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INTRODUCTIONTOLINEARALGEBRAFifth EditionMANUAL FOR INSTRUCTORSGilbert StrangMassachusetts Institute of .06video lectures: ocw.mit.edumath.mit.edu/ gswww.wellesleycambridge.comemail: [email protected] - Cambridge PressBox 812060Wellesley, Massachusetts 02482

Solutions to Exercises45Problem Set 3.1, page 131Note An interesting “max-plus” vector space comes from the real numbers R combinedwith . Change addition to give x y max(x, y) and change multiplication toxy usual x y. Which y is the zero vector that gives x 0 max(x, 0) x for everyx?1 x y 6 y x and x (y z) 6 (x y) z and (c1 c2 )x 6 c1 x c2 x.2 When c(x1 , x2 ) (cx1 , 0), the only broken rule is 1 times x equals x. Rules (1)-(4)for addition x 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(b) c(x y) is the usual (xy)c , while cx cy is the usual (xc )(y c ). Those are equal.With c 3, x 2, y 1 this is 3(2 1) 8. The zero vector is the number 1. 0 0 11 1 2 2 ; A and A .4 The zero vector in matrix space M is 20 01 1 2 2The smallest subspace of M containing the matrix A consists of all matrices cA.5 (a) One possibility: The matrices cA form a subspace not containing Bsubspace must contain A B I(b) Yes: the(c) Matrices whose main diagonal is all zero.6 When f (x) x2 and g(x) 5x, the combination 3f 4g in function space ish(x) 3f (x) 4g(x) 3x2 20x.7 Rule 8 is broken:If cf (x) is defined to be the usual f (cx) then (c1 c2 )f f ((c1 c2 )x) is not generally the same as c1 f c2 f f (c1 x) f (c2 x).8 If (f g)(x) is the usual f (g(x)) then (g f )x is g(f (x)) which is different. InRule 2 both sides are f (g(h(x))). Rule 4 is broken because there might be no inversefunction f 1 (x) such that f (f 1 (x)) x. If the inverse function exists it will be thevector f.9 (a) The vectors with integer components allow addition, but not multiplication by12(b) Remove the x axis from the xy plane (but leave the origin). Multiplication by anyc is allowed but not all vector additions : (1, 1) ( 1, 1) (0, 2) is removed.

Solutions to Exercises4610 The only subspaces are (a) the plane with b1 b2(d) the linear combinations of vand w(e) the plane with b1 b2 b3 0. a ba 11 (a) All matrices (b) All matrices 0 00ces.a0 (c) All diagonal matri-12 For the plane x y 2z 4, the sum of (4, 0, 0) and (0, 4, 0) is not on the plane. (Thekey is that this plane does not go through (0, 0, 0).)13 The parallel plane P0 has the equation x y 2z 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) Thesubspaces of D are D itself, the zero matrix by itself, and all the “one-dimensional”subspaces that contain all multiples of one fixed matrix : d1 0 for all c.c 0 d215 (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). Could be a line !(c) If x and y are in both S and T , x 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 Lis in the plane P) or R3 (when L is not in P).17 (a) The invertible matrices do not includeso they are not a subspace the zero matrix, (b) The sum of singular matrices 1 00 0 0001 is not singular: not a subspace.18 (a) True: The symmetric matrices do form a subspaceAT A do form a subspace(b) True: The matrices with(c) False: The sum of two unsymmetric matricescould be symmetric.19 The column space of A is the x-axis all vectors (x, 0, 0) : a line. The column spaceof B is the xy plane all vectors (x, y, 0). The column space of C is the line of vectors(x, 2x, 0).

Solutions to Exercises4720 (a) Elimination leads to 0 b2 2b1 and 0 b1 b3 in equations 2 and 3:Solution only if b2 2b1 and b3 b1(b) Elimination leads to 0 b1 b3in equation 3: Solution only if b3 b1 .21 A combinationof C is also a combination of the columns of the columns of A. ThenC 1 3 and A 1 2 have the same column space. B 2 62 4different column space. The key word is “space”.22 (a) Solution for every b1 23 6 has a(b) Solvable only if b3 0 (c) Solvable only if b3 b2 .23 The extra columnb enlargesthe column space unlessin the column space. b is already [A b] 1 0 10 0 1 (larger column space)(no solution to Ax b) 1 010 11 (b is in column space)(Ax b has a solution)24 The column space of AB is contained in (possibly equal to) the column space of A.The example B zero matrix and A 6 0 is a case when AB zero matrix has asmaller column space (it is just the zero space Z) than A.25 The solution to Az b b is z x y. If b and b are in C(A) so is b b .26 The column space of any invertible 5 by 5 matrix is R5 . The equation Ax b isalways solvable (by x A 1 b) so every b is in the column space of that invertiblematrix.27 (a) False:Vectors that are not in a column space don’t form a subspace.(b) True: Only the zero matrix has C(A) {0}. (c) True: C(A) C(2A).1 0 (or other examples).(d) False: C(A I) 6 C(A) when A I or A 0 0 1 1 01 1 211 2 0 28 A 1 0 0 and 1 0 1 do not have 1 in C(A). A 2 4 0 has 0 1 00 1 113 6 0C(A) line in R3 .29 When Ax b is solvable for all b, every b is in the column space of A. So that spaceis C(A) R9 .

Solutions to Exercises4830 (a) If u and v are both in S T , then u s1 t1 and v s2 t2 . So u v (s1 s2 ) (t1 t2 ) is also in S T . And so is cu cs1 ct1 : S T subspace.(b) If S and T are different lines, then S T is just the two lines (not a subspace) butS T is the whole plane that they span.31 If S C(A) and T C(B) then S T is the column space of M [ A B ].32 The columns of AB are combinationsof the columns of A. So all columns of [ A AB ] are already in C(A). But A 01 has a larger column space than A2 0 0For square matrices, the column space is Rn exactly when A is invertible.0 00 0 .Problem Set 3.2, page 142 1 2 2 41 (a) U 00 1 20 0 0 06 2 4 Free variables x2 , x4 , x5 (b) U 0 43 Pivot variables x , x 1300 02 Free x3 4 Pivot x , x1202 (a) Free variables x2 , x4 , x5 and solutions ( 2, 1, 0, 0, 0), (0, 0, 2, 1, 0), (0, 0, 3, 0, 1)(b) Free variable x3 : 1 2 0 0 3 R 0 0 1 2 0 0 0 0solution (1, 1, 1). Special solution for each free variable. 01 0 1 ,R 0 13 1 , R has the same nullspace as U and A. 00 004 (a) Special solutions (3, 1, 0) and (5, 0, 1) (b) (3, 1, 0). Total of pivot and free is n.5 (a) False: Any singular square matrix would have free variablesvertible square matrix has no free variables.(d) True (only m rows to hold pivots) 0 1 1 1 1 1 11 1 1 0 0 0 1 1 1 1 0 0 1 6 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 00 0 0(b) True: An in-(c) True (only n columns to hold pivots)1 1 11 1 10 0 10 0 01 0 0 0 1 1 0 0 0 0 1 0 0 0 0 10 0 0 01 10 10 00 01 1 0 0

Solutions to Exercises 11 01 1 14900 01 1001 1 1 0 0 1 1 1 1 0 0 0 0 0 1 0 1 1 1 , . Notice the identity7 0 0 0 0 0 0 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0matrix in the pivot columns of these reduced row echelon forms R.8 If column 4 of a 3 by 5 matrix is all zero then x4 is a free variable. Its special solutionis x (0, 0, 0, 1, 0), because 1 will multiply that zero column to give Ax 0.9 If column 1 column 5 then x5 is a free variable. Its special solution is ( 1, 0, 0, 0, 1).10 If a matrix has n columns and r pivots, there are n r special solutions. The nullspacecontains only x 0 when r n. The column space is all of Rm when r m. Allthose statements are important!11 The nullspace contains only x 0 when A has 5 pivots. Also the column space is R5 ,because we can solve Ax b and every b is in the column space.12 A [ 1 3 1 ] gives the plane x 3y z 0; y and z are free variables. Thespecial solutions are (3, 1, 0) and (1, 0, 1).13 Fill in 12the complete solution in R3 to x 3y z 12: then 3 then 1 to get x1231 y 0 y 1 z 0 one particular solution all nullspace solutions. z00114 Column 5 is sure to have no pivot since it is a combination of earlier columns. With4 pivots in the other columns, the special solution is s (1, 0, 1, 0, 1). The nullspacecontains all multiples of this vector s (this nullspace is a line in R5 ).15 To producespecial solutions(2, 2, 1, 0) and (3, 1, 0, 1) with free variables x3 , x4 : R 1 0 2 30 1 2 1 and A can be any invertible 2 by 2 matrix times this R.

Solutions to Exercises50 4 1 0 0 4 3 16 The nullspace of A 0 1 0 3 is the line through the special solution . 2 0 0 1 21 1 0 1/2101 17 A 1 3 2 has 1 and 3 in C(A) and 1 in N (A). Which other A’s? 5 1 3512 18 This construction is impossible for 3 by 3 ! 2 pivot columns and 2 free variables. 119 A 1 20 A 100 100 0 10 has (1, 1, 1) in C(A) and only the line (c, c, c, c) in N (A). 00 1 110 is not AT . has N (A) C(A). Notice that rref(AT ) 0 0021 If nullspace column space (with r pivots) then n r r. If n 3 then 3 2r isimpossible.22 If A times every column of B is zero,B is contained in the nullspace the column space of of A. An example is A 1 1 and B 1 1For B 0, C(B) is smaller than N (A).11 1 1 . Here C(B) equals N (A).23 For A random 3 by 3 matrix, R is almost sure to be I. For 4 by 3, R is most likelyto be I with a fourth row of zeros. What is R for a random 3 by 4 matrix? 0 11 0 shows that (a)(b)(c) are all false. Notice rref(AT ) .24 A 0 00 025 If N (A) line through x (2, 1, 0, 1), A has three pivots (4 columns and 1 special 1 00 2 solution). Its reduced echelon form can be R 0 1 0 1 (add any zero rows). 0 0 10 1 0 0 , R I. Any zero rows come after those rows.26 R [ 1 2 3 ], R 0 1 0

Solutions to Exercises 27 (a) 1 00 1 , 511000 , 1100 , 0 10 0 , 0 00 0 (b) All 8 matrices are R’s !28 One reason that R is the same for A and A: They have the same nullspace. (Theyalso have the same row space. They also have the same column space, but that is notrequired for two matrices to share the same R. R tells us the nullspace and row space.) y for y in R4 .29 The nullspace of B [ A A ] contains all vectors x y30 If Cx 0 then Ax 0 and Bx 0. So N (C) N (A) N (B) intersection. 1 1 11 0 0 0 rank 1 0 0 0 0 1 11 1 (c) R 0000 rank 1 000031 (a) R 010 (b) R 0 0 1 21200 3 rank 2 032 AT y 0 : y1 y3 y4 y1 y2 y5 y2 y4 y6 y4 y5 y6 0.These equations add to 0 0. Free variables y3 , y5 , y6 : watch for flows around loops.The solutions to AT y 0 are combinations of ( 1, 0, 0, 1, 1, 0) and (0, 0, 1, 1, 0, 1)and (0, 1, 0, 0, 1, 1). Those are flows around the 3 small loops.33 (a) and (c) are correct; (b) is completely false; (d) is false because R might have 1’sin nonpivot columns. 1 2 0 RA 34 RA 0 0 1 RB RA RARC 00 0 0 35 If all pivot variables come last then R 000RA Zero rows goto the bottom I . The nullspace matrix is N .00I 36 I think R1 A1 , R2 A2 is true. But R1 R2 may have 1’s in some pivots.

Solutions to Exercises5237 A and AT have the same rank r number of pivots. But pivcol (the column number) 0 1 0 is 2 for this matrix A and 1 for AT : A 0 0 0 . 0 0 038 Special solutions in N [ 2 4 1 0; 3 5 0 1 ] and [ 1 0 0; 0 2 1 ]. 1239 The new entries keep rank 1 : A 24 M abcbc/a 4 84 8 , 16 3 B 1 29 4.5 3 1.5 , 6 3 .40 If A has rank 1, the column space is a line in Rm . The nullspace is a plane in Rn(given by one equation). The nullspace matrix N is n by n 1 (with n 1 specialsolutions in its columns). The column space of AT is a line in Rn .i h h3 6 631 2 2 226421 41 1 2 2 1 and 1 1 3 2 14 8 8442 With rank 1, the second row of R is a zero row.43Invertible r by r submatricesUse pivot rows and columns S 1 31 4 and S [ 1 ] and S 1 00 11 3 2 .44 P has rank r (the same as A) because elimination produces the same pivot columns.45 The rank of RT is also r. The example matrix A has rank 2 with invertible S: 1 P 2 23 6 7 PT 1 2 23 6 7 ST 1 23 7 S 1 32 7 .46 The product of rank one matrices has rank one or zero. These particular matrices haverank(AB) 1; rank(AC) 1 except AC 0 if c 1/2.47 (uv T )(wz T ) u(v T w)z T has rank one unless the inner product is v T w 0.i

Solutions to Exercises5348 (a) By matrix multiplication, each column of AB is A times the corresponding columnof B. So if column j of B is a combination of earlier columns, then column j of ABis the same combination of earlier columns of AB. Then rank (AB) rank (B). No(b) The rank of B is r 1. Multiplying by A cannot increase this rank. The rank of AB stays the same for A1 I and B 11 11 . It drops to zero for A2 11 11 .new pivot columns!49 If we know that rank(B T AT ) rank(AT ), then since rank stays the same for trans-poses, (apologies that this fact is not yet proved), we have rank(AB) rank(A).50 We are given AB I which has rank n. Then rank(AB) rank(A) forces rank(A) n. This means that A is invertible. The right-inverse B is also a left-inverse: BA Iand B A 1 .51 Certainly A and B have at most rank 2. Then their product AB has at most rank 2.Since BA is 3 by 3, it cannot be I even if AB I.52 (a) A and B will both have the same nullspace and row space as the R they share.(b) A equals an invertible matrix times B,