## Solution Manual For: Linear Algebra By Gilbert Strang 3m ago
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Solution Manual for:Linear Algebraby Gilbert StrangJohn L. Weatherwax January 1, 2006IntroductionA Note on NotationIn these notes, I use the symbol to denote the results of elementary elimination matricesused to transform a given matrix into its reduced row echelon form. Thus when looking forthe eigenvectors for a matrix like 0 0 2A 0 1 0 0 0 2rather than say, multiplying A on the left by E33produces 1 0 0 0 1 0 1 0 1 0 0 2E33 A 0 1 0 0 0 0we will use the much more compact notation 0 0 20 0 2A 0 1 0 0 1 0 .0 0 20 0 0 [email protected]

Derivation of the decomposition (Page 170)Combining the basis for the row space and the basis for the nullspace into a common matrix Tto assemble a general right hand side x a b c dfrom some set of components Tc c1 c2 c3 c4we must have ac11 0 10c1 c2 0 1 01 c2 b A c3 1 0 1 0 c3 c dc40 1 0 1c4Inverting the coefficient matrix A by using the teaching code elim.m or augmentation andinversion by hand gives 1 0 101 0 1 01 .A 1 2 1 0 1 0 0 1 0 1So the coefficients of c1 , c2 , c3 , and c4 are given by ac1a c c2 1b 1 b d c3 A c 2 a cdc4b dAs verified by what is given in the book. Chapter 1 (Introduction to Vectors)Section 1.1 (Vectors and Linear Combinations)Problem 16 (dimensions of a cube in four dimensions)We can generalize Problem 15 by stating that the corners of a cube in four dimensions aregiven byn(1, 0, 0, 0) m(0, 1, 0, 0) l(0, 0, 1, 0) p(0, 0, 0, 1) ,for indices n, m, l, p taken from {0, 1}. Since the indices n, m, l, p can take two possible valueseach the total number of such vectors (i.e. the number of corners of a four dimensional cube)is given by 24 16.To count the number of faces in a four dimensional cube we again generalize the notionof a face from three dimensions. In three dimensions the vertices of a face is defined bya configuration of n, m, l where one component is specified. For example, the top face is

specified by (n, m, 1) and the bottom face by (n, m, 0), where m and n are allowed to takeall possible values from {0, 1}. Generalizing to our four dimensional problem, in countingfaces we see that each face corresponds to first selecting a component (either n, m, l, or p)setting it equal to 0 or 1 and then letting the other components take on all possible values.The component n, m, l, or p can be chosen in one of four ways, from which we have twochoices for a value (0 or 1). This gives 2 4 8 faces.To count the number of edges, remember that for a three dimensional cube an edge isdetermined by specifying (and assigning to) all but one elements of our three vector. Thusselecting m and p to be 0 we have (n, 0, 0) and (n, 0, 0), where n takes on all values from{0, 1} as vertices that specify one edge. To count the number of edges we can first specifyingthe one component that will change as we move along the given edge, and then specify acomplete assignment of 0 and 1 to the remaining components. In four dimensions, we canpick the single component in four ways and specify the remaining components in 23 8,ways giving 4 · 8 32 edges.Problem 17 (the vector sum of the hours in a day)Part (a): Since every vector can be paired with a vector pointing in the opposite directionthe sum must be zero.Part (b): We haveXvi i6 4X! v4 0 v4 v4 ,!1v1v1 v1 v1 0 ,222viall iwith v4 denoting the 4:00 vector.Part (c): We haveXi6 11vi v1 2Xall iviwith v1 denoting the 1:00 vector.Problem 18 (more clock vector sums)We have from Problem 17 that the vector sum of all the vi ’s is zero,Xvi 0 .i {1,2,.12}Adding twelve copies of (0, 1) ĵ to each vector givesX(vi ĵ) 12ĵ .i {1,2,.12}

But if in addition to the transformation above the vector 6:00 is set to zero and the vector12:00 is doubled, we can incorporate those changes by writing out the above sum and makingthe terms summed equivalent to the specification in the book. For example we have X (vi ĵ) (v6 ĵ) (v12 ĵ) 12ĵi6 {6,12} (vi ĵ) (0 ĵ) (2v12 ĵ) v6 v12 12ĵX(vi ĵ) (0 ĵ) (2v12 ĵ) (0, 1) (0, 1) 12(0, 1) 10(0, 1) .i6 {6,12} Xi6 {6,12} The left hand side now gives the requested sum. In the last equation, we have written outthe vectors in terms of their components to perform the summations.Problem 26 (all vectors from a collection )Not if the three vector are not degenerate, i.e. are not all constrained to a single line.Problem 27 (points in common with two planes)Since the plane spanned by u and v and the plane spanned by v and w intersect on the linev, all vectors cv will be in both planes.Problem 28 (degenerate surfaces)Part (a): Pick three vectors collinear, likeu (1, 1, 1)v (2, 2, 2)w (3, 3, 3)Part (b): Pick two vectors collinear with each other and the third vector not collinear withthe first two. Something likeu (1, 1, 1)v (2, 2, 2)w (1, 0, 0)

Problem 29 (combinations to produce a target)Let c and d be scalars such that combine our given vectors in the correct way i.e. 1431 dc812which is equivalent to the systemc 3d 142c d 8which solving for d using the second equation gives d 8 2c and inserting into the firstequation gives c 3(8 2c) 14, which has a solution of c 2. This with either of theequations above yields d 2.Section 1.2 (Lengths and Dot Products)Problem 1 (simple dot product practice)We haveu·vu·wv·ww·v .6(3) .8(4) 1.4 .6(4) .8(3) 03(4) 4(3) 2424 .Chapter 2 (Solving Linear Equations)Section 2.2 (The Idea of Elimination)Problem 1We should subtract 5 times the first equation. After this step we have2x 3y 11 6y 6or the systemThe two pivots are 2 and -6. 2 30 6

Problem 2the last equation gives y 1, then the first equation gives 2x 3 1 or x 2. Lets checkthe multiplication 132(1)( 1) (2) 11910If the right hand changes to 444 (2)then -5 times the first component added to the second component gives 44 20 24.Chapter 3 (Vector Spaces and Subspaces)Section 3.1Problem 5Part (a): Let M consist of all matrices that are multiples of 1 0.0 0Part (b): Yes, since the element 1 · A ( 1) · B I must be in the space.Part (c): Let the subspace consist of all matrices defined by 0 01 0 ba0 10 0Problem 6We have h(x) 3(x2 ) 4(5x) 3x2 20x.Problem 7Rule number eight is no longer true since (c1 c2 )x is interpreted as f ((c1 c2 )x) andc1 x c2 x is interpreted as f (c1 x) f (c2 x), while in general for arbitrary functions these twoare not equal i.e. f ((c1 c2 )x) 6 f (c1 x) f (c2 x).

Problem 8 The first rule x y y x is broken since f (g(x)) 6 g(f (x)) in general. The second rule is correct. The third rule is correct with the zero vector defined to be x. The fourth rule is correct if we define x to be the inverse of the function f (·), becausethen the rule f (g(x)) x states that f (f 1 (x)) x, assuming an inverse of f exists. The seventh rule is not true in general since c(x y) is cf (g(x)) and cx cy is cf (cg(x))which are not the same in general. The eighth rule is not true since the left hand side (c1 c2 )x is interpreted as (c1 c2 )f (x), while the right hand side c1 x c2 x is interpreted as c1 f (c2 f (x)) which are notequal in general.Problem 9Part (a): Let the vector xy 11 c 10 d 01 .For c 0 and d 0. Then this set is the upper right corner in the first quadrant of thexy plane. Now note that the sum of any two vectors in this set will also be in this set butscalar multiples of a vector in this set may not be in this set. Consider 1 1 1 21 ,2 12which is not be in the set.Part (b): Let the set consist of the x and y axis (all the points on them). Then for anypoint x on the axis cx is also on the axis but the point x y will almost certainly not be.Problem 10Part (a): YesPart (b): No, since c(b1 , b2 , b3 ) c(1, b2 , b3 ) is not in the set if c 21 .Part (c): No, since if two vectors x and y are such that x1 x2 x3 0 and y1 y2 y3 0 there isno guarantee that x y will have that property. Consider 01x 1 and y 0 11

Part (d): Yes, this is a subspace.Part (e): Yes, this is a subspace.Part (f): No this is not a subspace since if b1b b2 ,b3has this property then cb should have this property but cb1 cb2 cb3 might not be true.Consider 100b 10 and c 1 . 1Then b1 b2 b3 but cb1 cb2 cb3 is not true.Problem 11Part (a): All matrices of the formfor all a, b R. a b0 0 a a0 0 a 00 b Part (b): All matrices of the formfor all a R.Part (c): All matrices of the formor diagonal matrices.Problem 12 145Let the vectors v1 1 and v2 0 , then v1 v2 1 but 5 1 2( 2) 20 210 6 4 so the sum is not on the plane.

Problem 13 1The plane parallel to the previous plane P is x y 2z 0. Let the vectors v1 1 1 12 and v2 0 , which are both on P0 . Then v1 v2 1 . We then check that this1232point is on our plane by computing the required sum. We find that 2 1 2see that it is true.32 0, andProblem 14Part (a): Lines, R2 itself, or (0, 0, 0).Part (b): R4 itself, hyperplanes of dimension four (one linear constraining equation amongfour variables) that goes through the origin like the followingax1 bx2 cx3 dx4 0 .Constraints involving two linear equation like toe above (going through the origin)ax1 bx2 cx3 dx4 0Ax1 Bx2 Cx3 Dx4 0 ,which is effectively a two dimensional plane. In addition, constraints involving three equations like above and going through the origin (this is effectively a one dimensional line).Finally, the origin itself.Problem 15Part (a): A line.Part (b): A point (0, 0, 0).Part (c): Let x and y be elements of S T . Then x y S T and cx S T since xand y are both in S and in T , which are both subspaces and therefore x y and cx are bothin S T .Problem 16A plane (if the line is in the plane to begin with) or all of R3 .

Problem 17Part (a): LetA 1 00 1 and B which are both invertible. Now A B matrices is not a subspace. 1 00 1 , 0 0, which is not. Thus the set of invertible0 0Part (b): Let1 32 6 which are both singular. Now A B A of invertible matrices is not a subspace.and B 6 32 1 , 7 6, which is not singular, showing that the set4 6Problem 18Part (a): True, since if A and B are symmetric then (A B)T AT B T A B issymmetric. Also (cA)T cAT cA is symmetric.Part (b): True, since if A and B are skew symmetric then (A B)T AT B T A B (A b) and A B is skew symmetric. Also if A is skew symmetric then cA is also since(cA)T cAT cA. 0 11 3, which iswhich is unsymmetric and B Part (c): False since if A 0 02 5 1 2should be unsymmetric but its not. Thus the setalso unsymmetric then A B 2 5of unsymmetric matrices is not closed under addition and therefore is not a subspace. Problem 19 1 2If A 0 0 , then the column space is given by0 0 1 2 x1 2x2 0 0 x1 ,0x20 00

1 2which is a line in the x-axis (i.e. all combinations of elements on the x-axis. If B 0 2 0 0 x11 0 then the column space of B is 2x2 or the entire xy plane. If C 2 0 then Cx is00 0 x1 given by 2x2 or a line in the xy plane.0Problem 20Part (a): Consider the augmented matrix 142 284 1 4 2Let E21 be given byE21Then we find that 14284E21 2 1 4 2so that b2 2b1 and b1 b3 . b1b2 b3 1 0 0 2 1 0 ,1 0 1 b11 4 2b2 0 0 00 0 0b3 b1b2 2b1 ,b3 b 1Part (b): b114x1 2 b2 9 x2b3 1 4 Let E21 and E31 be given by E21Then we see that 1 0 0 2 1 0 0 0 1 14 29E31 E21 1 4and E31 1 0 0 0 1 0 ,1 0 1 b11 4 b2 0 1b30 0which requires that b1 b3 0, or b1 b3 . b1b2 2b1 ,b3 b 1

Problem 21A combination of the columns of B and C are also a combination of the columns of A. Thosetwo matrices have the same column span.Problem 22For the first system 1 1 1x1b1 0 1 1 x2 b2 ,0 0 1x3b3we see that for any values of b the system will have a solution. For the second system 1 1 1b1 0 1 1b2 0 0 0b3we see that we must have b3 0. For the 1 00which is equivalent to so we must have b2 b3 .third system b11 10 1b2 0 1b31 1 1 0 0 10 0 0 b1b2 ,b3 b2Problem 23 1 00Unless b is a combination of the previous columns of A. If A 0 1 with b 0 0 01 1 02 has a large column space. But if A 0 1 with b 0 the column space does not0 00change. Because b can be written as a linear combination of the columns of A and thereforeadds no new information to the column space.

Problem 24The columnin the andequals the column space of A. If possibly space of AB is contained 0 10 11 0which is of a smaller dimension than, then AB and B A 0 00 00 1the original column space of A.Problem 25If z x y is a solution to Az b b . If b and b are in the column space of A then so isb b .Problem 26Any A that is a five by five invertible matrix has R5 as its column space. Since Ax balways has a solution then A is invertible.Problem 27 1 210Part (a): False. Let A then x1 and x2 are each not in the1201 1column space but x1 x2 is in the column space. Thus the set of vectors not in the1column space is not a subspace.Part (b): True.Part (c): True.Part (d): False,the matrix add a full set of pivots (linearly independent rows). Let I can 1 00 0, then A has a column space consisting of the zero vector, with I A 0 10 0and 1 0A I ,0 1has all of R2 as its column space.

Problem 28 1 1 2 1 0 0 0 1 2or 1 1 2 1 0 1 0 1 1Section 3.2Problem 1Fpr the matrix (a) i.e let E21 be given by 1 2 2 4 6 1 2 3 6 9 0 0 1 2 3E21so thatNow let E33 be given by 1 2 2 4 6E21 A 0 0 1 2 3 .0 0 1 2 3E21So that 1 0 0 1 1 0 ,0 0 1 1 0 0 0 1 0 .0 1 1 1 2 2 4 6E33 E21 A 0 0 1 2 3 .0 0 0 0 0 Which has pivot variables x1 and x3 and free variables x2 , x4 and x5 . For the matrix (b) 2 4 2A 0 4 4 0 8 8let E32 be given byE32so that 1 0 0 0 1 0 ,0 2 1 2 4 2E32 A 0 4 4 U .0 0 0Then the free variables are x3 and the pivot variables are x1 and x2 .

Problem 2Since the ordinary echelon form for the matrix in (a) is 1 2 2 4 6U 0 0 1 2 3 ,0 0 0 0 0we find a special solution that corresponds to each free vector by assigning ones to each freevariable in turn and then solving for the pivot variables. For example, since the free variablesare x2 , x4 , and x5 we begin by letting x2 1, x4 0, and x5 0. Then our system becomes x1 1 2 2 4 6 1 0 0 1 2 3 x3 0 0 0 0 0 0 0 0or 1 2 2 0 1 x1 0 x30 00which has a solution x3 0 and x1 2. So our special solution in this case is given by 2 1 0 . 0 0For the next special solution let x2 0, x4 1, and x5 0. Then our special solution solves x1 1 2 2 4 6 0 0 0 1 2 3 x3 0 0 0 0 0 0 1 0or 1 2 x 41 0 1 x3 20 0Which requires x3 2 and x1 2( 2) 4 or x1 0. Then our second special solutionis given by 0 0 2 . 1 0

Our final special solution is obtained by setting x2 0, x4 0, and x5 1. Then oursystem is x1 1 2 2 4 6 0 0 0 1 2 3 x3 0 0 0 0 0 0 0 1which reduces to solving 1 2 6x1 0 1 3x30 0So that x3 3 and x1 6 2( 3) 0 is given by 0 0 3 . 0 1Lets check our calculations. Create a matrix N with columns consisting of the three specialsolutions found above. We have 2 00 100 ,0 2 3N 2 10 001And then the product of A times N should be zero. We see that 200 0 0 000 1 2 2 4 6 1 AN 0 0 1 2 3 0 2 3 0 0 0 ,0 0 00 0 0 0 0 0 2 1001as it should. For the matrix in part (b) we have that 2 4 2U 0 4 4 0 0 0then the pivot variables are x1 and x2 while the free variables are x3 . Setting x3 1 weobtain the system 2 4x1 2 ,0 4x2 4so that x2 1 and x1 2 (4)( 1)2 1, which gives a special solution of 1 1 .1

Problem 3From Problem 2 we have three special solutions 0 2 0 1 , v2 20v1 1 0 00 , v3 00 301 , then any solution to Ax 0 can be expressed as a linear combination of these specialsolutions. The nullsapce of A contains the vector x 0 only when there are no free variablesor there exist n pivot variables.Problem 4The reduced echelon form R has ones in the pivot columns of U. For Problem 1 (a) we have 1 2 2 4 6U 0 0 1 2 3 ,0 0 0 0 0 1 2 0 then let E13 0 1 0 , so that0 0 1 1 2 0 0 0E13 U 0 0 1 2 3 R0 0 0 0 0The nullspace of R is equal to the nullspace of U since row opperations don’t change thenullspace. For Problem 1 (b) our matrix U is given by 2 4 2U 0 4 4 0 0 0 1 1 0 so let E12 0 1 0 , so that0 0 1 2 0 2E12 U 0 4 4 .0 0 0 1/2 0 0 0 1/4 0 , thenNow let D 00 1 1 0 1DE12 U 0 1 1 .0 0 0

Problem 5For Part (a) we have that A then letting E21 1 0 2 1 Then since 1 02 1 ,we get thatE21 A 1E21 1 3 5 2 6 10 1 3 50 0 0 .we have thatA 1E21U 1 02 1 1 3 50 0 0 .Where we can define the first matrix on the right hand side of the above to be L. ForPart (b) we have that 1 3 5,A 2 6 7then letting E21 be the same as before we see that 1 3 5.E21 A 0 0 3so that a decoposition of A is given byA 1UE21 1 02 1 1 3 50 0 3 .Problem 6 1 3 5For Part (a) since we have that U so we see that x1 is a pivot variable and0 0 0x2 and x3 are free variables. Then two special solutions can be computed by setting x2 1,x3 0 and x2 0, x3 1 and solving for x1 . In the first case we have x1 3 0 or x1 3giving a special vector of 3 v1 1 .0In the second case we have x1 5 0 giving x1 5, so that the second special vector isgiven by 5v2 0 .1

Thus all special solutions to Ax 0 are contained in the set 35c1 1 c2 0 .01 1 3 5so we see that x1 and x3 are pivotFor Part (b) since we have that U 0 0 3variable while x2 is a free variables. To solve for the vector in the nullspace set x2 1 andsolve for x1 and x3 . This gives x1 1 3 5 1 0,0 0 3x3or the system 1 50 3 x1x3 30 .This gives x3 0 and x1 3. So we have a special vector given by 3 1 .0For an mxn matrix the number of free variables plus the number of pivot variables equals n.Problem 7For Part (a) the nullspace of A are all points (x, y, z) such that 3c1 5c2x c1 y ,c2zor the plane x 3y 5z. This is a plane in the xyz space. This space can also be describedas all possible linear combinations of the two vectors 35 1 and 0 .01 3For Part (b) the nullspace of A are all points that are multiples of the vector 1 which is03a line in R . Equating this vector to a point (x, y, z) we see that our line is given by x 3c,y c, and z 0 or equivalently x 3y and z 0.

Problem 8 1 0 1 3 5then we have that. Let D For Part (a) since we have that U 0 10 0 0 1 3 5,DU 0 00 which is in reduced row echelon form. The identity matrix in this case is simply the scalar1 giving 1 3 5DU 000where we have put a box around the “identity”in this case. For Part (b) since we have that 1 0 1 3 5so that defining D U we then have that0 310 0 3 1 3 5DU .0 01 1 5The let E13 and we then get that0 1 1 3 0,E13 DU 0 0 1for our reduced row echelon form. Our box around the identinty in the matrix R is aroundthe pivot rows and pivot columns and is given by 1 30001Problem 9Part (a): False.This depends on what the reduced echelonmatrixlooks like. Consider 1 11 1A . Then the reduced echelon matrix R is, which has x2 as a free1 10 0variable.Part (b): True. An invertible matrix is defined as one that has a complete set of pivots i.e.no free variables.Part (c): True. Since the number of free variables plus the number of pivot variables equalsn in the case of no free variables we have the maximal number of pivot variables n.Part (d): True. If m n, then by Part (c) the number of pivot variables must be lessthan n and this is equivalent to less than m. If m n then we have fewer equations thanunknowns and when our linear system is reduced to echelon form we have a maximal set ofpivot variables. We can have at most m, corresponding to the block identity in the reducedrow echelon form in the mxm position. The remaining n m variables must be free.

Problem 10Part (a): This is not possible since going from A to U involves zeroing elements belowthe diagonal only. Thus if an element is nonzero above the diagonal it will stay so for allelimination steps.Part (b): The real requirementto find a matrix A is that A have three linearly independent 1231 0 0 columns/rows. Let A 1 1 3 , then with E 1 1 0 we find that 1 2 21 0 1 1 2 3EA 0 1 0 .0 0 1 1 2 3Continuing this process let E ′ 0 10 then0 01 1 0 0E ′ EA 0 1 0 I .0 0 1Part (c): This is not possible and the reason is as follows.R must have zeros above each 1 1 1of its pivot variables. What about the matrix A which has no zero entries.2 2 2Then 1 1 11 0,A U 0 0 0 2 1which also equals R.Part (d): If A U 2R, then R 21 A 12 U so let 111 2 01 0 A U.R 0 12 0 222 2 0.so take A 0 2Problem 11Part (a): Consider 0 0 001000x000x100x010xxx0 xx x 0

Part (b): Consider 1 0 00x0000100xx00xx000010 00 1 1 0 0 00000000001000x000x100 xx 0 0 1000x000x100xx10xxx1xxxxPart (c): ConsiderProblem 12Part (a): Consider0 0R 00this is so that the pivot variables arex5 , and x6 we we have 1 0R 00 xx ,x xx2 , x4 , x5 , and x6 . For the free variables to be x2 , x4 ,x0000100xx00xx00xx000010 00 ,0 11000x00001000010xxx0xxx0 xx .x 0Part (b): Consider 0 0R 00Problem 13x4 is certainly a free variable and the special solution is x (0, 0, 0, 1, 0).Problem 14Then x5 is a free variable. The special solution is x (1, 0, 0, 0, 1).

Problem 15If an mxn matrix has r pivots the number of special solutions is n r. The nullspace containsonly zero when r n. The column space is Rm when r m.Problem 16When the matrix has five pivots. The column space is R5 when the matrix has five pivots.Since m n then Problem 15 demonstrates that the rank must equal m n.Problem 17 xIf A 1 3 1 and x y , the free variables are y and z. Let y 1 and z 0z 3 then x 3 giving the first special solution of 1 . The second special solution is given by0settingy 0andz 1,thenx 1 0orx 1, so we have a second special solution of 1 0 .1 Problem 18If x 3y z 12, then expressing the vector (x, y, z) in iterms of y and z we find x1231 y 0 y 1 z 0 .z001Problem 19For x in the nullspace of B means that Bx 0 thus ABx A0 0 and thus x is inthe nullspace of AB. The nullspace of B is contained in the nullspace of AB. An obviousexample when the nullspace of AB is larger than the nullspace of B is when 1 0B ,1 0

which has a nullspace given by the span of the vector 0 0, and has a nullspace given by the span of0 0 01,and10 0 00then AB . If A 0 01which is larger than the nullspace of B.Problem 20If A is invertible then the nullspace of AB equals the nullspace of B. If v is an element ofthe nullspace of AB then ABv 0 of Bv 0 by multiplying both sides by A 1 . Thus v isan element of the nullspace of B.Problem 21We see that x3 and x4 are free variables. To determine the special solutions we consider thetwo assignments (x3 , x4 ) (1, 0), and (x3 , x4 ) (0, 1). Under the first we have 1 0x12 0 1x22which give 1 0 20 1 2 x1 x2 0 .1In the same way under the second assignment we have x1 1 0 3 x2 0 .0 1 1x4when we combine these two results we find that x1 1 0 2 3 x2 0 , x3 0 1 2 1x4 so that A is given byA 1 0 2 30 1 2 1 .

Problem 22If x4 1 and the other variables are 1 0 0 10 0orsolved for we have 0x14 0x2 3 (1)1x32 x11 0 0 4 0 1 0 3 x2 0 x3 0 0 1 2x4 so that A is given by 1 0 0 4A 0 1 0 3 .0 0 1 2Problem 23We havewith a rank of two whichthat the nullity must be one. Let three equations means 1 0 a1A 1 3 b for some a, b, and c. Then if 1 is to be in the nullity of A we must5 1 c2have 11 0 a11 2aA 1 1 3 b 1 1 3 2b 0 .25 1 c25 1 2cWhich can be made true if we take a 12 , b case is 1 0A 1 35 1 2, and c 3. Thus our matrix A in this 1/2 2 . 3Problem 24The number of equations equals three and the rank is two. We are requiring that the nullspacebe of dimension two (i.e. spanned by two linearly independent vectors), thus m 3 andn 4. But the dimension of the vectors in the null space is three which is not equal to four.Thus it is not possible to find a matrix with such properties.

Problem 25 1 1 00We ask will the matrix A 1 0 1 0 , work? Then if the column space contains1 00 1(1, 1, 1) then m 3. If the nullspace is (1, 1, 1, 1) then n 4. Reducing A we see that 1 1 001 0 1 01 0 0 1A 0 1 1 0 0 1 1 0 0 1 0 1 .0 10 10 0 1 10 0 1 1So if Av 0, then x1 0 0 1 0 1 0 1 y 0 z 0 0 1 1w Implying that x w 0, y w 0, and z w 0, thus our vector v is given by 1x 1 y v z w 1 ,1wand our matrix A does indeed work.Problem 26A key to solving this problem is to recognize that if the column space of A is also its nullspacethen AA 0. This is because AA represents A acting on each column of A and this produces2zero since the column space is the nullspace. Thus we need a matrix A such that A 0. Ifa b, the requirement of A2 0 means thatA c d 20 0a bc ab bd. 0 0ac cd cb d2This gives four equations for the unknowns a,b,c, and d. To find one solution let a 1 thend 1 by considering the (1, 2) element. Our matrix equation then becomes 0 01 bc0. 0 00cb 1Now let 1 bc 0, which we can satisfy if we take b 1 and c 1. Thus with all of theseunknowns specified we have that our A is given by 11A . 1 1

r n-r 3-r122130Table 1: All possible combinations of the dimension of the column space and the row spacefor a three by three matrix.We can check this result. It is clear that A’s row space is spanned byis given by computing the R matrixR giving n 1 10 0 1 1 and its nullity, 1. 1Problem 27In a three by three matrix we have m 3 and n 3. If we say that the column space hasdimension r the nullity must then have dimension n r. Now r can be either 1, 2, or 3. Ifwe consider each possibility in tern we have Table 1, from which we see that we never havethe column space equal to the row space.Problem 28If AB 0 then the column space of B is contained in the nullity of A. For example theproduct AB can be written by recognizing this as the action of A on the columns of B. Forexample AB A b1 b2 · · · bn Ab1 Ab2 · · · Abn 0 , 1 1iwhich means that Ab 0 for each i. Let A which has nullity given by the1 1 11 2span of. Next consider B . From which we see that AB 0.11 2

Problem 29Almost sure to be the identity. With a randomend with 1 0 0 1R 0 00 0four by three matrix one is most likely to 00 .1 0Problem 30 11 1 1Part (a): Let A 1as its nullspace.has1 then A has 1 1 1, then x2Part (b): Let A 0 0 2 1 01T A 1 0 01 21 1 1 as its nullspace, but A is a free variable. Now 01 01 00 0 2 0 1 ,20 00 0which has no free variables. A similar case happens with 1 1 1A 0 0 2 ,0 0 0Then A has x2 as a free variable and AT has x3 as a free variable.Part (c): let A be given by Then 1 1 1A 0 0 0 .0 2 0 1 1 11 1 1 0 0 0 0 2 0 .0 2 00 0 0Which has x1 and x2 as pivot columns. While 1 0 0AT 1 0 2 ,1 0 0has x1 and x3 as pivot columns.T 1 11 1

Problem 31 I I . If B If A [II], then the nullspace for A is I. If C I, then the nullspace for C is 0.is I I I, then the nullspace for B0 0Problem 32x (2, 1, 0, 1) is four dimensional so n 4. The nullspace is a single vector so n r 1 or4 r 1 giving that r 3 so we have three pivots appear.Problem 33 2 3 We must have RN 0. If N 1 0 , then let R 1 2 3 . The nullity has0 1dimension of two and n 3 thereforeusing n r 2, we see that r 1. Thus we have only 0one nonzero in R. If N 0 the nullity is of dimension one and n 3 so from n r 11we conclude that r 2. Therefore we have two nonzero rows in R. 1 0 0.R 0 1 0If N [], we assume that this means that the nullity is the zero vector only. Thus the nullityis of dimension zero and n 3 still so n r 0 means that r 3 and have three nonzerorows in R 1 0 0R 0 1 0 .0 0 1Problem 34Part (a):R 1 00 0 1 11 00 10 0,,,,.0 00 10 00 0Part (b): 1 0 0 , 0 1 0 , 0 0 1 ,

and and 0 0 01 1 0 , 1 1 1 , , 1 0 1 , 0 1 1 .They are all in reduced row echelon form.Section 3.3 (The Rank and the Row Reduced Form)Problem 1Part (a): TruePart (b): FalsePart (c): TruePart (d): FalseProblem 5 A BIf R , then B is the rxr identity matrix, C D 0 and A is a r by n rC Dmatrix of zeros, since if it was not we would make pivot variables from them. The nullspaceIis given by N .0Problem 13Using the expression proved in Problem 12 in this section we have thatrank(AB) rank(A) .By replacing A with B T , and B with AT in the above we have thatrank(B T AT ) rank(AT ) .Now since transposes don’t affect the value of the rank i.e. rank(AT ) rank(A), by theabove we have thatrank(B T AT ) rank((AB)T ) rank(AB) rank(AT ) rank(A)proving the desired equivalence.

Problem 14From problem 12 in this section we have that rank(AB) rank(A) but AB I sorank(AB) I ntherefore we have that n rank(A), so equality must hold or rank(A) n. A then isinvertible and B must be its two sided inverse i.e. BA I.Problem 15From problem 12 in this section we know that rank(AB) rank(A) 2, since A is 2x3.This means that BA cannot equal the identity matrix I, which has rank 3. An e