91Ó°ÊÓ

Let's first assume that \(W\) is a subspace of vector space \(V\). As the span of \(W\) contains all linear combinations of vectors of \(W\), the whole subspace \(W\) is included in \(\operatorname{span}(W)\). So, we have \(W \subseteq \operatorname{span}(W)\). Since \(W\) is a subspace, by definition, it contains all linear combinations of its vectors, so it also contains \(\operatorname{span}(W)\). Thus, we have \(\operatorname{span}(W) \subseteq W\). Combining these two statements, we have \(W = \operatorname{span}(W)\).

Now, let's assume that \(\operatorname{span}(W) = W\). We need to verify the three properties of a subspace of a vector space to prove that \(W\) is a subspace. 1. Since \(\operatorname{span}(W)\) is always a subspace of \(V\), it must contain the zero vector. Therefore, if \(\operatorname{span}(W) = W\), it implies that the zero vector is in \(W\). 2. Let \(\vec{u}, \vec{v} \in W\). Since \(\operatorname{span}(W) = W\), we can write \(\vec{u}, \vec{v}\) as linear combinations of vectors in \(W\). Adding these two linear combinations gives another linear combination in \(W\), and since \(\operatorname{span}(W) = W\), the sum of these two vectors (\(\vec{u} + \vec{v}\)) must also be in \(W\). 3. Let \(\vec{u} \in W\) and let \(c \in \mathbb{F}\). Since \(\operatorname{span}(W) = W\), we can represent \(\vec{u}\) as a linear combination of vectors in \(W\). Multiplying this linear combination by the scalar \(c\) gives us another linear combination in \(W\), and since \(\operatorname{span}(W) = W\), the product \(c\vec{u}\) must also be in \(W\). As \(W\) satisfies all three properties of a subspace, we have proven that if \(\operatorname{span}(W) = W\), then \(W\) is indeed a subspace of \(V\). Thus, we have shown that a subset \(W\) of a vector space \(V\) is a subspace of \(V\) if and only if \(\operatorname{span}(W) = W\).