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Math 311: Topics in Applied Math 1 3: Vector Spaces 3.2: Subspaces Solution T A is not a subspace of R3 since x = 1 0 0 ∈A T but 2x = 2 0 0 ∈ / A because 2 + 0 6= 1. B is a subspace of R3 , geometrically a line in 3-D space T through the origin. For let x = x x x and T y= y y y be in B and α ∈ R. Then Summary Below, V is a vector space over a field F of scalars. T • B is nonempty since 0 = 0 0 0 ∈ B. T • αx = αx αx αx ∈ B. T • x+y = x+y x+y x+y ∈ B. • A nonempty subset S ⊂ V closed under scalar multiplication and vector addition is called a subspace of V . I.e., S is a subspace provided – 1. αx ∈ S whenever α ∈ F and x ∈ S, C is a subspace of R3 , geometrically a plane in space that T contains the origin. For let x = x1 x2 x1 + x2 and T be in C and α ∈ R. Then y = y1 y2 y1 + y2 – 2. x + y ∈ S whenever x, y ∈ S. • Every subpsace S of V is a vector space in its own right, having inherited the requisite stucture from V . Accordingly, every subspace of V must contain the zero vector of V . T • C is nonempty since 0 = 0 0 0 ∈ C. T • αx = αx1 αx2 αx1 + αx2 ∈ C. • x+y = • The null space N (A) of an m × n matrix A is the set T x1 + y1 x2 + y2 (x1 + y1 ) + (x2 + y2 ) ∈ C. of all solutions of the homogeneous system Ax = 0. n It forms a subspace of R , the vector space of all T n-element column vectors. D is not a subspace of R3 since x = 1 0 1 ∈D& T T y= 0 1 1 ∈ D but x + y = 1 1 2 ∈ /D n because 2 6= 1. • The set of all linear combinations ∑ αk vk where k=1 αk ∈ F and vk ∈ V is called the span of v1 , . . . , vn , written Span (v1 , . . . , vn ). It is a subspace of V . 131/3 Determine whether the following are subspaces of R2×2 , the set of all real 2 × 2 matrices over the real field R. • A set {v1 , . . . , , vn } is a spanning set of V if and only if every vector in V can be written as a linear combination of this spanning set. (a) The set D of all 2 × 2 diagonal matrices (c) The set L of all 2 × 2 lower triangular matrices (e) The set B of all 2 × 2 matrices B such that b11 = 0 Examples (g) The set S of all singular 2 × 2 matrices 131/2 Solution Determine whether the following sets form subspaces of R3 , the vector space of all 3-element real column vectors over the real field R. (To save vertical space, we’ll denote column vectors as transposes of row vectors.) n o T x1 x2 x3 A= x1 + x3 = 1 2×2 (a) The set D is asubspace of R . For let a11 0 b11 0 A= and B = 0 a22 0 b22 be in D and α ∈ R. Then 1 0 • D is nonempty since ∈ D. 0 2 αa11 0 • αA = ∈ D. 0 αa22 a11 + b11 0 • A+B = ∈ D. 0 a22 + b22 o T x = x = x 1 2 3 o T x3 x3 = x1 + x2 B= n x1 x2 x3 C= n x1 x2 D= n x1 x2 x3 o T x3 = x1 or x3 = x2 1 2×2 (c) The let set L is a subspace of R . For a11 0 b11 0 and B = A= a21 a22 b21 b22 be in L and α ∈ R. Then 1 0 • L is nonempty since ∈ L. 2 3 αa11 0 • αA = ∈ L. αa21 αa22 a11 + b11 0 • A+B = ∈ L. a21 + b21 a22 + b22 132/9a 1 0 . Determine the subspace S of R2×2 Let A = 0 −1 consisting of all 2 × 2 matrices that commute with A. Solution p q ∈ S. Then AB = BA, whence Let B = r s p q p −q = . Thus q = −q and −r = r. −r −s r −s In other words, 2q = 2r = 0, so q = r = 0. Accordingly, S is the set of all 2 × 2 matrices having the form p 0 , p, s ∈ R. 0 s 2×2 (e) The set B is a subspace of R . For let 0 a12 0 c12 A= and C = a21 a22 c21 c22 be in B and α ∈ R. Then 0 1 • B is nonempty since ∈ B. 2 3 0 αa12 • αA = ∈ B. αa21 αa22 0 a12 + c12 • A+C = ∈ B. a21 + c21 a22 + c22 R2×2 . 1 2 0 0 In other words, the set of 2 × 2 diagonal matrices. (g) S is not a subspace of For A = and 0 0 B= are in S since their determinants are zero 3 4 1 2 and hence they are singular. But A + B = ∈ /S 3 4 since its detminant is −2 6= 0 and thus it is nonsingular. 131/6b Recall that C [−1, 1] is the vector space of real continuous functions on [−1, 1] over R. Determine whether the set S of odd functions in C [−1, 1] is a subspace of C [−1, 1]. Solution The set S is a subspace of C [−1, 1]. For let f , g ∈ S and α ∈ R. Then • S is nonempty since the function h defined by h (x) = x on [−1, 1] is in S because h (−x) = −x = −h (x) for x ∈ [−1, 1]. • α f ∈ S since for x ∈ [−1, 1], we have (α f ) (−x) = α f (−x) = α (− f (x)) = − (α f (x)) = − (α f ) (x) , whence α f is an odd function. • f + g ∈ S since for x ∈ [−1, 1], we have ( f + g) (−x) = f (−x)+g (−x) = − f (x)−g (x) = − ( f (x) + g (x)) = − ( f + g) (x) , whence f + g is an odd function. 2