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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?

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Read about direct sums of vector spaces. –  Mariano Suárez-Alvarez Nov 11 '10 at 2:48
    
This book only defined direct sums for subspaces. But, I am reading otherwise now. –  a little don Nov 11 '10 at 2:56
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2 Answers

up vote 2 down vote accepted

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.

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This was very helpful. I admit the question was murky. I am grading these exams and students write strange things like $(w_0+W) + (u_0 + U) = (w_0+u_0) + (W \cup U)$ and I need to write little notes that explain why this is wrong. (In this case it is wrong, at least because we have no reason to believe that $W \cup U$ is a subspace. But, then I think "What if there is some higher maths structure that I don't know about? Better ask..." –  a little don Nov 11 '10 at 15:16
    
@a little don : in that case, instead of trying to grasp complex concepts, it's easier to go back to the definitions. That helps a lot, especially if you are trying to find counter-example. –  Djaian Nov 11 '10 at 15:28
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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.

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I see what I messed up. $u_0, w_0 \in V$ and I was thinking... non-zero. Doh! Sorry. –  a little don Nov 11 '10 at 15:25
    
I was looking at $u_0+U$ as the set of vectors u_0+u for u any vector in U. Then we can add the two sets to make the set of vectors which are the sum of one vector from the first set and one from the second. My reading of $u_0+U$ is the same as this with the first set having only one element. As all this is within V, it goes through. As Djaian says, this doesn't work as a representative of an equivalence class, but it does work for sets. –  Ross Millikan Nov 11 '10 at 15:27
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