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Let $V$ be a finite dimensional vector space over $\mathbb{R}$. Let $\mu, \nu: V \to \mathbb{R}$ be two linear functionals.

Define $M=\mu^{-1}(1)$ and $N=\nu^{-1}(-1)$ to be affine subspaces of $V$ -- note that they are translates of $\ker \mu$ and $\ker \nu$ respectively.

Then does one have that $\operatorname{span}(M \cap N) = \ker(\mu + \nu)$?

Attempt: If $v \in \ker(\mu + \nu)$, then $\mu(v)+\nu(v)=(\mu+\nu)(v)=0$, so in particular, $$\mu(v) = - \nu(v)\,. $$ Now if $v \in M \cap N$, then $\mu(v)=1$ and $\nu(v)=-1$, so $\mu(v)=-\nu(v)$. Thus $M \cap N \subset \ker(\mu + \nu)$. Since the $\operatorname{span}(M \cap N)$ is the smallest linear subspace containing $M \cap N$, and $\ker(\mu + \nu)$ is a linear subspace containing $M \cap N$, one must have that $\operatorname{span}(M \cap N) \subseteq \ker(\mu + \nu)$.

For the opposite inclusion, if $v \in \ker(\mu + \nu)$, then set $c = \mu(v)$. By the above, $\nu(v) = -c$. Thus $v \in c(M \cap N) \subset \operatorname{span}(M \cap N)$. Therefore $\ker(\mu + \nu) \subseteq \operatorname{span}(M \cap N)$.

Note: Even if the proof is wrong, at least the result seems plausible from dimension calculations: $M$ and $N$ are $(n-1)$-dimensional affine subspaces generically (i.e. when $\mu,\nu \not=0$), which intersect generically in an $(n-2)$-dimensional affine subspace, and then taking all scalings of an $(n-2)$-dimensional subspace will generically increase the dimension back up to $(n-1)$, the dimension of the kernel of a generic linear functional.

This result is probably obviously true or false. If it is obviously false, then it makes sense why I have never seen it before when I thought of it just now. If it is obviously true then I am not sure why I have never seen it before thinking of it just now, so I am therefore wary whether it can actually be true, hence why am I asking for verification.

For general linear transformations $S, T$, one has only that $\ker(S) \cap \ker (T) \subset \ker(S+T)$ and it is not possible to say more (I think). So if this proof were correct then it would have to depend crucially on the fact that the target space of $\mu, \nu$ is $\mathbb{R}$, which I believe it does.

In particular, only linear functionals are determined up to a non-zero scalar multiple by their kernels. (As a consequence of Rank-Nullity, I think.)

Motivation: I am trying to think of a "pictorial" way of adding "covectors" (i.e. linear functionals) using the hyperplane representation suggested in Lee's Introduction to Smooth Manifolds (p.280).

2 Answers 2

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It is false:

$V$ is finite dimensional, so we can think of it as $\mathbb R^n$, and $\mu$ and $\nu$ are represented by vectors $u,v$ such that $$\mu(z) = u^Tz \qquad \nu(z) = v^Tz.$$ so we can explicitly write your $M,N$ as $$ M = \frac{u}{\|u\|^2} + \{u\}^\perp, \qquad N = -\frac{v}{\|v\|^2} + \{v\}^\perp.$$ If $n=2$ you can imagine them well. Let $u=v=(0,1)$. You have $M=\{y=1\}$ and $N=\{y=-1\}$. They are disjoint, so the span of the intersection is only $\{0\}$, but $z=(1,0)$ is perpendicular to $u,v$ so $(\mu+\nu)(z)=0$.

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    This is a nice argument using the identification of $\mathbb{R}^n$ with its dual via the standard Euclidean inner product. Are the cases when one of $\mu, \nu$ is a scalar multiple (including $c=0$) of the other (as you showed above) the only instances when the argument fails? I think so but I would like to know your thoughts. – 2017-02-22
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    I think your argument fails always: If $n>2$ then there always exist a vector orthogonal to $\mu$ and $\nu$, and it won't probably be in the span on your intersection.. – 2017-02-22
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    Could you elaborate? I don't see how the existence of an orthogonal vector is relevant here -- in fact, I don't see how any choice of identification between the original and dual vector space is necessary. I also don't see how the argument depends on the dimension $n$. – 2017-02-22
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    If neither $\mu$ nor $\nu$ are scalar multiples of the other, then $\mu^{-1}(1)$ and $\nu^{-1}(-1)$ should both be $(n-1)$-dimensional affine subspaces which neither coincide nor are parallel, thus intersect in a $(n-2)$-dimensional affine subspace which does not contain the origin. This $(n-2)$-dimensional affine subspace should be contained in the kernel $(\mu+\nu)$ and can be used to define an $(n-1)$-dimensional subspace which is also contained in, thus equal to, the kernel (by rank-nullity). – 2017-02-22
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    Oh, yes, I think you're right – 2017-02-22
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    Even if that's true, the counterexample you gave was nevertheless very good -- it never even occurred to me consider that case or that the argument might fail then. The counterexample has given me much new insight into the problem. I appreciate your help greatly as a result. – 2017-02-22
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$\newcommand{\span}{\operatorname{span}}$This doesn't answer my question, and is just a technical note about a step in my "proof" above.

In general, for $k=1,\dots,n$, given a $(k-1)$-dimensional affine subspace $A \subseteq V$ disjoint from the origin, one won't have that $\mathbb{R}A:=\{cv: c\in \mathbb{R}, v \in A \}$ is a $k$-dimensional vector subspace.

The example I have in mind is $V = \mathbb{R}^3,\ A=\{y=0, z=1 \}$. One then has that $$\mathbb{R}A = \{ y=0 \} - \{ y=0,z=0, x \not=0 \} $$ since $0v=(0,0,0)$ for all $v \in A$.

However, we do have that the closure of this set is $\{y=0\} = \operatorname{span}(\{y= 0,z=1\})$ since, given $(a,0,0)$ in $\{ y=0,z=0, x \not=0 \}$, for all $\varepsilon >0$, one has that $(a,\pm \varepsilon, 0) \in \mathbb{R}A$.

Given linear functionals $\mu, \nu$, if one assumes that, generalizing from the above example, that the worst which can happen is that $\span(M \cap N) \not= \mathbb{R}(M \cap N)$ but $\span(M \cap N) = \operatorname{cl}(\mathbb{R}(M \cap N))$, then because $V$ is finite dimensional, $\mu+\nu$ is continuous, so the fact that $(\mu \cap \nu)$ equals $0$ on all of $\mathbb{R}(M \cap N)$ implies that $\mu \cap \nu$ equals $0$ on all of $\span(M \cap N)$, thus the claim that $\span(M \cap N) = \ker(\mu + \nu)$ still holds. (This is all provided that neither $\mu$ nor $\nu$ is a constant multiple of the other in light of the accepted answer.)

I believe that this is the worst which can happen, but still need to give a formal proof.

Attempt: For $k \in \{1, \dots, n \}$, let $W$ be a $(k-1)$-dimensional vector subspace of $V$. Let $0\not=v \in V$ be arbitrary. Then $v + W$ is an arbitrary $(k-1)$-dimensional affine subspace of $V$ disjoint from the origin. Now consider the set $$\mathbb{R}(v+W) = \{tv+tw:t\in \mathbb{R}, w \in W \} \,.$$ If $u \in \mathbb{R}(v+W)$, then for any $c \in \mathbb{R}$, $cu \in \mathbb{R}(v+W)$, so $\mathbb{R}(v+W)$ is closed under scalar multiplication.

Let $u_1, u_2 \in \mathbb{R}(v+W)$. Then $u_1 = t_1 v + t_1 w_1$, $u_2 = t_2v + t_2 w_2$, for $t_1, t_2 \in \mathbb{R},\ w_1, w_2 \in W$. Therefore $$u_1 + u_2 = (t_1 + t_2)v + t_1 w_1 + t_2 w_2 \,. $$ It follows that $u_1 + u_2 \in \mathbb{R}(v+ W)$ if and only if there exists a $w_3 \in W$ satisfying the condition: $$t_1w_1 + t_2w_2 = (t_1+t_2)w_3 \,.$$ Since $W$ is a vector subspace, thus closed under scalar multiplication and addition, this occurs whenever we have that $t_1 + t_2 \not= 0$, i.e. $t_1 \not= -t_2$.

In particular, since $\mathbb{R}(v+W)=\{\vec{0}\} \cup \{tv+tw: t\not=0, w \in W \}$, the worst that can happen is that $u_1 + u_2 = 0+ w$ for some $w \in W$. For any $\varepsilon >0$, $\varepsilon v + \varepsilon(\frac{1}{\varepsilon}w) \in \mathbb{R}(v+W)$ approximates $w$ arbitrarily closely, so we have that the span of $(v+W)$ is contained in the closure of $\mathbb{R}(v+W)$.

Of course, even that argument isn't entirely correct/rigorous, but it seems close enough to the truth to placate me for the moment.