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 $W_2$….

#### Solutions Collecting From Web of "Adding two subspaces"

$W_1+W_2$ 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 $x,y\in\Bbb R$. 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 $x$.

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 $x,y\in \mathbb R$?

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 $W_1\cup W_2$.