Sejam $W_1, \dots , W_n$ subespaços, mostre que $L(W_1, \dots , W_n) = W_1 + \dots + W_n$.

Definamos $L(W_1, \dots , W_n) = L(W_1 + \dots + W_n)$.

Obviamente $W_1 + \dots + W_n \subset L(W_1, \dots , W_n)$. ${\large (I)}$

Sejam $w$ um elemento de $L(W_1, \dots , W_n)$, e $a_i,\ i \in \mathbb{N}_m$ escalares,

$w = \displaystyle\sum_i a_i \displaystyle\sum_{j=1}^n w_j = \displaystyle\sum_{j=1}^n w_j \displaystyle\sum_i a_i,\ w_j \in W_j$.

Tomando quaisquer $a_i$'s tais que $\displaystyle\sum_i a_i = 1$, $w = \displaystyle\sum_{j=1}^n w_j$ que é um elemento de $\displaystyle\sum_{j=1}^n W_j$. Logo $L(W_1, \dots , W_n) \subset W_1 + \dots + W_n$. ${\large (II)}$

${\large (I)}\ \wedge\ {\large (II)}\ \Rightarrow\ L(W_1, \dots , W_n) = W_1 + \dots + W_n$

Quod Erat Demonstrandum.