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Math 333 - Practice Exam with Some Solutions
(Note that the exam will NOT be this long.)
1 Definitions
- (0 points) Let U be a subset of a vector space V. Let S = {v 1 , v 2 ,... , vn} be another subset of V. (a) Define “U is a subspace of V ”.
(b) Define “S is linearly independent”.
(c) Define “S generates V ”.
2 Vector Spaces and Subspaces
- (0 points) (a) Give three examples of 4-dimensional vector spaces.
(b) Give one example of an infinite dimensional vector space.
(c) Give an example of a zero-dimensional vector space.
- (0 points) Let S 1 and S 2 be subspaces of a vector space V. Prove that the union S 1 ∪ S 2 is a subspace of V if and only if one is contained in the other (that is, either S 1 ⊆ S 2 or S 2 ⊆ S 1 .)
Solution: (⇐=) S 1 and S 2 are subspaces. If S 1 ⊆ S 2 , then S 1 ∪ S 2 = S 2 is a subspace. If S 2 ⊆ S 1 , then S 1 ∪ S 2 = S 1 is a subspace. We’ve proved one direction.
(=⇒) S 1 and S 2 are subspaces, and suppose S 1 ∪ S 2 is a subspace. If S 1 ⊆ S 2 , then we are done. If S 1 * S 2 , then we need to show S 2 ⊆ S 1. Choose x ∈ S 2. Since S 1 * S 2 there must be some vector in S 1 that is not in S 2 , call it y. So y ∈ S 1 , but y /∈ S 2. Since S 1 ∪ S 2 is a subspace, it is closed under addition and
x + y must be in S 1 ∪ S 2 since x ∈ S 2 ⊆ S 1 ∪ S 2 and y ∈ S 1 ⊆ S 1 ∪ S 2. Thus we must have either x + y ∈ S 1 or x + y ∈ S 2. If x + y ∈ S 2 , then since x ∈ S 2 and S 2 is a subspace (i.e. closed under the operations) we have y = (x + y) − x ∈ S 2 , which contradicts the fact that y /∈ S 2. Thus x + y ∈ S 1. However, since y ∈ S 1 and S 1 is a subspace (i.e. closed under the operations) we have x = (x + y) − y ∈ S 1. Therefore, S 2 ⊆ S 1.
3 Linear Independence, Generating Sets, and Bases
- (0 points) Let S = {x^2 + 3x, x − 2 } be a subset of P 2 (R). (a) Explain why S is not a basis of P 2 (R).
(b) Is 13 x^2 + 2 in span(S)? Explain.
(c) Is 2x^2 + 5x + 4 in span(S)? Explain.
- (0 points) Consider the 3 vectors in R^3 given by v 1 = (1, 1 , −1), v 2 = (1, 1 , 1), and v 3 = (3, 5 , 7). Decide whether these 3 vectors provide a basis for R^3. Justify your answer.
- (0 points) Let W be the subspace of R^3 given by
W = {(x, y, z) | x + y + z = 0 and x − y − z = 0}.
Find a basis for W and the dimension of W.
- (0 points) Let S = {v 1 , v 2 ,... , vn} be a set of n vectors in a vector space V. Show that if S is linearly independent and the dimension of V is n, then S is a basis of V.
Solution: This is Corollary 2 (b) at the top of page 48 of the textbook. The proof is found there.
So 0 = ax + by + cz = (a+ 2 b− c)(x + y) + (b+c 2 − a)(y + z) + (c+a 2 − b)(z + x), and this means that a + b − c = b + c − a = c + a − b = 0 since {x + y, y + z, z + x} is linearly independent. Clearly from those equalities we have a = b = c = 0. Therefore, {x, y, z} is also linearly independent.
- (0 points) Let S 1 and S 2 be subsets of a vector space V over a field F. Prove that
span(S 1 ∩ S 2 ) ⊆ span(S 1 ) ∩ span(S 2 ).
Solution: Let x ∈ span(S 1 ∩ S 2 ). Then there exist vectors v 1 , v 2 ,... , vn ∈ S 1 ∩ S 2 and coefficients a 1 , a 2 ,... , an ∈ F such that x = a 1 v 1 + a 2 v 2 + · · · + anvn. But since v 1 , v 2 ,... , vn ∈ S 1 , we see that x ∈ span(S 1 ). Similarly, since v 1 , v 2 ,... , vn ∈ S 2 , we see that x ∈ span(S 2 ). Thus we have x ∈ span(S 1 ) ∩ span(S 2 ).
- (0 points) Consider the vector space V = P 1 (R). (a) Explain why you know that the set β = {1 + x, 1 − 2 x} is a basis of V.
Solution: Since neither vector is a multiple of the other, β is linearly independent. Since the dimension of V is 2 and β has 2 elements, it must be a basis.
(b) Express p(x) = 2x − 3 as a linear combination of β.
Solution: p(x) = 2x − 3 = (− 4 /3)(1 + x) + (− 5 /3)(1 − 2 x)
