Bad ideas: Adding vectors from different vector spaces? Let V be a vector space with a non-trivial subspaces U, W. $ U \neq W$. Consider $u_0 \in V$, with $u_0 \neq 0$ and $w_0 \in V$ with $w_0 \neq 0$ then we have $u_0 + U \in V/U$ and $w_0 + W \in V/W$. Is there any sensible way to define:
$(u_0 + U) + (w_0 + W)$? 
I'm thinking "no" $(2,4) + (2,6,7)$ makes no sense. So, why should this?
I mean, we would need to find some new vector space that had all of the elements in $V/U$ and in $V/W$ and their sums and scalar products. Maybe $V/U + V/W$ -- but then we need both $V/U$ and $V/W$ to be subspaces of some larger vector space. Is there a way to make sense of this at all?
 A: Well, let's go back to the definitions. I am sorry if I go too far into the details, but it is difficult to judge what you exactly understand based only on your question. So maybe I explain too much, but don't take this as an offense.
What exactly is $V/U$ ? I will explain this a little, since it might be the source of the problem. $V/U$ is the set of the class of equivalences of vectors from $V$. The equivalence relation is given by $x \sim y :\iff x-y \in U$. This set can naturally be considered as a vector space if you define:
$$[x] + [y] := [x+y] \text{ and } \lambda[x] := [\lambda x]$$

Let's take an example: $V = \mathbb{R}^4$, $U = \mathbb{R}^2$, $W = \mathbb{R}$. First, what does it mean to consider $\mathbb{R}^2$ and $\mathbb{R}$ as subspaces of $\mathbb{R}^4$ ? Well, we need to actually consider $\mathbb{R}^2 \times \{(0,0)\}$ and $\mathbb{R} \times \{(0,0,0)\}$. Thus, actually $U = \{ x = (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 | x_3 = x_4 = 0\}$ and $W = \{ x = (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 | x_2 = x_3 = x_4 = 0\}$.

So a first equivalence relation $\sim_A$ can be defined for $V/U$:
$$x \sim_A y :\iff x_3 = y_3 \text{ and } x_4 = y_4$$

Same for $W/U$:
$$x \sim_B y :\iff x_2 = y_2, x_3 = y_3, x_4 = y_4$$

Thus let's write, $V/U = \{[x]_A \}$ and $V/W = \{[x]_B \}$, where obviously $[x]_A = \{y \in \mathbb{R}^4 | y_3 = x_3, y_4 = x_4\}$ and $[x]_B = \{y \in \mathbb{R}^4 | x_2 = y_2, x_3 = y_3, y_4 = x_4\}$.

Note that $V/U$ has dimension 2 (a base is given by $\{[(0,0,1,0)]_A , [(0,0,0,1)]_A\}$) and that $V/W$ has dimension 3 (a base is given by $\{[(0,1,0,0)]_B , [(0,0,1,0)]_B, [(0,0,0,1)]_B\}$ ).

Now, how can we define an addition $[u]_A + [v]_B$ ? This would be trying to add sets. You can try to define $[u]_A + [v]_B$ in a natural way but you won't manage it: simply saying $[u]_A + [v]_B := (u_1+v_1, u_2 + v_2 , u_3 + v_3, u_4 + v_4)$ won't work, the addition will not be well-defined (because if you change the representant of $[u]_A$, $u_1$ and $u_2$ can be anything, and if you change representant $[v]_B$, $v_1$ can be anything).



Furthermore, your intuition should tell you that since both dimensions of $V/U$ and $V/W$ are 2 and 3, such an addition can only be defined properly in a vector space of at least dimension 5 (otherwise you lose information). Too bad $\mathbb{R}^4$ has only dimension 4. So you should be convinced that it is impossible to define a natural addition properly in $V$ for this example.


(Caution: I don't say it is impossible to define an addition inside $V$. Actually, it is possible to define several additions, but it won't be "natural")


 
Hence, the only way to do it properly is to consider $V/U \oplus V/W$. Of course, in special cases, where $\dim(V/U) + \dim(V/W) \leq \dim(V)$, then it is possible to rearrange things so that you consider $V/U \oplus V/W$ as a subspace of $V$.
Sorry for the long answer, that depending on your level of understanding might not be very clear.
A: Certainly you can add them as elements of $V$.  They should all have the same number of components as the dimension of $V$.  But if $u_0 \in U$ why isn't $u_0+U$ just $U$?  $U$ is a vector space, so is closed under addition.
