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MAT 242 | C LASS #90295 | FALL 2016 Problem Set #7 Solutions Due Wednesday, October 26 Problem 1: Let W be the subset of R3 consisting of all vectors (x, y, z) satisfying x = 5y. Is W a subspace of R3 ? Proof. To prove that W is a subspace, we need to show three things: (1) The set W is not the empty set. In other words, that W contains at least one element. (2) The set W is closed under addition. In other words, that given any two elements v, u ∈ W , their sum v + u is also an element of W . (3) The set W is closed under scalar multiplication. In other words, that given any elements v ∈ W and any scalar c ∈ R, the product c v is also an element of W . Let’s first prove (1). To do so, we simply note that (0, 0, 0) ∈ W since 0 = 5(0). Hence, W is nonempty. Next, let’s prove (2). Let v = (x, y, z) and u = (x 0 , y 0 , z 0 ) be arbitrary elements of W . We want to show that u + v = (x + x 0 , y + y 0 , z + z 0 ) is also an element of W . Which means we need to show that (x + x 0 ) = 5(y + y 0 ). Since v, u ∈ W , we know that x = 5y x 0 = 5y 0 . Therefore, x + x 0 = 5y + 5y 0 = 5(y + y 0 ). Since (x + x 0 ) = 5(y + y 0 ), we know that v + u = (x + x 0 , y + y 0 , z + z 0 ) is an element of W . Hence, W is closed under addition. Finally, let’s prove (3). Let v = (x, y, z) be an arbitrary element of W , and c ∈ R. We want to show that cv = (cx, c y, cz) is also an element of W . Which means we need to show that c x = 5(c y). Since v ∈ W , we know that x = 5y Therefore, c x = 5(c y). 1 Since cx = 5(c y), we know that cv = (c x, c y, c z) is an element of W . Hence, W is closed under scalar multiplication. • Problem 2: Let W be the subset of R2 consisting of all vectors (x, y) with |x| + |y| = 1. Is W a subspace of R2 ? Solution: The set W is not a subspace of R2 . There are a couple of ways to see this, but perhaps the easiest is to note that the zero vector (0, 0) is not an element of W since |0| + |0| 6= 1. Since any subspace of R2 must contain the zero vector, we know that W is not a subspace. We could also have shown that W is not closed under addition. Note that both (1, 0) and (0, 1) are both elements of W , but their sum (1, 1) is not. Similarly, W is not closed under scalar multiplication. Let c ∈ R with c 6= 1 and suppose (x, y) is an element in W . Then c(x, y) = (c x, c y) is not an element of W since |cx| + |c y| = c(|x| + |y|) = c 6= 1. • Problem 3: Let w = (7, 7, 9, 11). Is w a linear combination of the set S = {(2, 0, 3, 1), (4, 1, 3, 2), (1, 3, −1, 3)}? If it is, determine the linear combination of S giving w. If not, show that such a linear combination is impossible. Solution: We want to determine if there exist scalars such that 1 4 2 7 3 1 0 7 = x1 + x2 + x3 . −1 3 3 9 3 2 1 11 This is equivalent to solving the linear system whose augmented matrix is 2 4 1 7 0 1 3 7 . 3 3 −1 9 1 2 3 11 Putting this matrix in reduced echelon form, we get 1 0 0 6 0 1 0 −2 . 0 0 1 3 0 0 0 0 2 This means that w is a linear combination of S. Specifically, we have 2 4 1 7 0 1 3 7 6 −2 +3 = . 3 3 −1 9 11 1 2 3 • Problem 4: Is the following set S = {(3, 9, 0, 5), (3, 0, 9, −7), (4, 7, 5, 0)} linearly indpendent? Solution: To determine if these vectors are linear independent, we want to consider the linear system 3 3 4 0 9 0 7 0 x1 + x2 + x3 = . 0 9 5 0 5 −7 0 0 This homogeneous linear system has either one solution (the trivial solution (0, 0, 0)) or infinitely many solutions. If the only solution is the trivial solution, it means the vectors are linear independent. If there are infinitely many solutions, we know that the vectors are linearly dependent. Putting the augmented matrix associated to the above linear system into reduced echelon form, we get 1 0 7/9 0 0 1 5/9 0 . 0 0 0 0 0 0 0 0 We see that there are infinitely many solutions (there is one free variable), so the set S is linearly dependent. 3