Section 44 55. Proof Complete the proof of Theorem 4.7. THEOREM 4.7 Span(S) Is a

Question

Section 44 55. Proof Complete the proof of Theorem 4.7. THEOREM 4.7 Span(S) Is a Subspace of V If S = {v,, v2, . . . , 1s a set of vectors in a vector space V, then span(S) is a subspace of V. Moreover, span(S) is the smallest subspace of V that contains S, in the sense that every other subspace of V that contains S must contain span( PROOF To show that span(S), the set of all linear combinations of v,, V2, . . ., V,, is a subspace of V, show that it is closed under addition and scalar multiplication. Consider any two vectors u and v in span(S), where C1, c2.. . .,c and d,d are scalars. Then and which means that u + v and cu are also in span(S) because they can be written as linear combinations of vectors in S. So, span(S) is a subspace of V. It is left to you to prove that span(S) is the smallest subspace of V that contains S. (See Exercise 55.)

Explanation / Answer

Let W be a subspace of V that contains S and let W span(S). Let u be an arbitrary element of span(S). Then u is a linear combination of v1,v2,…,vn. Since W is a vector space, it is closed under vector addition so that u    W. This implies that span(S) W. Thus, W = span(S). This means that span(S) is the smallest subspace of V that contains S.

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