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### 5 Matching Questions

1. Theorem 4.5: Test for a Subspace
2. Definition of Spanning Set of a Vector Space
3. Definition of Basis
4. Theorem 4.8: A Property of Linearly Dependent Sets (Proof)
5. Theorem 4.12: Basis Tests in an n-dimensional Space
1. a A set of vectors S={v1, v2,..., vn} in vector space V is called a basis for V if the following conditions are true:
1. S space V.
2. S is linearly independent.
2. b 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.
3. c Let V be a vector space of dimension n.
1. if S={v1, v2,..., vk} is a linearly independent set of vectors in V, then S is a basis for V.
2. If S={v1, v2,..., vk} spans V, then S is a basis for V.
4. d 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.
5. e Assume S is a linearly dependent set. Then there exist scalars c1, c2, c3,..., ck (not all zero) that c1v1 + c2v2 + c3v3+...+ ckvk=0 at least one coefficient must be nonzero, so assume c1 is not 0, and solve for v1 as a linear combination of the other vectors. c1v1 = -c2v2 - c3v3-...- ckvk > v1 = -(c2/c1)v2 -(c3/c1)v3 -...-(ck/c1)vk. Conversely, suppose v1 in S is a linear combination of the other vectors v1 = c2v2 + c3v3 +...+ ckvk. Then the equation -v1 + c2v2 + c3v3 +...+ ckvk=0 has at least one nonzero coefficient -1. Therefore S is linearly dependent.

### 5 Multiple Choice Questions

1. 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.
2. A vector v in vector space V is called a linear combination of the vectors u1, u2,..., uk in V can be written in the form v=c1u1 + c2u2 +...+ ckuk, where c1, c2,..., ck are scalars.
3. 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.
4. 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.
5. If V and W are both subspaces of a vector space U, then the intersection of V and W (V^W) is also a subspace of U.

### 5 True/False Questions

1. Theorem 4.9: Uniqueness of Basis Representation (Proof)If S={v1, v2,..., vn} is a basis for a vector space V, then every vector in V can be written in one and only one way as a linear combination of vectors in S.

2. Definition of Dimension of a Vector SpaceLet 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.

3. 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.

4. Theorem 4.7: Span(S) is a Subspace of V (Proof)If 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).

5. Definition of Linear Dependence and Linear IndependenceA set of vectors S={v1, v2,..., vk} in a vector space V is called linearly independent if the vector equation c1v1 + c2v2 +...+ ckvk=0 has only the trivial solution c1=0, c2=0,..., ck=0. If there are also nontrivial solutions, then S is called linearly dependent.