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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$\begingroup$$W_1+W_2$ is by definition the set of all vectors $w_1+w_2$ such that $w_1\in W_1$ and $w_2\in W_2$. You have $$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\}\;,$$ so you’re looking at all vectors of the form $x(1,3)+y(2,0)$ for $x,y\in\Bbb R$. Every vector in $W_1$ can be written in this form (with $y=0$), and every vector in $W_2$ can be written in this form as well, but you can’t expect every vector in $W_1+W_2$ to belong to $W_1$ or $W_2$. In fact, this occurs if and only if one of the subspaces $W_1$ and $W_2$ is a subspace of the other.
Note: you must combine each vector in $W_1$ with every vector in $W_2$, so you need to allow the coefficients $x$ and $y$ to be different; that’s why I have $x$ and $y$ and not $x$ and $x$.
$\endgroup$ 8 $\begingroup$Carefully note that for any two sets (not only for subspaces) $S$ & $T$, $S+T=${$s+t:s\in S, t\in T$}. Thus your sample vector viz $(3,3)$ is just a single element of $W_1+W_2$. You need to accommodate all such in $W_1+W_2$. Thus what should be the general form of a vector in $W_1+W_2$? Isn't it $(x,3x)+(2y,0)$ for $x,y\in \mathbb R$?
$\endgroup$ 3 $\begingroup$Yes, that is the way. You have to add all pairs of $W_1$ and $W_2$. So, formally $$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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