If S is the set of all subspaces of a Vector space, does every subspace have an additive inverse?

If $V$ is a vector space over some field $F$. Let $S$ be the set of subspaces of $V$. Define addition by $$U_1 +U_2=\{u_1 +u_2:u_1\in U_1, u_2\in U_2\}.$$

Does every $U$ have an additive inverse? I am tempted to say that every $U$ would be an additive inverse of itself but am not sure if this works.

• Before we can speak of inverses, we need to agree what the neutral is - but that's quite clearly the 0-dimensional subspace $0$. But then $U_1+U_2=0$ only if $U_1=U_2=0$. – Hagen von Eitzen Oct 5 '17 at 22:21
• In fact it will be a commutative monoid. May be you should try to get its Grothendieck group. en.wikipedia.org/wiki/Grothendieck_group – Murphy Oct 5 '17 at 22:32
• $(S, +)$ acts more like an upper semi-lattice: $U + U = U$. (And it's bounded with bottom value $0$ and top value $V$. In fact, I think $(S, +, \cap)$ forms a lattice.) – Daniel Schepler Oct 5 '17 at 22:33
• Perhaps I'm overlooking something fundamental about adding subspaces, but once we have $U_1 + U_1$ in the form $(u_1 + u_1) + (u_2 +u_2)$ couldn't we rearrange elements such that the we pair each element with its additive inverse, thus giving us the $0$ subspace? – john fowles Oct 5 '17 at 23:05

Before we can speak of inverses, we need to agree what the neutral element is - but that's quite clearly the 0-dimensional subspace $\{0\}$.
But then $U_1+U_2=\{0\}$ if and only if $U_1=U_2=\{0\}$.
The point is that $U_1+U_2$ contains all possible combinations of $u_1+u_2$ where $u_i$ runs through the whole $U_i$. In particular, as both $U_i$ contain the $0$ vector, $U_i\subseteq U_1+U_2$ for both $U_i$'s.
More specifically, this operation defines a semilattice (which is a commutative and idempotent monoid), in that $U_1+U_2$ is the smallest subspace that contains both $U_i$'s: lowest upper bound
• But the lattice isn't distributive in general - for example, with $U = \langle (1, 0) \rangle$, $V = \langle (1, 1) \rangle$, $W = \langle (0, 1) \rangle$, then $U \cap (V + W) \ne (U \cap V) + (U \cap W)$. – Daniel Schepler Oct 5 '17 at 23:17