I am working through Axler’s “Linear Algebra Done Right”, and it asks the reader to verify that if
$$
U_1,…,U_m
$$
are subspaces of $V$, a vector space over $F$, a field either $\Bbb C$ or $\Bbb R$, then the sum of the subspaces
$$
U_1 + \dots + U_m = \{u_1 + \dots + u_m : u_1 \in U_1, \dots , u_m \in U_m \}
$$
is a subspace of $V$:
I feel intuition for why this should be true, but am unable to feel confident that I can rigorously prove that it is a the sum is a subspace. Could someone post a link to a proof (preferable) or prove it:
I. the sum is a subset of $V$;
II. the sum contains the additive identity (I think I have this one; each subset will have an additive identity, since it is a vector space, and there will be a possible sum of only additive identities, thus the sum will contain the additive identity of $V$.);
III. the sum is closed under addition;
IV. the sum is closed under scalar multiplication,
and thus proving the sum is a subspace of $V$.
Best Answer
I. $U_1+...+U_m \subseteq V$ by the vector space (additive) closure axiom for $V$
II. For each $U_i$, $0 \in U_i$ since $U_i$ is a subspace, so $0+...+0=0 \in U_1+...+U_m$.
III. Take $u_1+...+u_m, v_1+...+v_m \in U_1+...+U_m$. Then: $u_1+...+u_m+v_1+...+v_m=(u_1+v_1)+...+(u_m+v_m)$ by commutativity. $u_i + v_i \in U_i$ by additive closure of $U_i$, since $U_i$ is a subspace, thus $(u_1+v_1)+...+(u_m+v_m) \in U_1+...U_m$.
IV. Take $a \in F$. Then $au_1+...+au_m \in U_1+...+U_m$, since $au_i \in U_i$ since $U_i$ is a subspace. Then by definition $au_1+...+au_m=a(u_1+...+u_m) \in U_1+...+U_m$.