \ker (\varphi _j) , }$$ since LHS is $${ \ker(\rho _{N+1}) \cap \ker(\rho) }$$ and RHS is $${ \bigcap _1 ^N (\ker(\rho _{N+1}) \cap \ker(\rho _j) ). }$$
By induction hypothesis, there are $${ \lambda _1, \ldots, \lambda _N \in K }$$ such that $${ \varphi = \lambda _1 \varphi _1 + \ldots + \lambda _N \varphi _N }$$ on $${ \ker(\rho _{N+1}) }.$$ So $${ \rho - ( \lambda _1 \rho _1 + \ldots + \lambda _N \rho _N ) = 0 }$$ on $${ \ker(\rho _{N+1}) }.$$ Equivalently, $${ \ker( \rho - (\sum _1 ^N \lambda _j \rho _j ) ) }$$ $${ \supseteq \ker(\rho _{N+1}). }$$
Now from $${ n = 1 }$$ case proved above, $${ \rho - (\sum _1 ^N \lambda _j \rho _j ) \in \text{span}(\rho _{N+1}) ,}$$ giving $${ \rho \in \text{span}(\rho _1, \ldots, \rho _{N+1}) }$$ as needed.