30
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ "Walter Rudin: Functional Analysis, Lemma 3.9." (from a deleted answer, for future reference of others) $\endgroup$ Feb 22, 2017 at 20:19
  • 1
    $\begingroup$ @Chill2Macht Exactly what I was looking for. Thanks! $\endgroup$
    – PtF
    Dec 4, 2021 at 12:26
  • 1
    $\begingroup$ So every non-injective linear functional is a scalar multiple of any injective one? $\endgroup$
    – John Mars
    Feb 20 at 0:15

3 Answers 3

21
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ Hello Daniel, do you by any chance know of a proof that uses Hahn-Banach? I'd greatly appreciate it. $\endgroup$ Jan 12, 2019 at 22:46
  • $\begingroup$ @GuillermoMosse Lemma 3.2 of Brezis's Functional Analysis book. $\endgroup$
    – John Mars
    Jan 5 at 3:32
2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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). $\endgroup$
    – Jochen
    Mar 13, 2014 at 7:54
  • $\begingroup$ @Jochen: Do you mind telling me which part of Daniel's proof used Hahn-Banach? $\endgroup$ Nov 28, 2020 at 12:54
  • $\begingroup$ ...we can extend $\hat L$ to all of $K^n$... $\endgroup$
    – Jochen
    Nov 29, 2020 at 14:45
1
$\begingroup$

[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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.