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I have two subspaces:

$$W_1 = \{(x, 3x) : x\in \Bbb R \}$$ and

$$W_2 = \{(2x, 0): x\in \Bbb R \}$$

How do I get $W_1 + W_2$? I tried simply adding a sample vector from each, i.e. $$ (1, 3) + (2, 0) = (3, 3)$$ but I don't think this makes sense since this new vector doesn't fit it $W_1$ nor $W_2$....

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up vote 7 down vote accepted

$W_1+W_2$ is by definition the set of all vectors $w_1+w_2$ such that $w_1\in W_1$ and $w_2\in W_2$. You have $$W_1=\big\{(x,3x):x\in\Bbb R\big\}=\big\{x(1,3):x\in\Bbb R\big\}$$ and $$W_2=\big\{(2x,0):x\in\Bbb R\big\}=\big\{x(2,0):x\in\Bbb R\big\}\;,$$ so you’re looking at all vectors of the form $x(1,3)+y(2,0)$ for $xy\in\Bbb R$. Every vector in $W_1$ can be written in this form (with $y=0$), and every vector in $W_2$ can be written in this form as well, but you can’t expect every vector in $W_1+W_2$ to belong to $W_1$ or $W_2$. In fact, this occurs if and only if one of the subspaces $W_1$ and $W_2$ is a subspace of the other.

Note: you must combine each vector in $W_1$ with every vector in $W_2$, so you need to allow the coefficients $x$ and $y$ to be different; that’s why I have $x$ and $y$ and not $x$ and $x$.

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So is there any way to explicitly describe $W_1 + W_2$ as its own entity or can you only describe it as the sum of two other spaces? – CodyBugstein Dec 9 '12 at 3:12
@Imray: In this case it has a very simple description. The vectors $(1,3)$ and $(2,0)$ are linearly independent, so $W_1+W_2$ is ... ? – Brian M. Scott Dec 9 '12 at 3:21
.......$\Bbb R^2$ ? – CodyBugstein Dec 9 '12 at 3:25
@Imray: That’s right: it’s the whole space. – Brian M. Scott Dec 9 '12 at 3:32
:) What about if these did not span a whole space, i.e. two subspaces of $\Bbb R^3$ that didn't span the whole space? – CodyBugstein Dec 9 '12 at 3:37

Carefully note that for any two sets (not only for subspaces) $S$ & $T$, $S+T=${$s+t:s\in S, t\in T$}. Thus your sample vector viz $(3,3)$ is just a single element of $W_1+W_2$. You need to accommodate all such in $W_1+W_2$. Thus what should be the general form of a vector in $W_1+W_2$? Isn't it $(x,3x)+(2y,0)$ for $x,y\in \mathbb R$?

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So the space $W_1+W_2$ is $(x+2y, 3x)$ ?? – CodyBugstein Dec 9 '12 at 3:13
Yes. But for all $x,y\in \mathbb R$. – Sugata Adhya Dec 9 '12 at 3:14
In the example I gave, what is $W1∪W2$? – CodyBugstein Dec 9 '12 at 3:42

Yes, that is the way. You have to add all pairs of $W_1$ and $W_2$. So, formally $$W_1+W_2=\{w_1+w_2\mid w_1\in W_1\text{ and }w_2\in W_2\}.$$ For example the sum of two lines (both containing the origo) in the space is the plane they span.

Anyway, it is worth to mention, that $W_1+W_2$ is the smallest subspace that contains $W_1\cup W_2$.

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Aren't $W_1+W_2$ and $W_1\cup W_2$ the same thing? – CodyBugstein Dec 9 '12 at 3:06
Absolutely not. Note that $W_1\bigcup W_2$ is not necessarily a subspace but $W_1+W_2$ is. – Sugata Adhya Dec 9 '12 at 3:08
In the example I gave, what is $W_1\cup W_2$? – CodyBugstein Dec 9 '12 at 3:15
It's just the two lines $y=3x$ & the $x$-axis whereas $W_1+W_2$ is the entire $\mathbb R^2$. Can you give the reason behind it? – Sugata Adhya Dec 9 '12 at 4:30

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