91Ó°ÊÓ

Let's take an arbitrary vector v in span(S). By definition, v can be written as a linear combination of the vectors in S. That is, we can write: \(v = \sum_{i=1}^n c_i s_i\), where \(c_i\) are scalar coefficients and \(s_i\) are the vectors in S. Now, let's take an arbitrary vector u in S^\perp, which implies that every vector in S is orthogonal to u. So, we have: \(\langle u, s_i\rangle = 0\) for all \(s_i\) in S. Now, we will compute the inner product of u with v: \(\langle u, v\rangle = \langle u, \sum_{i=1}^n c_i s_i\rangle = \sum_{i=1}^n c_i \langle u, s_i\rangle\) Since \(\langle u, s_i\rangle = 0\), we have: \(\langle u, v\rangle = \sum_{i=1}^n c_i (0) = 0\) This implies that the vector v is orthogonal to every vector in S^\perp, and therefore v ∈ (S^\perp)^\perp, which is S^{00}. Since v was an arbitrary vector in span(S), we have proven that span(S) ⊆ S^{00}.

Let's take an arbitrary vector w in S^{00}. By definition, w is orthogonal to every vector in S^\perp. In other words, for any vector u in S^\perp, we have: \(\langle w, u\rangle = 0\) By definition of the orthogonal complement, this means that the vector w is orthogonal to every vector u that is orthogonal to every vector v in S. Hence, w must lie in the set span(S). Thus, we have shown that for an arbitrary vector w in S^{00}, w ∈ span(S), and therefore S^{00} ⊆ span(S).

Since we have proven that span(S) ⊆ S^{00} and S^{00} ⊆ span(S), we can now conclude that the linear span of a subset S in V is equal to its double orthogonal complement: \(\operatorname{span}(S) = S^{00}\)