To show that $S + T = \text{Span}\{u_1, \dots, u_p, v_1, \dots, v_q\}$, let's proceed by proving that each side of the equality is contained in the other.

Step 1: Show that $S + T \subseteq \text{Span}\{u_1, \dots, u_p, v_1, \dots, v_q\}$.

1. By definition, $S = \text{Span}\{u_1, \dots, u_p\}$. This means that every vector in $S$ can be written as a linear combination of $u_1, \dots, u_p$.

2. Similarly, $T = \text{Span}\{v_1, \dots, v_q\}$, so every vector in $T$ can be written as a linear combination of $v_1, \dots, v_q$.

3. The sum $S + T$ is defined as: $S + T = \{ \mathbf{w} \in V \mid \mathbf{w} = \mathbf{s} + \mathbf{t} \text{ for some } \mathbf{s} \in S \text{ and } \mathbf{t} \in T \}.$

4. Since $\mathbf{s} \in S$ can be written as a linear combination of $u_1, \dots, u_p$, say $\mathbf{s} = a_1 u_1 + \dots + a_p u_p$, and $\mathbf{t} \in T$ can be written as a linear combination of $v_1, \dots, v_q$, say $\mathbf{t} = b_1 v_1 + \dots + b_q v_q$, we have: $\mathbf{w} = \mathbf{s} + \mathbf{t} = a_1 u_1 + \dots + a_p u_p + b_1 v_1 + \dots + b_q v_q.$

5. This shows that any vector $\mathbf{w} \in S + T$ can be expressed as a linear combination of $u_1, \dots, u_p, v_1, \dots, v_q$. Therefore, $\mathbf{w} \in \text{Span}\{u_1, \dots, u_p, v_1, \dots, v_q\}$.

6. Hence, $S + T \subseteq \text{Span}\{u_1, \dots, u_p, v_1, \dots, v_q\}$.

Step 2: Show that $\text{Span}\{u_1, \dots, u_p, v_1, \dots, v_q\} \subseteq S + T$.

1. Let $\mathbf{w}$ be any vector in $\text{Span}\{u_1, \dots, u_p, v_1, \dots, v_q\}$.

2. By definition, this means $\mathbf{w}$ can be written as: $\mathbf{w} = a_1 u_1 + \dots + a_p u_p + b_1 v_1 + \dots + b_q v_q$ for some scalars $a_1, \dots, a_p, b_1, \dots, b_q$.

3. Let $\mathbf{s} = a_1 u_1 + \dots + a_p u_p$ and $\mathbf{t} = b_1 v_1 + \dots + b_q v_q$. Then $\mathbf{s} \in S$ and $\mathbf{t} \in T$ by the definitions of $S$ and $T$.

4. Therefore, $\mathbf{w} = \mathbf{s} + \mathbf{t}$ with $\mathbf{s} \in S$ and $\mathbf{t} \in T$, which means $\mathbf{w} \in S + T$.

5. This shows that $\text{Span}\{u_1, \dots, u_p, v_1, \dots, v_q\} \subseteq S + T$.

Conclusion

Since we have shown both $S + T \subseteq \text{Span}\{u_1, \dots, u_p, v_1, \dots, v_q\}$ and $\text{Span}\{u_1, \dots, u_p, v_1, \dots, v_q\} \subseteq S + T$, it follows that: $S + T = \text{Span}\{u_1, \dots, u_p, v_1, \dots, v_q\}.$

This completes the proof.

Would you like any more details on specific steps?

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Here are some related questions to deepen your understanding:

1. What are the conditions for $S \cap T$ in terms of the spanning sets?
2. How does this result change if $S$ and $T$ are not subspaces?
3. What is the dimension of $S + T$ in terms of the dimensions of $S$ and $T$?
4. How does the span of two subspaces relate to their direct sum?
5. Can $S + T$ ever equal $V$ under certain conditions?

Tip: Remember that the sum of two subspaces $S + T$ captures all possible linear combinations of vectors from each subspace, allowing you to understand how they combine to form larger subspaces.