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### 5 Matching Questions

1. Definition of Spanning Set of a Vector Space
2. Theorem 4.6: The Intersection of Two Subspaces is a Subspace (Proof)
3. Theorem 4.10: Basis and Linear Dependence
4. Theorem 4.9: Uniqueness of Basis Representation (Proof)
5. Definition of a Subspace
1. a Let S={V1, V2,..., Vk} be a subset of a vector space V. The set S is called a spanning set of V if every vector in V can be written as a linear combination of vectors in S. In such cases it is said that S spans V.
2. b A nonempty subset W of a vector space V is called a subspace of V if W is a vector space under the operations of addition and scalar multiplication defined in V.
3. c If S={v1, v2,..., vn} is a basis for a vector space V, then every set containing more than n vectors in V is linearly dependent.
4. d Because V and W are both subspaces of U, you know that both contain the zero vector, so V^W is nonempty. Let v1 and v2 be any two vectors in V^W. THen bcause V and W are both subspaces of U, you know that both are closed under addition. Because v1 and v2 are both in V, their sum v1+v2 is in W because v1 and v2 are both in W. This implies that v1+v2 is in V^W, and therefore V^W is closed under addition.
5. e Because S spans V, we know tat an arbitrary vector u in V can be expressed as u=c1v1 + c2v2+...+cnvn. Suppose u has another representation u=b1v1 + b2v2 +...+bnvn. Subtracting the second from the first we get u-u=(c1-b1)v1 + (c2-b2)v2 +...+(cn-bn)vn=0. But because S is linearly independent, the only solution to this equation is the trivial solution. c1-b1=0, c2-b2=0,..., cn-bn=0 > ci=bi for all i. Therefore u has only one representation.

### 5 Multiple Choice Questions

1. If a vector space V has a basis consisting of n vectors, then the number n is called the dimension of V, denoted by dim(V)=n. If V consists of the zero vector alone, the dimension of V is defined as zero.
2. If W is a nonempty subset of a vector space V, then W is a subspace of V if and only if the following closure conditions hold:
1. If u and v are in W, then u+v are in W.
2. If u is in W and c is any scalar, then cu is in W.
3. A set S={v1, v2,..., vk}, k>=2, is linearly dependent if and only if at least one of the vectors vj can be written as a linear combination of the other vectors in S.
4. To show that span(S) 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). u=c1v1 + c2v2 +...+ckvk, v=d1v1 + d2v2 +...+dkvk where c1, c2,..., ck and d1, d2,..., dk are scalars. Then u+v=(c1+d1)v1 + (c2+d2)v2 +...+ (ck+dk)vk and cu=(cc1)v1 + (cc2)v2 +...+(cck)vk. 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.
5. Let S1={V1, V2,...,Vn} be the basis for V and let S2={U1, U2,...,Un} be any other basis for V. Because S1 is a basis and S2 is linearly independent, Theorem 4.10 implies that m<=n. Similarly, n<=m because S1 is linearly independent and S2 is a basis. Consequently n=m.

### 5 True/False Questions

1. Theorem 4.7: Span(S) is a Subspace of VIf S={V1, V2,..., Vk} is 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 contains S must contain span(S).

2. Definition of Linear Combination of VectorsIf a vector space V has a basis consisting of n vectors, then the number n is called the dimension of V, denoted by dim(V)=n. If V consists of the zero vector alone, the dimension of V is defined as zero.

3. Theorem 4.9: Uniqueness of Basis RepresentationBecause S spans V, we know tat an arbitrary vector u in V can be expressed as u=c1v1 + c2v2+...+cnvn. Suppose u has another representation u=b1v1 + b2v2 +...+bnvn. Subtracting the second from the first we get u-u=(c1-b1)v1 + (c2-b2)v2 +...+(cn-bn)vn=0. But because S is linearly independent, the only solution to this equation is the trivial solution. c1-b1=0, c2-b2=0,..., cn-bn=0 > ci=bi for all i. Therefore u has only one representation.

4. Definition of the Span of a SetIf S={V1, V2,..., Vk} is a set of vectors in a vector space V, then the span of S is the set of all linear combinations of the vectors in S, span(S)={c1V1 + c2V2 +...+ ckVk}: c1, c2,..., ck are real numbers}. If span(S) = V, it is said that V is spanned by {V1, V2,..., Vk}, or that S spans V.

5. Theorem 4.8: A Property of Linearly Dependent Sets (Proof)A set S={v1, v2,..., vk}, k>=2, is linearly dependent if and only if at least one of the vectors vj can be written as a linear combination of the other vectors in S.