Definition
If $S$ is a subset of a vector space $\mathscr{V}$ then the subspace $L(S)$ generated by $S$ is the smallest vector subspace of $\mathscr{V}$ containing $S$.
Proposition
The subspace $L(S)$ generated by $S$ is the set of all finite linear combination of vectors of $S$.
Definition
If $W_1$ and $W_2$ are vector subspaces of $\mathscr{V}$ then their sum $W_1+W_2$ is the subspace $L(W_1\cup W_2)$ generated by $(W_1\cup W_2)$.
Proposition
If $W_1$ and $W_2$ are vector subspaces of $\mathscr{V}$ then
$$
W_1+W_2=\{w_1+w_2:w_1\in W_1,\,w_2\in W_2\}
$$Definition
The sum of two vector subspaces $W_1$ and $W_2$ of $\mathscr{V}$ is direct if $W_1\cap W_2={0}$. In particular the finite sum of a collection $\mathcal{W}:=\{W_i: i\in\ I\}$ vector subspace is said direct if $W_i\cap W_j=\{0\}$ for each $i,j\in I$.
So with the previous formalism I’m asked to prove that the sum of subspaces is commutative and associative and the direct sum too. In the case of the simple sum I thought that this could trivially follow by the fact that
$$
W_1+W_2=\{w_1+w_2:w_1\in W_1,\,w_2\in W_2\}=\{w_2+w_1:w_2\in W_2,\,w_1\in W_1\}=W_2+W_1
$$
and
$$
W_1+(W_2+W_3)=\{w_1+(w_2+w_3):w_i\in W_i, i=1,2,3\}=\{(w_1+w_2)+w_3:w_i\in W_i, i=1,2,3\}=(W_1+W_2)+W_3
$$
but I am not sure about the correctedness of the arguement. Then I think that I cant implement the first argument to show that
$$
W_1\oplus W_2=W_2\oplus W_1
$$
but I still am not sure about this and finally if the second argument shows that the sum is associative I would have only prove that $(W_1\oplus W_2)\cap W_3=\{0\}$ but I don't be able to do this unfortunately. So could someone help me, please?
Best Answer
The fact that $W_1+W_2=W_2+W_1$ is fairly obvious, because $$ W_1+W_2=L(W_1\cup W_2)=L(W_2\cup W_1)=W_2+W_1 $$ by the very definition.
What about associativity? In this case you use the proposition: if $W_1,W_2,W_3$ are subspaces, $X=W_1+W_2$ and $Y=W_2+W_3$, you want to prove that $$ X+W_3=W_1+Y $$ Let $x\in X,w_3\in W_3$; then, by the proposition, $x=w_1+w_2$, with $w_1\in W_1$, $w_2\in W_2$; then $$ x+w_3=(w_1+w_2)+w_3=w_1+(w_2+w_3)\in W_1+Y $$ because $w_2+w_3\in Y$. Thus $X+W_3\subseteq W_1+Y$. The reverse inclusion follows similarly.
About direct sums there is a big misunderstanding. While the definition of “direct sum” in the case of two subspaces is correct, it is incorrect to say that the sum of more than two subspaces is direct when $W_i\cap W_j=\{0\}$ for $i\ne j$.
The condition is stricter, namely that $$ W_i\cap\sum_{j\ne i}W_j=\{0\},\qquad i=1,2,\dots,n $$ at least if one wants to stick with the common terminology and one of the most important properties of direct sums, namely that $$ \dim(W_1\oplus W_2\oplus\dots\oplus W_n)=\dim W_1+\dim W_2+\dots+\dim W_n $$ in case of finite dimensional spaces.
For instance, the enclosing vector space being $\mathbb{R}^3$, if $W_1$ is generated by $(1,0,0)$, $W_2$ by $(0,1,0)$ and $W_3$ by $(1,1,0)$, it is true that $W_1\cap W_2=\{0\}$, $W_1\cap W_3=\{0\}$, $W_2\cap W_3=\{0\}$, but $$ \dim(W_1+W_2+W_3)=2\ne\dim W_1+\dim W_2+\dim W_3 $$
In any case, since a direct sum is a sum of subspaces to begin with, proving associativity (once the definition is fixed) and commutativity is not a problem, because it has already been done.
Let's tackle associativity of direct sum. Suppose $W_1,W_2,W_3$ are independent subspaces (meaning that their sum is direct). Then, by definition, $$ W_1\cap(W_2+W_3)=\{0\}=(W_1+W_2)\cap W_3 $$ Therefore also $W_1\cap W_2=\{0\}=W_2\cap W_3$; hence $W_1+W_2=W_1\oplus W_2$ and $W_2+W_3=W_2\oplus W_3$. Hence $$ W_1+(W_2+W_3)=W_1\oplus(W_2+W_3)=W_1\oplus(W_2\oplus W_3) $$ $$ (W_1+W_2)+W_3=(W_1\oplus W_2)+W_3=(W_1\oplus W_2)\oplus W_3 $$ But these are equal by the previous argument.