Intersection of kernels and linear dependence of functionals

Question

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

Answer 1

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

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

Answer 2

[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.

Answer 3

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.