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Linear Algebra 1 Quiz 7 Solutions 7/3/3 Linear Algebra 1 Quiz 7 Solutions 7/3/3 Question 1 Explain why each of the following sets is or is not a vector space, with the given operations of ”addition” and ”scalar multiplication”: • The set X of all non-negative real numbers under ordinary addition, with ”scalar multiplication” by the absolute value of the scalar: a +X b = a + b, sXa = |s|a. If this is a vector space then 0 is the zero vector, since a+X 0 = 0+X a = a, for any a in X. In a vector space if v is a vector, then there is a unique vector w, called the additive inverse of v, such that v + w = 0. Here 1 is a positive real number, so it is in the space X. But if w is any element of X, then 1 +X w = 1 + w is strictly positive and is never zero. So the additive inverse of 1 does not exist. So X is not a vector space with the given operations. • The set Y of all two by two matrices with ”addition” being ordinary matrix multiplication and with scalar multiplication the standard mul- tiplication of a matrix by a scalar: A +Y B = AB, sYA = sA. In a vector space the addition of vectors is commutative. But multiplication of two by two matrices is not commutative, so the operation of multiplication will not do as an operation of vector addi- tion for the set Y. Specifically we may multiply the following matrices: 0 1 0 0 A = B = 0 0 1 0 1 0 0 0 AB = BA = 0 0 0 1 Since AB 6= BA, we have A +Y B 6= B +Y A, so the set Y is not a vector space with the given operations. • The set Z of all points [x, y, z] on the line x = 3t, y = 7t, z = −2t, as t ∈ R varies, under ordinary vector addition and scalar multiplication. This is a vector space. To show this, by the subspace theorem, we just have to show that the operations are well-defined on Z and then Z is automatically a subspace of R3. Let A and B be two points of Z. Then A = [3s, 7s, −2s] and B = [3t, 7t, −2t], for some real s and t. Then we have: A + B = [3s + 3t, 7s + 7t, −2s − 2t] = [3(s + t), 7(s + t), −2(s + t)] = [3u, 7u, −2u]. Here u = s + t. So A + B lies in Z. Next for any real r, we have: rA = r[3s, 7s, −2s] = [3rs, 7rs, −2rs] = [3w, 7w, −2w]. Here w = rs. So rA lies in Z. So the vector space operations of R3 are well-defined on Z. So Z is a subspace of R3 and, in particular, Z is a vector space. • The collection U of all polynomials in the variable x, of degree in x no more than four, which have the same value at x = −1 and x = 1, with the usual rules for addition and scalar multiplication of polynomials. We see that U is a vector space. We know that the space of all real polynomials of degree not more than four forms a vector space, P4. So, by the subspace theorem, we just have to show that U is closed under the operations of P4. If f(x) and g(x) are in U, so f(1) = f(−1) and g(1) = g(−1), put h(x) = (f + g)(x) = f(x) + g(x) ∈ P4. Then we have: h(1) = f(1) + g(1) = f(−1) + g(−1) = h(−1). So h ∈ U. If s ∈ R, put k(x) = (sf)(x) = sf(x). Then we have: k(1) = sf(1) = sf(−1) = k(−1). So k ∈ U. So U is closed under addition and scalar multiplication in P4, so U is a subspace of P4 and is vector space, as required. Question 2 Let A be the set of all (x, y, z) ∈ R3 such that x − y + 3z = 0. • Show that A is a vector space, with the usual rules for vector addition and scalar multiplication. The set A is given as the solution space of a homogeneous linear sys- tem, so forms a vector space: If we write any such system as Mv = 0, with v ∈ Rn and M a fixed ma- trix, then if Ma = 0 and Mb = 0, then M(a+b) = Ma+Mb = 0+0 = 0, so the solution space is closed under addition. Also, if Ma = 0 and s is a real number, then M(sa) = s(Ma) = s(0) = 0, so the solution space is closed under scalar multiplication. So the solution space has well-defined operations of addition and scalar multiplication and then by the subspace theorem, the other axioms of a vector space follow from those of Rn. • Find the general solution of the homogeneous system x − y + 3z = 0 and show that every element of the vector space A can be written as some linear combination of two fixed vectors in the space A. The linear system has the reduced echelon form: 1 −1 3 0 There is one pivot, corresponding to the variable x and two free vari- ables, the variables y and z. So we solve by putting y = s and z = t. Then x = y − 3z = s − 3t, so we get: (x, y, z) = (s − 3t, s, t) = s(1, 1, 0) + t(−3, 0, 1), where s and t are arbitrary real numbers. This shows that the general solution is a linear combination of the two vectors B = (1, 1, 0) and C = (−3, 0, 1). But each of these vectors solve the given linear system, as is easily checked. So the general solution is a linear combination of two vectors of A, namely B and C, as required..
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