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Mathematics 206 Solutions for HWK 13a Section 4.3 p184 Section 4.3 p184 Problem 5. For W the set of all functions that are continuous on [0, 1] and V the set of all functions that are integrable on [0, 1], verify that W is a subspace of V . Solution. We know that continuous functions on [0, 1] are also integrable, so each function belonging to W does belong to V . In other words, W is at least a subset of V . There are lots of continuous functions on [0, 1], so W is certainly nonempty. We show that W is both closed under addition and closed under scalar multiplication. Closed under addition. Assume that f and g belong to W . In other words, assume that f and g are continuous on [0, 1]. A theorem of calculus tells us that f + g is continuous at x if f and g are, so f + g is also continuous on [0, 1]. (This is essentially a consequence of the definition of continuity together with the fact that the limit of a sum is the sum of the limits, provided those limits exist.) In other words f + g belongs to W , as required. Closed under scalar multiplication. Assume that f belongs to W and c is a real number. Let g = cf . Here, too, a theorem of calculus tells us that if f is continuous at x, then so is the function we’ve called g. Therefore g belongs to W , as required. Section 4.3 p184 Problem 9. For W the set of all nonnegative functions in C(−∞, ∞), give a specific example to show that W fails the subspace test and is therefore not a vector subspace of C(−∞, ∞). Solution. Although W is both nonempty and closed under addition, it fails to be closed under scalar multiplication. One easy example that could be used to show this is to let f be, for instance, either the constant function 1 or the function that takes output to be the square of input (i.e. f (x) = x2 ) and take c to be a negative real number, for instance c = −2. Then we have f ∈ W but cf ∈ / W. Section 4.3 p184 Problem 11. For W the set of all n × n matrices with zero determinant, give a specific example to show that W fails the subspace test and is therefore not a vector subspace of Mn,n. Solution. In this case, W is closed under scalar multiplication but fails to be closed under addition. Once n is specified, easy to · it’s pretty ¸ · cook¸up an example to show this. With n = 2, for 0 0 2 1 instance we could put A = and B = . Then |A| = |B| = 0 but |A + B| = 1 6= 0. 1 1 0 0 In other words, A ∈ W, B ∈ W, but A + B ∈ / W. Page 1 of 4 A. Sontag March 25, 2001 Math 206 HWK 13a Solns contd 4.3 p184 Section 4.3 p184 Problem 13. Which of the following subsets of C(−∞, ∞) are subspaces of C(−∞, ∞)? (a) The set of all nonnegative functions (i.e. the set of all functions f satisfying f (x) ≥ 0 for every x). This set is not closed under scalar multiplication, hence NOT a subspace. (b) The set of all even functions (i.e. the set of all functions f satisfying f (−x) = −f (x) for every x) is a subspace. [Proof. We know even functions exist. Suppose f and g are even and c is a real number. Then, for every x, we have (fg )(−x) = f (−x) + g(−x) = f (x) + g(x) = (f + g)(x) and (cf )(−x) = c(f (−x)) = c(f (x)) = (cf )(x). ] (c) The set of all odd functions (i.e. the set of all functions f satisfying f (−x) = −f (x) for every x) is a subspace. The reasoning is similar to that for (b). (d) The set of all constant functions is a subspace. Constant functions exist, the sum of two constant functions is also constant, and every scalar multiple of a constant function is a constant function. (e) The set, W , say, of all functions f such that f (0) = 0. This set is a subspace. The constant function zero satisfies the required condition and therefore belongs to W . If f, g ∈ W , then (f + g)(0) = f (0) + g(0) = 0 + 0 = 0 so f + G ∈ W . Similarly, if f ∈ W and c is a scalar, then (cf )(0) = c(f (0) = c(0) = 0 so the function cf also belongs to W . (f) The set, W , say, of all functions f such that f (0) = 1. This set is NOT a subspace. Although it’s nonempty (the constant function 1 belongs), it is neither closed under addition nor closed under scalar multiplication. For instance, if f, g ∈ W , then (f + g)(0) = f (0) + g(0) = 1 + 1 = 2 6= 1 so f +g ∈ / W. Page 2 of 4 A. Sontag March 25, 2001 Math 206 HWK 13a Solns contd 4.3 p184 Section 4.3 p184 Problem 17. Given W = {(a, b, a + 2b) : a, b ∈ R}, decide whether W is a vector subspace of R3 (assuming the usual operations). Solution. Another way to describe W is as follows: W = {(x, y, z) ∈ R3 : z = x + 2y}. It’s easy to see that (0, 0, 0) ∈ W . If (x, y, z) ∈ W and (x0 , y 0 , z 0 ) ∈ W , then we have (x, y, z) + (x0 , y 0 , z 0 ) = (x+x0 , y+y 0 , z+z 0 ) and z+z 0 = (x+2y)+(x0 +2y 0 ) = (x+x0 )+2(y+y 0 ), so (x, y, z)+(x0 , y 0 , z 0 ) ∈ W . In other words, W is closed under addition. Similarly, if (x, y, z) ∈ W and c is a scalar, then c(x, y, z) = (cx, cy, cz) and cz = c(x + 2y) = cx + 2(cy) so c(x, y, z) ∈ W . In other words, W is also closed under scalar multiplication. Conclusion: yes, W is a subspace of R3 . Section 4.3 p184 Problem 19. Given W = {(x1 , x2 , x1 x2 ) : x1 , x2 ∈ R}, decide whether W is a vector subspace of R3 (assuming the usual operations). Solution. We have W = {(x, y, z) ∈ R3 : z = xy}. Although W is obviously nonempty, it is neither closed under addition nor closed under scalar multiplication. For instance, (1, 1, 1) ∈ W but (1, 1, 1) + (1, 1, 1) = (2, 2, 2) ∈ / W and 2(1, 1, 1) = (2, 2, 2) ∈ / W. Section 4.3 p184 Problem 23. (Guided Proof.) Let W be a nonempty subset W of a vector space V . Prove that W is a subspace of V iff ax + by ∈ W for all scalars a and b and all vectors x, y ∈ W . Proof. (=⇒). Assume that W is a subspace of V . Then assume that x, y ∈ W and a, b ∈ R. As a subspace, W is closed under scalar multiplication, so ax ∈ W and by ∈ W . Then, since W is also closed under addition, we see that ax + by ∈ W , as required. (⇐=). Now assume, for all x, y ∈ W and all scalars a, b, that ax + by ∈ W . To show that W is a subspace, it suffices to show (we already know W is nonempty) that W is both closed under addition and closed under scalar multiplication. Assume x, y ∈ W . Applying our hypothesis with this same x and y and with a = 1, b = 1, we see that x + y must belong to W . Since x and y were chosen arbitrarily, it follows that W is closed under addition. Finally, assume that x ∈ W and c is a scalar. Applying our hypothesis with x as chosen, y unspecified, a = c and b = 0, we see that ax + by = cx + 0 = cx must belong to W . Therefore W is also closed under scalar multiplication. Page 3 of 4 A. Sontag March 25, 2001 Math 206 HWK 13a Solns contd 4.3 p184 · ¸ 1 Section 4.3 p184 Problem 25. Let A be a fixed 2×3 matrix. Let W := {x ∈ R : Ax = }. 2 Prove that W is not a subspace of R3 . 3 · ¸ · ¸ 2 1 Proof. Assume that x ∈ W . Then A(2x) = 2Ax = 6= } so 2x ∈ / W . Since W is not 4 2 closed under scalar multiplication, it can’t be a subspace. (Alternatively, you could use a similar argument to show that W is not closed under addition. Note that W might or might not be nonempty, depending on the particular matrix A. A more general result is that the solution set for a nonhomogeneous system of linear equations will NEVER be a vector space under the usual operations.) Section 4.3 p184 Problem 26. Let A be a fixed m × n matrix. Let W := {x ∈ Rn : Ax = 0}. Prove that W is a subspace of Rn . Proof. The zero vector for Rn obviously belongs to W . Suppose that x, y ∈ W and let c be a scalar. Then A(x + y) = Ax + Ay = 0 + 0 = 0 and A(cx) = cAx = c0 = 0 so x + y and cx both belong to W . We’ve shown that W is nonempty, closed under addition, and closed under scalar multiplication. Therefore W is a subspace. (So what you’ve just shown is that the solution set for a homogenous system of linear equations is ALWAYS a vector space, under the usual operations.) Section 4.3 p184 Problem 29. Let A be a fixed 2 × 2 matrix. Let W := {X ∈ M2,2 : XA = AX} is a subspace of M2,2 . (W is sometimes referred to as the set of 2 × 2 matrices that commute with the given matrix A.) Prove that W is a subspace of M2,2 . Proof. It is easy to see that the matrix 02,2 belongs to W , since 02,2 A = 02,2 = A02,2 . Suppose X, Y ∈ W . The distributivity of matrix multiplication give us (X + Y )A = XA + Y A and A(X + Y ) = AX + AY . The hypothesis on X and Y gives AX = XA and AY = Y A. Therefore (X + Y )A = XA + Y A = AX + AY = A(X + Y ), which shows that X + Y belongs to W , as required. Now suppose that X ∈ W as before and that c is a scalar. Then A(cX) = c(AX) = c(XA) = (cX)A. This shows that cX must belong to W . Page 4 of 4 A. Sontag March 25, 2001