Suppose \({latex.inline[V_{1}, ...,V_{m}](V_{1}, ...,V_{m})} are subspaces of V. Then \){latex.inlineV{1} + ... + V{m}} is the smallest containing subspace containing ${latex.inlineV{1}, ..., V{m}}.
We first verify that the sum of the subspaces is itself a subspace. Then we show it is the smallest one containing all of the individual subspaces.
To show the sum is a subspace, we use the criteria set out in 1753142531 - Axler 1.34 Conditions for a subspace|1.34:
0 is in the sum We know that in each \({latex.inline[0 \in V_{i}](0 \in V_{i})} per subspace criteria. Then 0 in the sum, which we call \){latex.inlineV_{s}} from now on, can be written as: \({latex.inline[0 = 0_{1} + ... + 0{m}](0 = 0_{1} + ... + 0{m})} where each 0 on the RHS is the 0 from \){latex.inlineV_{i}}.
the sum is closed under addition Suppose we have two elements \({latex.inline[v_{1s}, v_{2s} \in V_{s}](v_{1s}, v_{2s} \in V_{s})}. Then by definition of the sum of subspaces, we have that \){latex.inlinev{1s} = v{1} + ... + v_{m}}. But \({latex.inline[v_{2s} = w_{1} + ... + w_{m}](v_{2s} = w_{1} + ... + w_{m})}. So when we add \){latex.inlinev_{1s}} and \({latex.inline[v_{2s}](v_{2s})} together, we get: \){latex.inlinev{1s} + v{2s} = (v{1} + w{1}) + ... + (v{m} + w{m})}. But, since each \({latex.inline[V_{i}](V_{i})} is a vector space, each \){latex.inlinev{i} + w{i} \in V_{i}} and we get that addition is closed.
the sum is closed under scalar multiplication This is essentially a repeat of the addition proof.
Now we show that the sum is the smallest containing subspace. Note that each individual subspace is contained in the sum. This is obvious if you consider: \({latex.inline[0 + 0 + v_{i} + 0 + 0](0 + 0 + v_{i} + 0 + 0)} for each \){latex.inlinev_{i}}.
Conversely, every subspace of V containing \({latex.inline[V_{1}, ..., V_{m}](V_{1}, ..., V_{m})} must also contain \){latex.inlineV{1} + ... + V{m}} because subspaces must contain all finite sums of their elements. Thus, we get the conclusion.
Suppose one of the Vs is not a subspace. Then the addition may just not be defined because we have different operations etc.