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Solutions to Exercises 27 Problem Set 3.1, page 127 1 x C y ¤ y C x and x C .y C z/ ¤ .x C y/ C z and .c1 C c2 /x ¤ c1 x C c2 x. 2 When c.x1 ; x2 / D .cx1 ; 0/, the only broken rule is 1 times x equals x. Rules (1)-(4) for addition x C y still hold since addition is not changed. 3 (a) cx may not be in our set: not closed under multiplication. Also no 0 and no !x 4 5 6 7 8 9 10 11 (b) c.x C y/ is the usual .xy/c , while cx C cy is the usual .x c /.y c /. Those are equal. With c D 3, x D 2, y D 1 this is 3.2 C 1/ D 8. The zero vector is the number 1. ! " ! " ! " 0 0 1 1 !1 !2 2 The zero vector in matrix space M is I 2A D and !A D . 0 0 1 !1 !2 2 The smallest subspace of M containing the matrix A consists of all matrices cA. (a) One possibility: The matrices cA form a subspace not containing B (b) Yes: the subspace must contain A ! B D I (c) Matrices whose main diagonal is all zero. When f .x/ D x 2 and g.x/ D 5x, the combination 3f ! 4g in function space is h.x/ D 3f .x/ ! 4g.x/ D 3x 2 ! 20x. Rule 8 is broken: If cf .x/ is defined to be the usual f .cx/ then .c1 C c2 /f D f ..c1 C c2 /x/ is not generally the same as c1 f C c2 f D f .c1 x/ C f .c2 x/. If .f C g/.x/ is the usual f .g.x// then .g C f /x is g.f .x// which is different. In Rule 2 both sides are f .g.h.x///. Rule 4 is broken there might be no inverse function f !1 .x/ such that f .f !1 .x// D x. If the inverse function exists it will be the vector !f . (a) The vectors with integer components allow addition, but not multiplication by 12 (b) Remove the x axis from the xy plane (but leave the origin). Multiplication by any c is allowed but not all vector additions. The only subspaces are (a) the plane with b1 D b2 (d) the linear combinations of v and w (e) the plane with b1 C b2 C b3 D 0. ! " ! " a b a a (a) All matrices (b) All matrices (c) All diagonal matrices. 0 0 0 0 12 For the plane x C y ! 2z D 4, the sum of .4; 0; 0/ and .0; 4; 0/ is not on the plane. (The key is that this plane does not go through .0; 0; 0/.) 13 The parallel plane P0 has the equation x C y ! 2z D 0. Pick two points, for example .2; 0; 1/ and .0; 2; 1/, and their sum .2; 2; 2/ is in P0 . 14 (a) The subspaces of R2 are R2 itself, lines through .0; 0/, and .0; 0/ by itself (b) The subspaces of R4 are R4 itself, three-dimensional planes n " v D 0, two-dimensional subspaces .n1 " v D 0 and n2 " v D 0/, one-dimensional lines through .0; 0; 0; 0/, and .0; 0; 0; 0/ by itself. 15 (a) Two planes through .0; 0; 0/ probably intersect in a line through .0; 0; 0/ (b) The plane and line probably intersect in the point .0; 0; 0/ (c) If x and y are in both S and T , x C y and cx are in both subspaces. 16 The smallest subspace containing a plane P and a line L is either P (when the line L is in the plane P) or R3 (when L is not in P). 17 (a) The invertible matrices do not!include so they are not a subspace " the! zero matrix, " 1 0 0 0 (b) The sum of singular matrices C is not singular: not a subspace. 0 0 0 1 Solutions to Exercises 28 18 (a) True: The symmetric matrices do form a subspace 19 20 21 22 23 (b) True: The matrices with AT D !A do form a subspace (c) False: The sum of two unsymmetric matrices could be symmetric. The column space of A is the x-axis D all vectors .x; 0; 0/. The column space of B is the xy plane D all vectors .x; y; 0/. The column space of C is the line of vectors .x; 2x; 0/. (a) Elimination leads to 0 D b2 ! 2b1 and 0 D b1 C b3 in equations 2 and 3: Solution only if b2 D 2b1 and b3 D !b1 (b) Elimination leads to 0 D b1 C 2b3 in equation 3: Solution only if b3 D !b1 . A combination of" C is also a combination of the columns ! " of the columns ! ! of A. " Then 1 3 1 2 1 2 C D and A D have the same column space. B D has a 2 6 2 4 3 6 different column space. (a) Solution for every b (b) Solvable only if b3 D 0 (c) Solvable only if b3 D b2 . The extra column b enlarges the column space unless b is already in the column space. ! " ! " 1 0 1 (larger column space) 1 0 1 (b is in column space) ŒA b! D 0 0 1 (no solution to Ax D b) 0 1 1 (Ax D b has a solution) 24 The column space of AB is contained in (possibly equal to) the column space of A. 25 26 27 28 29 30 31 32 The example B D 0 and A ¤ 0 is a case when AB D 0 has a smaller column space than A. The solution to Az D b C b" is z D x C y. If b and b" are in C .A/ so is b C b" . The column space of any invertible 5 by 5 matrix is R5 . The equation Ax D b is always solvable (by x D A!1 b/ so every b is in the column space of that invertible matrix. (a) False: Vectors that are not in a column space don’t form a subspace. (b) True: Only the zero matrix has C .A/ D f0g. ! (c) "True: C .A/ D C .2A/. 1 0 (d) False: C .A ! I / ¤ C .A/ when A D I or A D (or other examples). 0 0 " # " # " # 1 1 0 1 1 2 1 2 0 A D 1 0 0 and 1 0 1 do not have .1; 1; 1/ in C .A/. A D 2 4 0 0 1 0 0 1 1 3 6 0 has C .A/ D line. When Ax D b is solvable for all b, every b is in the column space of A. So that space is R9 . (a) If u and v are both in S C T , then u D s1 C t 1 and v D s2 C t 2 . So u C v D .s1 C s2 / C .t 1 C t 2 / is also in S C T . And so is cu D cs1 C ct 1 : a subspace. (b) If S and T are different lines, then S [ T is just the two lines (not a subspace) but S C T is the whole plane that they span. If S D C .A/ and T D C .B/ then S C T is the column space of M D Œ A B !. The columns of AB are combinations ! "of the columns of A. So all columns of Œ!A AB"! 0 1 0 0 are already in C .A/. But A D has a larger column space than A2 D . 0 0 0 0 n For square matrices, the column space is R when A is invertible.