In Axler's proof of the dimension of sum formula (page 47 of Linear Algebra Done Right), there is a step that requires showing that $u_1,…,u_m,v_1,…,v_j,w_1,…w_k$ is a basis of $U_1+U_2$.
Now, I understand that first I have to show that this set of vectors spans $U_1+U_2$. However, he says:
"Clearly span($u_1,…,u_m,v_1,…,v_j,w_1,…w_k$) contains $U_1$ and $U_2$, and hence equals $U_1+U_2$."
Why does that chain of logic lead to $U_1+U_2 =$ span($u_1,…,u_m,v_1,…,v_j,w_1,…w_k$)? Isn't it supposed to lead to $U_1+U_2 \subseteq $ span($u_1,…,u_m,v_1,…,v_j,w_1,…w_k$)? What about showing that span($u_1,…,u_m,v_1,…,v_j,w_1,…w_k$) $\subseteq$ $U_1+U_2$? What does it exactly mean to say that $u_1,…,u_m,v_1,…,v_j,w_1,…w_k$ spans $U_1+U_2$?
Best Answer
I'll add a little more. You are right that one technically has to show both inclusions.
The since every vector $u_{i}$, $v_{j}$, $w_{k}$ is in $U_{1}+U_{2}$, the span of them is in $U_{1}+U_{2}$ because it's a vector subspace (closed under vector addition and scalar multiplication). Now every vector $v\in U_{1}+U_{2}$ is the sum $v_{1}+v_{2}$ for $v_{i}\in U_{i}$. Now the $u_{i}$ with the $v_{j}$ give a basis for $V_{1}$ so $v_{1}=\sum a_{i}u_{i} +\sum b_{j}v_{j}$ and similarly, $v_{2} = \sum\tilde{a}_{i}u_{i}+\sum\tilde{b}_{i}w_{j}$. Puting these together yeilds $$ \begin{align}v&=\sum a_{i}u_{i} +\sum b_{j}v_{j}+ \sum\tilde{a}_{i}u_{i}+\sum\tilde{b}_{i}w_{j}\\ &=\sum (a_{i}+\tilde{a}_{i})u_{i}+\sum b_{j}v_{j}+\sum\tilde{b}_{i}w_{j} \end{align} $$
So you have the reverse inclusion as well.