Addition of Vector Spaces I'm working through Linear Algebra Done Right by Axler, and there is one question where the following is stated.

Assume $U$ is a subspace of $V$. What is $U+U$?

The answer (after having to refer to a solutions manual as I didn't even have a definition of addition of vector spaces) is $U + U \subset U$. And the manual arrives at that conclusion by looking at the fact that that $U$ is closed under addition, thus any $v, w \in U \implies v+w \in U$. 
But this is done before Axler even defines addition between vector spaces. And subspaces of vector spaces are themselves vector spaces. What's more the addition between vector spaces above seem to be simply the Minkowski Sum of two sets, and the Minkowski Sum would make sense, since vector spaces are sets, but it doesn't seem to work for vector spaces as I outline below.
It doesn't seem to me that addition of two vector spaces even be well defined, (assuming we are working with the Minkowski Sum definition), as if $V$ and $W$ are two arbitrary vector spaces, their dimensions might not be similar, $\dim(V) \neq \dim(W)$, and furthermore, $V$ and $W$ might be vector spaces with respect to different fields $F_1$ and $F_2$, leading me to believe that no such definition of addition of vector spaces exist.
Is there a definition of addition of vector spaces? If not, and addition is not defined for arbitrary vector spaces, how can they be defined for specific vector spaces like in the problem above, or is it just that the problem is misleading?
 A: The Minkowski sum is intended here. You can't sum two unrelated vector spaces in general, that's perfectly right, but that does not happen here. Given that both summands are subsets of the same vector space $V$, they are inherently over the same field and same "dimension"* and indeed inherit the $+$ operation that's simply applied to each tuple of elements.
Note that in general the problem with summing two unrelated sets would be even more serious than for the reasons you list: $+$ would not be defined between their elements.
*) Meaning the number of components or such. $\dim A$ and $\dim B$ can differ in $A+B$: for example, you can add the zero-dimensional space $\{0\} \subseteq V$ to any other subspace $U \subseteq V$. Note that the quotation marks are necessary because in general it's not possible to "divine" the dimension of the underlying space by looking at the subspace: consider, for example, the subspace of constant polynomials in the space of polynomials up to degree $k$.
A: As $(U,+,.)$ is a vector space over some field say $F$ then you should remember that $(U,+)$ is a group under addition .
Also $+ $ is defined to be the addition of $U$ as a group
A: You don't have an addition of vector spaces, but you have an addition of subspaces of a vector space. Namely

if $U_1$ and $U_2$ are subspaces of the vector space $V$, then $U_1+U_2=\{x_1+x_2:x_1\in U_1, x_2\in U_2\}$ is a subspace of $V$

The proof is just verifying the required properties.


*

*$0\in U_1+U_2$ by taking $x_1=0$ and $x_2=0$

*If $v,w\in U_1+U_2$ then $v=x_1+x_2$ and $w=y_1+y_2$ for some $x_1,y_1\in U_1$ and $x_2,y_2\in U_2$; then $v+w=(x_1+y_1)+(x_2+y_2)\in U_1+U_2$ since $x_1+y_1\in U_1$ and $x_2+y_2\in U_2$

*If $v\in U_1+U_2$ and $\alpha$ is a scalar, then $v=x_1+x_2$ for some $x_1\in U_1$ and $x_2\in U_2$; then $\alpha v=(\alpha x_1)+\alpha x_2)\in U_1+U_2$.
Actually

$U_1+U_2$ is the smallest subspace of $V$ (with respect to set inclusion) that contains $U_1$ and $U_2$

Indeed, if $W$ is a subspace of $V$, $U_1\subseteq W$ and $U_2\subseteq W$, then, for each $x_1\in U_1$ and each $x_2\in U_2$, we have $x_1,x_2\in W$ so $x_1+x_2\in W$. Therefore $U_1+U_2\subseteq W$.
In the case $U_1=U_2=U$, the smallest subspace of $V$ containing $U$ is $U$ itself, so $U+U=U$.
You can also prove $U+U=U$ directly from the definition; the inclusion $U+U\subseteq U$ is clear. If $u\in U$, then $u=u+0\in U+U$.
