Can a subspace be its own additive identity? This problem is for Linear Algebra Done Right 3rd Edition:
Problem 18/Chapter 1:
Does the operation of addition on the subspaces of V have an additive identity? Which subspaces have additive inverses?
My doubt is in my attempt of answer for the first question:
Clearly for a Subspace $U$ of a vector space $V$ we have that the sum with the subspace $\{0\}$ implies:
$$U+\{0\}=U$$ (which is the answer given in previous posts)
However, problem 15 in the same chapter gives the result that $U+U=U$
We know that the additive identity is unique, then, Do the later results imply a contradiction?  Since the addition of $U$ or $\{0\}$ gave the same result(both act as additive identity) for $U$.
I would appreciate if someone can illuminate this basic question
 A: If $U\ne \{0\}$ then it's not an additive identity. It would have to satisfy $W + U = W$ for all subspaces $W$, not only for the case $W=U$.
There's no such thing as "identity for U".
A: The collection of all subspaces of $V$ is a set.  The author, Axler,has defined an operation for combining two such subspaces, namely $+$.
Remember:

*

*The additive identity is "the thing that always leaves the original thing unchanged."  In this case $\{0\}+U=U$.  Since $U$ does not leave the original thing (namely $\{0\}$) unchanged, $U$ cannot be the additive identity.


*An element's additive inverse is "the thing that gets the given element back to the additive identity." So if  $U$ has an additive inverse $\overline U$, then $\overline U$ is "the thing that gets $U$ back to the additive identity. (See previous bullet.)
So, "No, the facts you have noticed above are not a contradiction."  In fact, what you have noticed is that if $U$ is a subspace of $V$, then both the equations you wrote above are in fact true, i.e. $U+\{0\}=U$ and $U=U+U$.  So, if $U$ has an additive inverse, $\overline U$, the fact that $\{0\}+U=U+U$ means that by adding $\overline U$ to both sides we get $\{0\}+(U+\overline U) =U+(U+ \overline U)$. This simplifies to tell us $\{0\}=U$. (Reason: Adding the additive identity to both sides of an equation must leave the original term on each side unchanged.)
A: It's a feature of groups that if $a+b=a$ for some $a$, then $a+b=a$ for all $a$, making $b$ the additive identity. Subspaces with subspace addition don't form a group, though, just a monoid. And in monoids, an element $b$ satisfying $a+b=a$ for one single $a$ is not necessarily an additive identity.
For instance, in the monoid $(\mathbb Z,\cdot)$ we have $0\cdot0=0$, but $0$ is not the neutral element since, for instance, $0\cdot1\neq1$. Your case is just another example of this behavior.
