MATH 218 Problems Solution Manual Chapter 3.3, Fall 2011

3.3 Basis and Dimension ; 71 Exercise Set 3.3 In Section 3.3 of the text, the connection between a spanning set of a vector space and linear independence is completed. The minimal spanning sets, minimal in the sense of the number of vectors in the set, are those that are linear independent, A basis for a vector space V is a set B such that B is linearly independent and span(B) = V. For example, : e B= {e1,€2,...,en} is a basis for R™ 1 0 . , -B={\ 5 off of [! of le 1 | ts bosis for Mavs ¢ B= ({l,2,x",...,0"} is a basis for Py. Every vector space has infinitely many bases. For example, if c #0, then B = {cey,eg,... ,©n} is another basis for R”, But all bases for a vector space have the same number of vectors, called the dimension of the vector space, and denoted by dim(V). As a consequence of the bases noted above: e dim(R") =n © dim(M2x2) = 4 and in general dim(Mmnxn) = mn e dim(P,) =n+1, If S = {v1,V2,..., Vin} is a subset of a vector space V and dim(V) = n recognizing the following possibilities will be useful in the exercise set: . e If the number of vectors in S exceeds the dimension of V, that is, m > n, then $ is linearly dependent and hence, can not be.a basis, e Ifm > n, then span(S) can equal V, but in this case some of the.vectors are linear combinations of others and the set S can be trimmed down to a basis for V. e Ifm <n, then the set S can be either linearly independent of linearly dependent. e Ifm <n, then § can not be a basis for V, since in this case span(S) 4 V. e Ifm <n and the vectors in $ are linearly independent, then S' canbe expanded to a basis for V. e Ifm =n, then 9 will be a basis for V if either $ is linearly independent or span(S) = V. So in this case it is sufficient to verity only one of the conditions, 1 3 : The two vectors v1 = | —1 | andvg= | —1 | are linearly independent but can not be a basis for IR? since 2 2 ~ all bases for R? must have three vectors. To expand to a basis start with the matrix 1 3 10 07. 1 3 100 -l- -1 0 1 0 | reduce to the echelon form| 0 2 1 1 0 2 2 001 000 2 1 q The pivots in the echelon form matrix are located in’columns one, two and four, so the corresponding column vectors in the original matrix form the basis. So . L 3 0 B= —-l],)/ -1 ],] 4 is a basis for R°, 2 2 0 To trim a set of vectors that span the space to a basis the procedure is the same. For example, the set 0 2 0 3 7 S= —-1},)2},}/2],] -1 4 -1 1 2 -1 72 Chapter 3 Vector Spaces is not a basis for R® since there are four vectors and hence, S$ is linearly dependent. Since 2 reduces to | 0 ; , 3, -1 2°-1 0 3 then the span of the four vectors is R°, The pivots in the reduced matrix are in columns one, two and four, 0 2 0 8 -1 2 2 -1 -1 12 -1 0 2 3 so a basis for R? is -1],/ 2 ],] -1 —1 1 —1 @ Solutions to Exercises 1. Since dim(IR*) = 3 every basis for IR® has three vectors. Therefore, since S has only two vectors it is not a basis for R®. 3. Since the third vector can be written as the sum of the first two, the set S is linearly depen- dent and hence, is not a a basis for R3. 5. Since the third polynomial is a linear combi- nation of the first two, the set S is linearly de- pendent and hence is not a basis for P3. 7. The two vectors in S$ are not scalar multi- ples and hence, the set S is linearly independent. Since every linearly independent set of two vec- tors in R? is a basis, the set 9 is a basis, 9. Since 1 0 0 1 0 0 vl -2 2 reduces t 0 -2 2 ; 1 -3 -2 0 0 -5 S is a linearly independent set of three vectors in. R3 and hence, $ is a basis. 11. Since 11 0 17 110 1 0 1 1 O07} cancer to! 2 : 1 0 1-1 -1 0 | SEP oo 1 Ht 00 2 1 000 8 the set S is a linearly independent set of four ma- trices in Mgx2. Since dim(Me,y2) = 4, then S is a basis. ‘ 2. Since dim(R?) = 2 every basis for R? has two vectors. Therefore, since S has three vectors it is linearly dependent. and hence it is not a basis for R?, 4, Since dim(P3) = 4 every basis has four poly- nomials and hence S is not a basis. 6. Since the matrix — 2 -3 1-0 a} 0 1 0 0 [i a]=2[o a} e[o o][t ol]: the set S is linearly dependent and hence, is not a basis. 1.61 8. Since det (| 3° 4 }) = —2, the set S is linearly independent and hence, is a’basis for IR, 10. Since -1 2 1 ‘T-1 2 1 ~Il -1 1 |reducesto}| 0 -3 0], —— ——b 0 -3 2 y 0 0 2 S is a linearly independent set of three vectors in R? and hence, S$ is a basis, 12, Since 1 2 0 1 2 0 01 41 reduces to} 0 1 1], a 10 1° 0041 S is a linearly independent set of three polynomi- als in Py and hence, is a basis. 3.3 Basis.and Dimension 73 13, Since -1 1 ii -l1 1 oi 2 0 1 reduces tg 0 2 8 ; 1 o1it 0 oOo -1 the set S' is a linearly independent set of three vectors in R°, so is a basis. 15, Since 1 2 2 -1 1 2 2 -1 1 14 2 educes 0 -1 2 8 132 0 reduces ti 001 1 1-15 8 0 0 0 0 the homogeneous linear systern has infinitely many solutions so the set S' is linearly dependent and is therefore not a basis for R4. 14, Since Dan eR ow | the set 5 is a linearly independent set of three vectors in R®, so is a basis, 16, Since 2 5 3 2 —2 1 1 | reduces tg | 0 121 - 0 -1 2 1 | -1 2 1 2 1 1 3 1 0 3 4 8 | go -1 1 1 pteduceste} 9 og 7 6 | 1 2 -1 2 | 0 0 0 32 the set $ is a linearly independent set of four vec- tors in R4, so is a basis, 17, Notice that 4(2a° +2 + 2-+ 2(-a? + x) — 2(1)) =a and 2x? + @-+ 2+ (—a? + z) — 2a —2 = 2", so the span of $ is Pz. Since dim(P2) = 3, the set S is a basis. 18, Since 11-2 0 1 1 -2.0 01 1 0} ances to} 2 2 1 9 oo 1 9 |Res too 1 0]? 00 1 2 090 0 2 the set. S is a linearly independent set of four matrices in Moy, so is a basis. 8+ 2t —st+t 19, Every vector in S can be written as ' t 1 2 1 =s] —l | +t] 1 |. Since the vectors | —1 | and 0 ft. 0 2 1 2 1 | are linear independent a basis for S is B = -l | : 1 | 1 0 1 | and dim(S) = 2. 20, Since every matrix in S' can be written in the form a+d d cord). 11 01 al i 0) i] and the two matrices on.the right hand side are linearly independent, a basis for S$ is 1 01 \ B= {| 10 , 11 |}: Consequently, dim(S')) = 2, 21, Every 2 x 2 symmetiic matrix has the form and dim(S) = 3. 22. Every 2 x 2 skew symmetric matrix has the form B= {| on \} with dim($) = 1. a >}_ ft 0},,f9 1],,J0 b dl} ~*1o of F811 af tho 0 1 three matrices on-the right are linearly independent a basis for S is B =: { 1 , , antl 74 Chapter 3 Vector Spaces 23. Since every polynomial p(x) in S satisfies p(0) = 0, we have that p(x) = ax+ba?. Therefore, a basis for Sis B = {%,2*} and dim(S) = 2. 24, If p(x) = av +ba?+cx+d and p(0) = 0, then d = 0, so p(z) has the form p(x) = az? + ba + cx. If in addition, p(1) = 0, then a+b+¢ = 0, so ¢= —a—b and hence. p(t) = aa®+be?+(—a—b)x = a(e3—a)+b(a*—2), Since 2° — a and x? — « are linear independent, then a basis for S is {v3 ~ x, 2 —z}, so dim($) = 2. 2 2 1 25. Since det 2 0 2 = —4, the set $ is already a basis for-IR® since it is a linearly independent ~1 -2 1 set of three vectors in R%, 2 —2 4 26. Since | 0 | = L | +] —1 |, the vectors in S are linearly dependent. Since the first two vectors 5 3 2 —2 4 7 are not scalar multiples of each other a basis for span(S) is 1 },} -1 3 2 27, The vectors can not be a basis since a set, of four vectors in R® is linearly dependent, To trim the set down to a basis for the span row reduce the matrix with column vectors the vectors in S. This gives 2 0 -1 2 20 -1 2 ~3 2 -1 3 | reducestg| 0 2 — 8 6 0 2 0 =-1 00 2° -7 | A basis for the span consists of the column vectors in the original matrix corresponding to the pivot columns of the row echelon matrix. So a basis for the span of $' is given by : 2 0 —1 B= -31,/ 2],] -1 . Observe that span(.S) = R°. 0 2 0 28. The vectors can not be a basis since a set of four vectors in R® is linearly dependent. To trim the set down to a basis for the span row reduce the matrix with column vectors the vectors in S$, This gives —2 1 -3 1 2 0 -1 2 0 OO -3 2 | reduces to; 0 -2 -5 -1 2 -3 -2 -2 0 0 -8 2 A basis for the span consists of the column vectors in the original matrix corresponding to the pivot columns of the row echelon matrix. So a basis for the span of $ is given by —2 1 —3 B= Oo }|,] 0 |],} -8 . Observe that span(S) = R°, 2 —3 —2 29, The vectors can not be a basis since a set of four vectors in R? is linearly dependent. To trim the set down to a basis for the span row reduce the matrix with column vectors the vectors in S$, This gives 20 2 4 202 4 -32 -1 0f/—]o022 6]. 02 2 4 000 -2. A basis for the span consists of the column vectors in the original matrix corresponding to the pivot columns of the row echelon matrix. So a basis for the span of $' is given by 3.3 Basis and Dimension 75 2 0 4 B= -~3],/ 2],] 0 . Observe that span(S) = R°. 0 2 4 30. The vectors can not be a basis since a set of four vectors in R° is linearly dependent. To trim the set down to a basis for the span row reduce the matrix with column vectors the vectors in S. This gives 2 1 £0 2 2 1 £0 2 2 ~1 2 38 | reduces tg 0 -2 2 1 0 0 21 0 0 2.1 A basis for the span consists of the column vectors in the original matrix corresponding to the pivot columns of the row echelon matrix, So a basis for the span of S is given by 2 1 0 B= 2),/ -1],] 2 . Observe that span(S) = R°. 0 0 2 31, Form the 3 x 5 matrix with first two column vectors the vectors in S and then augment the identity matrix. Reducing this matrix, we have that 21100 ‘21 1 0 0 —-1 0 0 1 O |} reducestg] 0 1 1 2 0 3.2001 00 -2 -1 1 A basis for R® consists of the column vectors in the original matrix corresponding to the pivot columns of the 2 ] 1 row echelon matrix. So a basis for IR® containing S$ is B = -~l},/ 07]7,] 0 : 3 2° 0 32. Form the 3 x 5 matrix with first.two column vectors the vectors ii S and then augment the identity matrix. Reducing this matrix, we have that : -l1 1 1°00 -1 11 0 0 1 1010 reduces tg 0 21 1 «0 3 1001 0 01 -2 1 A basis for R° consists of the column vectors in the original matrix corresponding to the pivot columns of the —1 t 1 row echelon matrix. So a basis for R® containing S is B = 1 >| 1 ],} 0 3 1 0 33. A basis for R* containing 9 is 34, A basis for R4 containing 9 is 1 3 1 0 —1 1 1 fo -J{-1 1 0 0 7 1 3 ~2 0 B=V) oo lela lelolela B=y) a jet ea fo] -1 fe] o 4 2 0 0 -1 2 3 L 0 35. -A basis for R® containing S is 36. A basis for R® containing 9 is ~1 1 1 ; 2 1 7 -1 1] B= 1 },f1},fo 5. B= 2 },{ -1 },] 0 3 1 0 -1] [3 0 | 37, Let ey denote the n x n matrix with a 1 in the row i, column i component and 0 in all other locations. Then B = {ey | 1 <4 <n} is a basis for the subspace of all n x n diagonal matrices. 76 Chapter 3 Vector Spaces 38, Consider the equation ¢1(¢v1) + ca(eva) ees + en(eVn) = 0. Then e(crvi + cove + +++ + CrVn) = 0 and since c 4 0, we have that cyva + cove + +++ + Cnn = 0. Now since S is a basis, it is linearly independent, so C1 = Cg = +++ = Cy, = 0 and hence S$" is a. basis. 39. It is sufficient to show that the set S” is linearly independent, Consider the equation ca Avy + coAve +++) + CpAvn = 0. This is equivalent to A(crvi + c2V2 +e + + CnVn) = 0. Multiplying both sides of this equation by A~ gives the equation civ1 +¢2v2+++++CnVn = 0. Since S is linearly independent, then cy = C2 =+++C, = 0, so that 9" is linearly independent, 40. To solve the homogeneous equation Ax = 0 consider the matrix 38.3 ~=«S 3 1 0 il 0 -1 0 -1 1 | that reduces to} 0.1 2/3 0]. 0 0 202 1 oO ot ~% -1 ‘ 2 2 So a vector is a solution provided it has the form x = ° = 2: { . Hence, a basis for S$ is 0 0 -1 2 3 1 0 41. Since H is a subspace of V, then H CV. Let S = {vi,v2,..., Vn} be a basis for H, so that S' is a linearly independent set of vectors in V. Since dim(V) = n, then S' is also a basis: for V. Now let v be a vector in'V, Since $' is a basis for V, then there exist scalars cy, c2, ...Cy such that cyv_ + cove +++) + CnVn = V. Since S is a-basis for H, then v is a linear combination of vectors in H and is therefore, also in H. Hence, V C H and we have.that H = V. 42. Since S = {ax + ba? +cx | a,b,c € R}, then dim(S) = 3. A polynomial g(z) = ax* + ba? + ex +d is in T if-and only ifa+b+e¢+d = 0, that isd = -a-—b—c. Hence, a polynomial in T has the form q(x) = a(x® — 1) + b(a* — 1) + c(a — 1), so that dim(T) = 3. Now SNOT = {ax? + ba? +cx| a+b+c=0}. Hence, a polynomial g(x) is in SMT if and only if q(x) = a(z3 — x) + b(a — x) and hence, dim($ MT) = 2. 43, Every vector in W can be written as a linear combination of the form 2s+t+3r 2 1 3 8s—t+2r | =s] 3 —lL ptr] 24. 1 | 2 +t s+t+2r 1 3 ‘ 2 1 ‘ 2 1 3 But | 2 + and hence, span 3 1,] 1 = span 3 ],; -l ],] 2 . 2 1 1 1 1 2 u . Since B = 3 is linear independent, B is a basis for W, so that dim(W) = 2. ° T 1 0 44, Since S$ = s|° +t 4 s,t ER >, then dim(S) = 2. Since 0 0 0 0 T= s , +t I s,t € R }, we also have that dim(TZ') = 2. For-the intersection, since 0 0