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Solutions to Tutorial 3 Simon Rose March 21, 2013 Recall that a subset W ⊂ V is a subspace if 1. The zero vector ~0 ∈ W . 2. For every w, v ∈ W we have that w + v ∈ W . 3. For every w ∈ W and for every α ∈ R, we have that α · w ∈ W Remember also that the zero vector of R2 is ~0 = (0, 0). Problem 1 (a): S = {(x, y) ∈ R2 | 6x + 8y = 0}. Let us check the three conditions: (a) Let us check that (x, y) = (0, 0) satisfies the given condition. 6x + 8y = 6 · 0 + 8 · 0 = 0 + 0 = 0 and so ~0 ∈ S. (b) Suppose that w, v ∈ S. Then w = (x1 , y1 ) such that 6x1 + 8y1 = 0 v = (x2 , y2 ) such that 6x2 + 8y2 = 0 Then w + v = (x1 + x2 , y1 + y2 ) (by definition of addition). Let us consider the defining condition for S: we want that 6(x1 + x2 ) + 8(y1 + y2 ) = 0. Let us compute this: 6(x1 + x2 ) + 8(y1 + y2 ) = 6x1 + 6x2 + 8y1 + 8y2 = 6x1 + 8y1 + 6x2 + 8y2 | {z } | {z } =0 =0 =0+0=0 and so w + v ∈ S, meaning that this also satisfies condition 2. (c) Lastly, let us consider multiplication. Again, consider some w = (x1 , y1 ) ∈ S. Because w, ∈ S, we have that 6x1 + 8y1 = 0. So choose α ∈ R, and consider the product α · w = α · (x1 , y1 ) = (αx1 , αy1 ) 1 Then we have that 6(αx1 ) + 8(αy1 ) = α(6x1 ) + α(8y1 ) = α(6x1 + 8y1 ) | {z } =0 =α·0=0 and so α · w satisfies the defining condition of the subset S. It follows that S satisfies all of the conditions to be a subspace of V . Problem 1 (g): S = {(x, y) ∈ R2 | x2 + y 2 = 1} This is not a subspace, since the very first condition fails: 02 + 02 6= 1, and so ~0 ∈ / S. Problem 1 (h): S = {(x, y) ∈ R2 | xy = 0} The first and the third conditions are satisfied (This implies that this set is what mathematicians call a cone), which you can and should verify for practice. The second condition, however, fails. To show this, we want to choose w, v ∈ S such that w + v ∈ / S. So let us consider w = (1, 0) and v = (0, 2). Note that both of these satisfy the condition that xy = 0. Then w + v = (1, 0) + (0, 2) = (1, 2) Now, the defining condition for S is that the product of the two coordinates is 0. However, 1 · 2 = 1 6= 0, and so this does not satisfy that condition. Hence it is not in S as claimed, and so S is not a subspace of R2 . Problem 4: Show that if W is a subspace of (V, +, ·), then (W, +, ·) is also a vector space. Proof We first note that since addition and multiplication are simply those from V , they automatically satisfy the following properties (why should this be so? Think about this and convince yourself!): (a) Associativity of + (b) Commutativity of + (c) Associativity of · (d) Distributivity of +, · (both versions) (e) 1 · x = x Thus it only remains to verify the following two properties. (a) There exists a zero element in W (b) For all v ∈ W , there is an element (−v) ∈ W such that v + (−v) = ~0 2 The first property is true by assumption that W is a subspace; in particular, it included the assumption that ~0 ∈ W . For the last, choose some v ∈ W . Since W is a subspace, and since (−1) ∈ R, it follows that (−1) · v ∈ W . But v + (−1) · v = 1 + (−1) · v = 0 · v = 0 as we have seen earlier, and so this last condition is satisfied. Thus (W, +, ·) is a vector space as claimed. Q.E.D. Remark Alternatively, we could have used the earlier Theorem from class which stated that (−1) · v = (−v) 3