LINEAR ALGEBRA. 4th STEPHEN H.FRIEDBERG, ARNOLD J.INSEL, LAWRENCE E.SPENCE. Exercises Of Chapter 1-4

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1 LINEAR ALGEBRA 4th STEPHEN H.FRIEDBERG, ARNOLD J.INSEL, LAWRENCE E.SPENCE Exercises Of Chapter 1-4 http : //math.pusan.ac.kr/caf e home/chuh/

2 Contents 1. Vector Spaces Introduction Vector Spaces Subspaces Linear Combinations and Systems of Linear Equations Linear Dependence and Linear Independence Bases and Dimension Maximal Linearly Independent Subsets Linear Transformations and Matrices Linear Transformations, Null Spaces, and Ranges The matrix representation of a linear transformation Composition of Linear Transformations and Matrix Multiplication Invertibility and Isomorphisms The change of Coordinate Matrix Dual Spaces Homogeneous Linear Differential Equations with Constant Coefficients Elementary Matrix Operations and Systems of Linear Equations Elementary Matrix Operations and Elementary Matrices The Rank of a Matrix and Matrix Inverse Systems of Linear equations - Theoretical aspects Systems of Linear equations - Computational aspects

3 4. Determinants Determinants of Order Determinants of Order n Properties of Determinants Summary-Important Facts about Determinants A characterization of the Determinants

4 Vector Spaces 1.1. Introduction 1. Only the pairs in (b) and (c) are parallel (a) x = (3, 1, 2) and y = (6, 4, 2) 0 t R s.t. y = tx (b) (9, 3, 21) = 3( 3, 1, 7) (c) (5, 6, 7) = 1( 5, 6, 7) (d) x = (2, 0, 5) and y = (5, 0, 2) 0 t R s.t. y = tx 2. (a) x = (3, 2, 4) + t( 8, 9, 3) (b) x = (2, 4, 0) + t( 5, 10, 0) (c) x = (3, 7, 2) + t(0, 0, 10) (d) x = ( 2, 1, 5) + t(5, 10, 2) 3. (a) x = (2, 5, 1) + s( 2, 9, 7) + t( 5, 12, 2) (b) x = ( 8, 2, 0) + s(9, 1, 0) + t(14, 7, 0) (c) x = (3, 6, 7) + s( 5, 6, 11) + t(2, 3, 9) (d) x = (1, 1, 1) + s(4, 4, 4) + t( 7, 3, 1) 1 PNU-MATH

5 x = (a 1, a 2,, a n ) R n, i = 1, 2,, n 0 = (0, 0,, 0) R n s.t. x + 0 = x, x R n 5. x = (a 1, a 2 ) tx = t(a 1, a 2 ) = (ta 1, ta 2 ) 6. A + B = (a + c, b + d), M = ( a+c 2, b+d 2 ) 7. C = (v u) + (w u) + u = v + w u OD + 1 DB = OA + 1 AC 2 2 i.e. w + 1(v w) = 1(v + w) = u + 1 (v + w u u) PNU-MATH

6 Vector Spaces 1. (a) T (b) F (If 0 s.t. x + 0 = x, x V, then 0 = = = 0, 0 = 0) (c) F (If x = 0, a b, then a 0 = b 0 but a b) (d) F (If a = 0, x y, then a x = 0 = a y but x y) (e) T (f) F (An m n matrix has m rows and n cilumns) (g) F (h) F (If f(x) = ax + b, g(x) = ax + b, then degf=degg=1, deg(f + g)=0) (i) T (p.10 Example4) (j) T (k) T (p.9 Example3) M 13 = 3, M 21 = 4, M 22 = 5 4. (a) ( 6 3 ) PNU-MATH

7 (b) 3 5 ( 3 8 ) (c) (d) (e) 2x 4 + x 3 + 2x 2 2x + 10 (f) x 3 + 7x (g) 10x 7 30x x 2 15x (h) 3x 5 6x x U = 3 0 0, D = 3 0 0, T = ( ) The total number of crossings Fall Spring Winter Brook trout Rainbow trout Brown trout M = , 2M A = suites were sold during the June sale 4 PNU-MATH

8 We are going to show that f(t) = g(t) and (f + g)(t) = h(t), t S = {0, 1} (i) f(0) = 1 = g(0), f(1) = 3 = g(1) (ii) (f + g)(0) = 2 = h(0), (f + g)(1) = 6 = h(1) In F(S, R), f = g and f + g = h 8. VS 7, 8 9. (a) Exercise 1(b) (b) If y s.t. x + y = 0, then x + y = 0x + y By the theorem 1.1, y = y (c) a 0 + a 0 = a(0 + 0) = a 0 = a By the theorem 1.1, a 0 = V = D(R), s R (1) f, g V, f + g = g + f (f + g)(s) = f(s) + g(s) = g(s) + f(s) = (g + f)(s) f + g = g + f (2) f, g, h V, (f + g) + h = f + (g + h) ((f + g) + h)(s) = (f + g)(s) + h(s) = f(s) + g(s) + h(s) = f(s) + (g + h)(s) = (f + (g + h))(s) (f + g) + h = f + (g + h) 5 PNU-MATH

9 1.2. (3) 0 V s.t. f + 0 = f, f V (f + f )(s) = f(s) + f (s) = 0(s) f (s) = 0(s) f(s) = (0 f)(s) = ( f)(s) f = f (5) f V, 1 f = f (1 f)(s) = 1(f(s)) = f(s) 1 f = f (6) a, b F, (ab)f = a(bf) ((ab)f)(s) = (ab)f(s) = a(bf(s)) = a(bf)(s) (ab)f = a(bf) (7) a F, a(f + g) = af + ag a(f + g)(s) = a(f(s) + g(s)) = af(s) + ag(s) = (af + ag)(s) a(f + g) = af + ag (8) a, b F, (a + b)f = af + bf (a + b)f(s) = af(s) + bf(s) = (af + bf)(s) (a + b)f = af + bf 11. V = {0}, a, b F 12. (1) f, g V, t R (f + g)( t) = f( t) + g( t) = f(t) + g(t) = (f + g)(t) f + g V 6 PNU-MATH

10 1.2. (g + f)( t) = (g + f)(t) g + f V f + g = g + f (2) f, g, h V, (f + g) + h = f + (g + h) (3) 0 V s.t. f + 0 = f, f V (4) f V s.t. f + f = 0, f V, f = f (5) f V, 1f = f (6) a, b F, (ab)f = a(bf) ((ab)f( t) = (ab)f(t), abf V ) (7) a F, f, g V, a(f + g) = af + ag (8) a, b F, f V, (a + b)f = af + bf 13. No, (VS 4) fails (VS 3) (0, 1) V s.t. (a 1, a 2 ) + (0, 1) = (a 1, a 2 ), (a 1, a 2 ) V (VS 4) If a 2 = 0, then (b 1, b 2 ) V s.t. (a 1, 0) + (b 1, b 2 ) = (0, 1) 14. Yes ( R C) 15. No ( C R) α F = C, αx / V = R n, x V 16. Yes ( Q R) 7 PNU-MATH

11 No, (VS 5) fails (VS 5) If a 2 0, then 1(a 1, a 2 ) = (a 1, 0) (a 1, a 2 ) 18. No, (VS 1) fails (VS 5) If a 1 b 1, then it fails to hold (VS 1) 19. No, (VS 8) fails (VS 8) If c 1 + c 2 0, c 1 0, c 2 0, then it fails to hold (VS 8) 20. (VS 1) {a n }, {b n } V {a n } + {b n } = {a 1 + b 1, a 2 + b 2, } = {b 1 + a 1, b 2 + a 2, } = {b n } + {a n } (VS 2) ({a n } + {b n }) + {c n } = {a n } + ({b n } + {c n }) (VS 3) {0} s.t. {a n } + {0} = {a n } (VS 4) { a n } s.t. {a n } + { a n } = {a n } {a n } = {0}, {a n } V (VS 5) {1} s.t. {1}{a n } = {a n }, {a n } V (VS 6) α, β F, (αβ){a n } = α(β{a n }) (VS 7) α F, {a n }, {b n } V, α({a n } + {b n }) = α{a n } + α{b n } (VS 8) α, β F, (α + β){a n } = α{a n } + β{a n } mn 8 PNU-MATH

12 Subspaces 1. (a) F (p.1 Definition of subspace) (b) F (0 / ) (c) T (V and { } are subspaces of V ) (d) F (p.19 Theorem 1.4) (e) F (f) F (p.18 Example 4) (g) F ((0, 0, 0) W, but (0, 0, 0) / R 2 ) 2. (b), (c), (e), (f), (g) are not square matrices (a) -5, (d) 12, (h) A, B M m n (F ), a, b F (1 i m, 1 j n) (aa + bb) t ij = (aa + bb) ji = (aa) ji + (bb) ji = a(a) ji + b(b) ji = aa t ij + bbij t = (aa t + bb t ) ij (aa + bb) t = aa t + bb t 4. (A t ) t ij = (A t ) ji = A ij 5. (A + A t ) t = A t + (A t ) t = A t + A = A + A t A + A t is symmetric 9 PNU-MATH

13 tr(aa + bb) = n (aa + bb) ii = n (aa) ii + n (bb) ii = atr(a) + btr(b) i=1 i=1 a A = a 22 O O a 33 At = A a 44 A is symmetric i=1 8. (a) Yes (b) No ((0, 0, 0) / W 2 ) (c) Yes (d) Yes (e) No ((0, 0, 0) / W 5 ) (f) No x + y / W 6 ), x, y W 6 9. (1) W 1 W 3 = {0} is a subspace of R 3 (2) W 1 W 4 = W 1 is a subspace of R 3 (3) W 3 W 4 = {(a 1, a 2, a 3 ) R 3 3a 1 = 11a 2, 3a 3 = 23a 2 } is a subspace of R (i) W 1 is a subspace of F n (ii) W 2 is not a subspace of F n ( (0, 0, 0) / W 2 ) 10 PNU-MATH

14 No (The given set is not closed under addition) (Example) f, g W Let f(x) = a n x n + a n 1 x n a 0, degf = n g(x) = b n x n + b n 1 x n b 0, degg = n If b n = a n, then deg(f + g) = n 1 f + g / W W is not a subspace of V 15. Yes 17. ( ) Theorem 1.3 ( ) W is a subspace of V (i) 0 W (a = 0 F : field ) (ii) ax W (iii) x + y W 18. ( ) Theorem 1.3 ( ) W is a subspace of V (i) 0 W (ii) ax W (y = 0 W, by (i) ) (iii) x + y W (a = 1 F : field) 11 PNU-MATH

15 ( ) If W 1 W 2, then W 1 W 2 = W 2 is a subspace of V If W 2 W 1, then W 1 W 2 = W 1 is a subspace of V W 1 W 2 is a subspace of V ( ) If a / W 1, a W 2 b / W 2, b W 1, then ab W 1 W 2 But if ab W 1, then a = (ab)b 1 inw 1 if ab W 2, then b = (ba)a 1 inw 2 It s a contrdiction W 1 W 2 or W 2 W Induction on n In case of n = 2, it s clear Assume that this holds for n = k 1 (k > 2) By the induction hypothesis, k 1 a i w i + a k w k W, w k W, a k F i=1 n w i W, a i w i W, where a i F, i = 1, 2,, n i=1 21. (i) {0} 0, {0} W (ii) lim{a n } = a, lim{b n } = b lim({a n } + {b n }) = lim{a n } + lim{b n } = a + b {a n } + {b n } W 12 PNU-MATH

16 1.3. (iii) lim c{a n } = c lim{a n } c{a n } W W is a subspace of V 22. Let W 1 = {g F (F 1, F 2 ) g( t) = g(t), for each t F 1 } W 2 = {g F (F 1, F 2 ) g( t) = g(t), for each t F 1 } (1) g 1, g 2 W 1, c F (i) 0 W 1 (ii) (g 1 + g 2 )( t) = g 1 ( t) + g 2 ( t) = g 1 (t) + g 2 (t) = (g 1 + g 2 )(t) g 1 + g 2 W 1 (iii) cg 1 ( t) = c(g 1 (t)) = cg 1 (t) cg W 1 W 1 is a subspace of V (2) g 1, g 2 W 2, c F (i) 0 W 2 (ii) (g 1 + g 2 )( t) = g 1 ( t) + g 2 ( t) = g 1 (t) g 2 (t) = (g 1 + g 2 )(t) g 1 + g 2 W 2 (iii) cg 1 ( t) = c( g 1 (t)) = cg 1 (t) cg W 2 W 2 is a subspace of V 13 PNU-MATH

17 (a) W 1 + W 2 = {x + y x W 1 and y 2 } (i) 0 = W 1 + W 2 (ii) x 1 + y 1, x 2 + y 2 W 1 + W 2 (x 1 + y 1 ) + (x 2 + y 2 ) = (x 1 + x 2 ) + (y 1 + y 2 ) W 1 + W 2 (iii) x 1 + y 1 W 1 + W 2, c F c(x 1 + y 1 ) = cx 1 + cy 1 W 1 + W 2 (b) W as a subspace of V s.t. W 1 W, W 2 W x W 1 W, y W 2 W x + y W 1 + W 2 W 24. V = F n (i) W 1 W 2 = {(0, 0,, 0)} (ii) W 1 + W 2 V is clear v = (a 1, a 2,, a n ) = (a 1, a 2,, a n 1, 0) + (0, 0,, 0, a n ) W 1 + W 2 V = W 1 + W (a) W 1 W 2 = {0} (b) V = {f(x) P (F ) f(x) = a 0 + a 1 x + } ( ) Since W 1 V and W 2 V, W 1 + W 2 V is clear f V, f = a 0 + a 2 x + + a 1 x + a 3 x 3 + W 1 + W 2 V = P (F ) = W 1 W (a) W 1 W 2 = {A M m n A ij = 0 i, j} = {0} 14 PNU-MATH

18 1.3. (b) W 1 + W 2 V is clear a 11 a 12 a 1n a 21 a 22 a 2n A =.... M m n, a m1 a m2 a mn a 11 a 12 a 1n a 22 a 2n A = a W 1 + W a mn a m1 a m(n 1) 0 V = P (F ) = W 1 W (a) W 1 W 2 = {A M m n A ij = 0 i, j} = {0} (b) W 1 + W 2 V is clear a 11 a 12 a 1n 0 a 22 a 2n A =..... M m n, 0 0 a mn a a 12 a 1n 0 a 22 0 A = a 2n..... W 1 + W a mn 0 0 a (n 1) n V = P (F ) = W 1 W (a) W 1 W 2 = {0} (b) W 1 + W 2 V is clear A V, A = ( A At 2 ) + ( A+At 2 ) W 1 + W 2 V W 1 + W 2 15 PNU-MATH

19 1.3. (cf) char(f ) 2 { 1 2 (A At )} t = 1 2 (At A) = { 1 2 (A At )} 1 2 (A At ) W 1 { 1 2 (A At )} t = 1 2 (A + At ) 1 2 (A + At ) W W 1 : strictly lower triangular matrices W 2 : all symmetric matrices (a) W 1 W 2 = {0} a 11 a 12 a 1n a 11 a 12 a 1n a 21 a 22 a 2n (b) Let A =.... V and a 12 a 22 a 2n A =.... W 2 a n1 a n2 a nn a 1n a 2n a nn then A = (A A ) + A W 1 + W 2 V W 1 + W 2 V = W 1 + W ( ) If x = x 1 + x 2 = x 3 + x 4, x 1, x 3 W 1 and x 2, x 4 W 2 then x 1 x 3 = x 4 x 2 W 1 W 2 = {0} x 1 = x 3, x 2 = x 4 ( ) (i) V = W 1 + W 2 is clear (ii) If w W 1 W 2, then w = w + 0 = w = 0 W 1 W 2 = {0} 16 PNU-MATH

20 (a) ( ) v W v + W = W v + W is a subspace of V ( ) v + w 1, v + w 2 W v + w 1 + v + w 2 = v + (w 1 + w 2 + v) v + W w 1 + w 2 + v W v W (b) v 1 v 2 W v 1 v 2 + W = W v 1 + W = v 2 + W (c) Since v 1 v 1 W and v 2 v 2 W, (v 1 v 1) + (v 2 v 2) = (v 1 + v 2 ) (v 1 + v 2) W (v 1 + v 2 ) + W = (v 1 + v 2) + W (v 1 + W ) + (v 2 + W ) = (v 1 + W ) + (v 2 + W ) Since a(v 1 v 1) W, a(v 1 + W ) = a(v 1 + W ) (d) S = V/W = {v + W v V } is a vector space (VS 1) (v 1 +W )+(v 2 +W ) = (v 1 +v 2 )+W = (v 2 +v 1 )+W = (v 2 +W )+(v 1 +W ) (VS 2) {(v 1 + W ) + (v 2 + W )} + (v 3 + W ) = (v 1 + v 2 ) + W + v 3 + W = (v 1 + v 2 + v 3 ) + W = v 1 + (v 2 + v 3 ) + W = {v 1 + W } + {(v 2 + v 3 ) + W } = (v 1 + W ) + {(v 2 + W ) + (v 3 + W )} (VS 3) 0 + W s.t. (v + W ) + (0 + W ) = (v + 0) + W = v + W (VS 4) v + W, (1 + W )(v + W ) = v + W (VS 5) v + W s.t. (v + W ) + ( v + W ) = 0 + W 17 PNU-MATH

21 1.3. (VS 6) a, b F, (ab)(v+w ) = abv+w = a(bv)+w = a(bv+w ) = a(b(v+w )) (VS 7) a F, a(v 1 + W + v 2 + W ) = av 1 + av 2 + W = av 1 + W + av 2 + W = a(v 1 + W ) + a(v 2 + W ) (VS 8) (a + b)(v + W ) = (a + b)v + W = av + bv + W = a(v + W ) + b(v + W ) 18 PNU-MATH

22 Linear Combinations and Systems of Linear Equations 1. (a) T (0v = 0, v V ) (b) F (p.30 span( ) = {0}) (c) T (p.30 Theorem 1.5) (d) F (p.27) (e) T (p.27) (f) F (p.33 2(b)) 2. (a) {r(1, 1, 0, 0) + s( 3, 0, 2, 1) + (5, 0, 4, 0) r, s R} (b) ( 2, 4, 3) (c) There are no solutions (d) {r( 8, 3, 1, 0) + ( 16, 9, 0, 2) r R} (e) {r(0, 3, 1, 0, 0) + s( 3, 2, 0, 1, 0) + ( 4, 3, 0, 0, 5) r, s R} (f) (3, 4, 2) 3. (a) yes ( 2, 0, 3) = 4(1, 3, 0) + ( 3)(2, 4, 1) (b) Yes (1, 2, 3) = 5( 3, 2, 1) + 8(2, 1, 1) (c) No (d) Yes (2, 1, 0) = 4(1, 2, 3) + 6 (1, 3, 2) 5 5 (e) No (f) Yes ( 2, 2, 2) = 4(1, 2, 1) + 2( 3, 3, 3) 19 PNU-MATH

23 (a) Yes (x 3 3x + 5) = 3(x 3 + 2x 2 x + 1) + ( 2)(x 3 + 3x 2 1) (b) No (c) Yes 4, 3 (d) Yes 2, 5 (e) No (f) No 5. (a) Yes (2, 1, 1) = 1(1, 0, 2) + ( 1)( 1, 1, 1) span(s) (b) No (c) No (d) Yes 2, 1 (e) Yes 1, 3, 1 (f) No (g) Yes 3, 4, 2 (h) No 6. Let span{(1, 1, 0), (1, 0, 1), (0, 1, 1)} = W v = (a 1, a 2, a 3 ) F 3, v = r(1, 1, 0)+s(1, 0, 1)+t(0, 1, 1) s.t. r = 1 2 (a 1+a 2 a 3 ), s = 1 2 (a 1 a 2 +a 3 ), t = 1 ( a a 2 + a 3 ) W V W Since W F 3 is clear, W = F 3 20 PNU-MATH

24 v = (a 1, a 2,, a n ) F n v = a 1 e 1 + a 1 e a n e n, a i F, i = 0, 1,, n 8. f(x) = a 0 + a 1 x + + a n x n P n (F ) f(x) = a a 1 x + + a n x n, a i F, i = 0, 1,, n ( ) a11 a 9. A = 12 M a 21 a 2 2 (F ) ( ) 22 ( ) ( ) A = a 11 + a a ( ) a ( ) a c 10. A = M c b 2 2 (F ) ( ) ( ) ( ) A = a + b + c span{m , M 2, M 3 } 11. If x 0, then span({x}) = {ax a F } is the line through the origin in R 3 Otherwise span({x}) = {ax a F } is the origin 12. ( ) By the theorem 1.5, span(w ) is a subspace of V W is a subspace of V ( ) v W, v = 1 v span(w ) W span(w ) v span(w ), v = a 1 v a n v n 21 PNU-MATH

25 1.4. Since a i v i W, v = n W span(w ) W i=1 13. (i) S 1 S 2 span(s 2 ) span(s 1 ) span(s 2 ) (ii) By (i) span(s 1 ) = V span(s 2 ) V span(s 2 ) = V 14. S 1, S 2 V, S 1 = {x 1, x 2,, x m }, S 2 = {x m+1, x m+2,, x n } (i) Since S 1, S 2 S 1 S 2, span(s 1 ), span(s 2 ) span(s 1 S 2 ) If v = m a i x i + n a i x i span(s 1 ) + span(s 2 ), a i F i=1 i=m+1 then v span(s 1 S 2 ) (ii) If v = n a i x i span(s 1 S 2, a i F i=1 then v = m a i x i + i=1 n i=m+1 a i x i span(s 1 ) + span(s 2 ) 15. Since S 1 S 2 S 1 span(s 1 ) and S 1 S 2 S 2 span(s 2 ), span(s 1 S 2 ) span(s 1 ) span(s 2 ) 16. v span(s), suppose v = a 1 v a n v n = b 1 v b n v n then (a 1 b 1 )v (a n b n )v n = 0 (a 1 b 1 ) = = (a n b n ) = 0 a 1 = b 1,, a n = b n 22 PNU-MATH

26 W must be a finite set (i) F is an infinite field If 0 w W, then {aw a F } W = {0} (ii) F is a finite field If β = {w 1,, w n }, then W = F β dim W < 23 PNU-MATH

27 Linear Dependence and Linear Independence 1. (a) If S is a linearly dependent set, then each vector in S is a linear combination of other vector in S. Ans : F (Example) V = R 3, S={e 1 = (1, 0, 0), e 2 = (0, 1, 0), e 3 = (0, 0, 1), e 4 = (1, 1, 0)} {e 1, e 2, e 3 } : linearly independent {e 1, e 2, e 4 } : linearly dependent (b) Any set containing the zero vector is linearly dependent. Ans : T ( ) a F, 0 = a 0, a 0 (c) The empty set is linearly dependent. Ans : F ( ) linearly dependent set must be non-empty. (d) Subsets of linearly dependent sets are linearly dependent. Ans : F ( ) theorem 1.6 (e) Subsets of linearly independent sets are linearly independent. Ans : T ( ) the corollarly from theorem 1.6 (f) If a 1 x 1 + a 2 x a n x n = 0 and x 1, x 2,, x n are linearly independent, then all the scalars a i are zero. Ans : T ( ) from the definition. 2. (a), (d), (e), (g), (h), (j) : linearly independent (b), (c), (f), (i) : linearly dependent 24 PNU-MATH

28 In M 2 3 (F ), prove that the set , 1 1, 0 0, 1 0, is linearly dependent. ( ) a + d = 0, b + d = 0, c + d = 0, a + e = 0, b + e = 0, c + e = 0 a = d, b = d, c = d, d, e = d the given set is linearly dependent. 4. In F n, let e j denote the vector whose jth coordinate is 1 and whose other coordinates are 0. Prove that {e 1, e 2,, e n } is linearly independent. ( ) (a 1, a 2,, a n ) = a 1 e 1 + a 2 e a n e n = (0,, 0) a 1 = a 2 = = a n = 0 {e 1, e 2,, e n } is linearly independent. 5. Show that the set {1, x, x 2,, x n } is linearly independent in P n (F ). ( ) If a 0 + a 1 x 1 + a 2 x a n x n = x x x n, then a i = 0, 1 i n. 25 PNU-MATH

29 In M n n (F ), let E ij denote the matrix whose only nonzero entry is 1 in the ith row and ith column. Prove that {E ij : 1 i m, 1 j n} is linearly independent. ( ) A M n n (F ), If A = a 11 E 11 + a 12 E a nn E nn = 0 then, a 11 a 12 a 1n a 21 a 22 a 2n... = a n1 a n2 a nn a ij = 0, where 1 i m, 1 j n 7. Recall from Example 3 in section 1.3 that the set of diagonal matrices in M 2 2 (F ) is a subspace. Find a linearly independent set that generates this subspace. ( ) ( ) ( ), ( 0 ) 0 ( 0 1 ) ( ) a = a + b, a, b F 0 b Let S = {(1, 1, 0), (1, 0, 1), (0, 1, 1)} be a subset of the vector space F 3. (a) Prove that if F = R, then S is linearly independent. (b) Prove that if F has characteristic 2, then S is linearly independent. ( ) (a) a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (0, 0, 0), a, b, c F 26 PNU-MATH

30 1.5. then a + b = 0, a + c = 0, b + c = 0 a = b = c = 0 (b) If a = b = c = 1, then a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (2, 2, 2) = (0, 0, 0) {(1, 1, 0), (1, 0, 1), (0, 1, 1)} is linearly dependent 9. Let u and v be distinct vectors in a vector space V. Show that {u, v} is linearly dependent if and only if u or v is a multiple of the other. ( ) u v, {u, v} : linearly independent ( ) If au+bv= 0, (1)a 0 u = a b v (2)a = 0 b 0 v = a b u ( )u = kv 1u kv = Give an example of three linearly dependent vectors in R 3 such that none of the three is a multiple of another. (Example) {( 1, 0, 1), (1, 1, 0), (0, 1, 1)} 11. Let S = {u 1, u 2,, u n } be a linearly independent subset of a vector space V over the field Z 2. How many vectors are there in span(s)? span(s) = F n, when the set is linearly independent. * what if the given set is not linearly independent? 27 PNU-MATH

31 1.5. then the number of vectors in the set is smaller than F n. (Example) a i = {0, 1}, v 1 = ( 1, 0, 1), v 2 = (1, 1, 0), v 3 = (0, 1, 1) span(s)={(0, 0, 0), (0, 1, 1), (1, 1, 0), (1, 0, 1)} 12. Prove Theorem 1.6 and its corollary. Let S 1 = {u 1,, u n }, S 2 = {u 1,, u n, v} (i) a 1 u 1 + a 2 u a n u n + a n+1 v = 0, not all a i 0 (n 1) then v = b 1 u b n u n, where b i = a i a n+1 S 2 is linearly independent F (ii) a 1 u a n u n + 0v = 0 a 1 = = a n = 0 S 1 is linearly independent 13. Let V be a vector space over a field of characteristic not equal to two. (a) Let u and v be distinct vectors in V. Prove that {u, v} is linearly independent if and only if {u + v, u v} is linearly independent. ( )( ) Suppose a(u + v) + b(u v) = 0, a, b F (a + b)u + (a b)v = 0 a = b = 0 ( ) {u + v, u v} is linearly independent. au + bv = ( a+b 2 a b )(u + v) + ( )(u v) = PNU-MATH

32 1.5. ( a+b a b ) = 0 and ( ) = a = b = 0 (b) ( ) a(u + v) + b(u + w) + c(v + w) = 0 (a + b)u + (a + c)v + (b + c)w = 0 a + b = a + c = b + c = 0 a = b = c = 0 ( ) au + bv + cw = ( a+b 2 ( a+b 2 ) = ( a+c 2 ) = ( b+c 2 ) = 0 a = b = c = 0 )(u + v) + ( a+c 2 b+c )(u + w) + ( )(v + w) ( ) By the exercise 1(b), if S = {0}, then S is linearly dependent If v = a 1 u a n u n, then a 1 u a n u n 1 v = 0 S = {u 1,, u n, v} is linearly dependent ( ) If S {0}, a i 0 s.t. a 1 u a n u n + a n+1 v = 0 Let a n+1 0 v = b 1 u 1 + b n v n span({u 1,, u n }), where b i = a i a n+1 F 15. ( ) By the theorem 1.6 ( ) If u 1 = 0, then it s clear So we may assume u PNU-MATH

33 1.5. Let k 0 be the first integer s.t. u 1,, u k linearly independent and {u 1,, u k, u k+1 } linearly dependent So a 1 u a k u k + a k+1 u k+1 = 0 for some scalar a 1, a 2,, a k+1 (not all zero) If a k+1 = 0, then a 1 u a k u k + a k+1 u k+1 = a 1 u a k u k = 0 a 1 = = a k = a k+1 = 0 It s a contradiction Thus u k+1 = b 1 u b k u k span(u 1,, u k ), where b i = a i a k ( ) By the corollary of theorem 1.6 ( ) S S is linearly independent 17. Let M (1) = (a 11, 0,, 0) t, M (2) = (a 12, a 22, 0,, 0) t,, M (n) = (a 1n, a 2n,, a nn ) t M span{m (1), M (2),, M (n) a ii 0} Suppose k 1 M (1) + + k n M (n) = 0 k 1 = k 2 = = k n = 0, a ii 0 S is linearly independent 18. f 0 (x) = a 0 f 1 (x) = a 0 + a 1 x. f n (x) = a 0 + a 1 x + + a n x n k 0 f 0 (x) + k 1 f 1 (x) + + k n f n (x) = 0 30 PNU-MATH

34 1.5. k n a n = 0 (k n 1 + k n )a n 1 = 0. (k k n )a 0 = 0 k i = (a 1 A a k A k ) t = (0) t a 1 A t a k A t k = 0 a 1 = = a k = 0 {A t 1,, A t k } is linearly independent 20. ae rt + be st = 0, r s a + be (s r)t = 0 Since e (s r)t 0, a = b = 0 {e rt, e st } is linearly independent 31 PNU-MATH

35 Bases and Dimension 1. (a) F ( ) is a basis for the zero vector space. * span{ } = {0} and is linearly independent. (b) T ( ) Theorem 1.9 ; If a vector space V is generated by a finite set S, then some subset of S is a basis for V. (c) F (Counterexample) {1, x, x 2, } is a basis for P (F ) (d) F ( ) Corollary 2 (c) from Theorem 1.10 Every linearly independent subset of V can be extended to a basis for V. (e) T ( ) Corollary 1 from Theorem 1.10 Let V be a vector space having a finite basis. Then every basis for V contains the same number of vectors. (f) F ( ) {1, x, x 2,, x n } is a basis for P n (F ) (g) F ( ) The dimension of M m n (F ) is m n (h) T ( ) Replacement theorem. (i) F ( ) Theorem 1.8. Let V be a vector space and β = {u 1, u 2,, u n } be a subset of V. Then β is a basis for V if and only if each v V can be uniquely expressed as a linear combination of vectors of β. (Example) V = R 2, S = {v 1 = (1, 0), v 2 = (0, 1), v 3 = (1, 1)} (a, b) = av 1 + bv 2 + 0v 3 = 0v 1 + (b a)v 2 + av 3 32 PNU-MATH

36 1.6. (j) T Theorem Let W be a subspace of a finite-dimensional vector space V. Then W is finite-dimensional and dim(w) dim(v). Moreover, if dim(w)=dim(v), then V = W. (k) T (i) The vector space {0} has dimension zero and 0 is unique element in V. So V has exactly one subspace with dimension 0. (ii) From the theorem 1.11, W V as a subspace If dim(w )=dim(v ), then V = W So V has exactly one subspace with dimension n. (l) T ( ) If S is linearly independent, Let W be a space spanned by S. Then S is a basis for W (the corollary 2 from 1.10) And dimw = n W = V S is a basis for V ( ) If S is a generating set for V that contains n vectors, then by the corollary 2 from 1.10, S is linearly independent. 2. (a) {(1, 0, 1), (2, 5, 1), (0, 4, 3)} : a basis for R 3 ( ) 0 = a (1, 0, 1) + b (2, 5, 1) + c (0, 4, 3) = (a + 2b, 5b 4c, a + b + 3c) 33 PNU-MATH

37 1.6. a = b = c = 0 (*) = 27 0 {(1, 0, 1), (2, 5, 1), (0, 4, 3)} is linearly independent. (b) {(2, 4, 1), (0, 3, 1), (6, 0, 1)} : a basis for R 3 ( ) 0 = a (2, 4, 1) + b (0, 3, 1) + c (6, 0, 1) = (2a + 6c, 4a, a b c) a = b = c = 0 (c) {(1, 2, 1), (1, 0, 2), (2, 1, 1)} : a basis for R 3 (d) {( 1, 3, 1), (2, 4, 3), ( 3, 8, 2)} : a basis for R 3 (e) {(1, 3, 2), ( 3, 1, 3), ( 2, 10, 2)} ( ) 0 = a (1, 3, 2) + b ( 3, 1, 3) + c ( 2, 10, 2) = (a 3b 2c, 3a + b 10c, 2a + 3b 2c) a = 2b, c = 1 2 b (4, 2, 1) (0, 0, 0) (*) = = 0 {(1, 3, 2), ( 3, 1, 3), ( 2, 10, 2)} is linearly dependent. 34 PNU-MATH

38 (a) { 1 x + 2x 2, 2 + x 2x 2, 1 2x + 4x 2 } ( ) 0 = a ( 1 x + 2x 2 ) + b (2 + x 2x 2 ) + c (1 2x + 4x 2 ) = (1 + 2a 2b + 4c)x 2 ) + ( a + b 2c)x + ( a + 2b + c) a = 5c, b = 3c ( 5, 3, 1) (0, 0, 0) (*) = = 0 { 1 x + 2x 2, 2 + x 2x 2, 1 2x + 4x 2 } is linearly dependent. (b) {1 + 2x + x 2, 3 + x 2, x + x 2 } : a basis for P 2 (R) (c) {1 2x 2x 2, 2 + 3x x 2, 1 x + 6x 2 } : a basis for P 2 (R) (d) { 1 + 2x + 4x 2, 3 4x 10x 2, 2 5x 6x 2 }: a basis for P 2 (R) (e) {1 + 2x x 2, 4 2x + x 2, x 9x 2 } ( ) 0 = a ( 1 x + 2x 2 ) + b (4 2x + x 2 ) + c ( x 9x 2 ) = ( a + b 9c)x 2 ) + (2a 2b + 18c)x + (a + 4b c) a = 11c, b = 2c ( 11, 2, 1) (0, 0, 0) 4. No. {x 3 2x 2 + 1, 4x 2 x + 3, 3x 2} = 3 and dim(p 3 (R)) = 4 The generating set for V contains at least 4 vectors. 35 PNU-MATH

39 No. Any n + 1 or more vectors in V are linearly dependent. Since dim(r 3 ) = 3, every linearly independent set contains at most 3 vectors. 6. {( ) ( ) ( ) ( )} ,,, {( ) ( ) ( ) ( )} ,,, {( ) ( ) ( ) ( )} ,,, Select any nonzero vector in the given set, say u 1, to be a vector in the basis. Since u 3 = 4u 1, the set {u 1, u 3 } is linearly dependent. Hence we don t include u 3 On the other hand, the set {u 1, u 2 } is linearly independent. Thus we include u 2. And the set {u 1, u 2, u 4 } is linearly dependent. So we exclude u 4. {u 1, u 2, u 5 } is a basis for R 3. (*) = = = 0 36 PNU-MATH

40 1.6. {u 1, u 2, u 4 } is linearly dependent. 8. {u 1, u 3, u 4, u 8 } is a basis for W. 9. a, b, c, d F, (a 1, a 2, a 3, a 4 ) = au 1 + bu 2 + cu 3 + du 4 = (a, a + b, a + b + c, a + b + c + d) a = a 1, b = a 2 a 1, c = a 3 a 2, d = a 4 a 3 (a 1, a 2, a 3, a 4 ) = a 1 u 1 + (a 2 a 1 )u 2 + (a 3 a 2 )u 3 + (a 4 a 3 )u (a) f 0 (x) = 1 3 (x2 1), f 1 (x) = 1 2 (x2 + x 2), f 2 (x) = 1 6 (x2 + 3x + 2) g(x) = 2 b i f i (x) = 4x 2 x + 8 (b) i=0 f 0 (x) = 1 35 (x2 4x + 3), f 1 (x) = 1 10 (x2 + x 12), f 2 (x) = 1 14 (x2 + 3x 4) g(x) = 2 b i f i (x) = 3x + 12 (c) i=0 f 0 (x) = 1 15 (x3 3x 2 x + 3), f 1 (x) = 1 8 (x3 2x 2 5x + 6), f 2 (x) = 1 12 (x3 7x 2 6), f 3 (x) = 1 40 (x3 + 2x 2 x 2) g(x) = 3 b i f i (x) = x 3 + 2x 2 + 4x 5 (d) i=0 f 0 (x) = 1 12 (x3 + x 2 2x), f 1 (x) = 1 6 (x3 + 2x 2 3x), f 2 (x) = 1 6 (x3 + 4x 2 + x 37 PNU-MATH

41 1.6. 6), f 3 (x) = 1 12 (x3 + 5x 2 + 6x) g(x) = 3 b i f i (x) = 3x 3 6x 2 + 4x + 15 i=0 11. (i) We need to show that {u + v, au} is linearly independent k 1, k 2 are scalars, If k 1 (u + v) + k 2 (au) = (k 1 + ak 2 )u + (k 1 v) = 0 Since {u, v} is a basis for V, k 1 + ak 2 = 0, k 1 = 0 (a is a nonzero scalar) k 1 = k 2 = 0 (ii) If k 1 (au) + k 2 (bv) = (ak 1 )u + (bk 2 )v = 0 Since a 0, b 0 and {u, v} is a basis for V. k 1 = k 2 = 0 {au, bv} is a basis for V. 12. k 1 (u + v + W ) + k 2 (v + w) + k 3 (w) = 0, k 1, k 2, k 3 F k 1 u + (k 1 + k 2 )v + (k 1 + k 2 + k 3 )w = 0 k 1 = k 2 = k 3 = {a(1, 1, 1) a R} is a solution set {(1, 1, 1)} is a basis for the given system 38 PNU-MATH

42 {(0, 0, 1, 0, 0), (0, 0, 0, 1, 0), (0, 1, 0, 0, 0), (0, 0, 0, 0, 1)} : a basis for W 1 {(1, 0, 0, 0, 1), (0, 1, 1, 1, 0)} : a basis for W 2 dim(w 1 )=4, dim(w 2 )=2 (*) The dimension of the solution space AX = 0 is equal to n ranka (n is the number of rows of A) (i) a 1 a 3 a 4 = 0 i.e. a 1 + 0a 2 a 3 a 4 + 0a 5 = 0 (*) a 1 a 2 A = (1, 0, 1, 1, 0) 1 5, X = a 3 a 4 a 5 (*) is equal to AX = 0 The dimension of the solution space AX = 0 is equal to n ranka = 5 1 = 4 (ii) a 2 = a 3, a 2 = a 4, a 1 + a 5 = A = rank(a)=3 The dimension of the solution space AX = 0 is equal to n ranka = 5 3 = 2 39 PNU-MATH

43 (i) {( n E ij, where i j) and ( )} : a basis for W. i,j=1 ( ) = {(1, 1, 0,, 0), (0, 1, 1,, 0),, (0,, 1, 1, 0), (0,, 1, 1)} that is, ith component of the element in ( ) is 1 (ii) Dim(W )=n 2 n + (n 1) = n 2 1 ( ) n 2 : the dim of M n n (F ), n : the number of vectors consist of diagonal, (n 1) : the number of vectors consist of ( ) * When charf =2, if A t = A, then a ij = a ji and a ii = a ii a ii = 0 (Example) W = {(a 1, a 2, a 3, a 4, a 5 ) n a 1 = 0} R 5 is a subspace of V, dim(w ) = 4 i=1 β = {v 1 = (1, 1, 0, 0, 0), v 2 = (1, 0, 1, 0, 0), v 3 = (1, 0, 0, 1, 0), v 4 = (1, 0, 0, 0, 1)} (or β = {(1, 1, 0, 0, 0), (0, 1, 1, 0, 0), (0, 0, 1, 1, 0), (0, 0, 0, 1, 1)}) From a 1 + a 2 + a 3 + a 4 + a 5 = 0, a 5 = (a 1 + a 2 + a 3 + a 4 ) (a 1, a 2, a 3, a 4, a 5 ) = (a 1, a 2, a 3, a 4, (a 1 + a 2 + a 3 + a 4 ) = (a 1, 0, 0, 0, a 1 ) + (0, a 2, 0, 0, a 2 ) + (0, 0, a 3, 0, a 3 ) + (0, 0, 0, a 4, a 4 ) 16. A W : the set of all upper triangular n n matrices 40 PNU-MATH

44 1.6. a 11 a 12 a 1n 0 a 22 a 2n a nn (i) {E ij E ij = 0, i > j} is a basis for W (ii) dim(w ) = n = 1 n(n + 1) A W : the set of all skew-symmetric n n matrices 0 a 12 a 13 a 1n a 21 0 a 23 a 2n a 31 a 32 0 a 3n, a ij = a ji..... a n1 a n2 a n3 0 (i) A basis for W , (ii) dim(w ) = (n 1) = 1 n(n 1) 2 (iii) In case of charf =2 if A t = A, then a ij a ji ( ) if a ii = a ii a ii = 0 in char(f )=2,, PNU-MATH

45 V consists of all sequences {a n } in F that have only a finite number of nonzero terms a n Let {e i } = {0,, 0, 1, 0, 0} i.e. the ith term is 1 then {e i i = 1,, n} is a basis for V 19. ( ) Suppose each v V can be uniquely expressed as a linear combination of vectors of β, then clearly β spans V. If 0 = a 1 u 1 + a 2 u a n u n And we also have 0 = 0 u u u n By hypothesis, the representation of zero as a linear combination of the u i is unique. Hence each a i = 0, and the u i are linearly independent. 20. (a) Prove that there is a subset of S that is a basis for V. (Be careful not to assume that S is finite.) If V = (0) ( S) a basis. So we may assume that V (0) Let C = {B B S, B is linearly independent} then B C, B n(= dim V ) (i) Choose B C with maximal element i.e. B C and B C, B B (ii) Claim S span(b ) 42 PNU-MATH

46 1.6. ( ) If not, S span(b ), then v S, v / span(b ) By the theorem 1.7(p.39), B {v} is linearly independent Since B B {v} S, this contradicts to the maximality of B And V = span(s) span(span(b )) = span(b ) V = span(b ) Therefore B is a basis for V (b) Let Q V s.t. span(q) = V and Q < n From (a), we can find a subset Q of Q is a basis for V. This contradicts to the following fact : If β = {v 1, v 2,, v n } is a basis for V, then any set of n + 1 vectors in V is linearly dependent Any spanning set S for V must contain at least n vectors. 21. (i) Suppose dim V = V has an infinite set of linearly independent (ii) By the Replacement theorem 22. dim(w 1 W 2 ) = dim(w 1 ) W 1 W 2 = W 1 W 1 W (a) dim(w 1 ) = dim(w 2 ) if and only if v span{v 1, v 2,, v k } = W 1 (b) If dim(w 1 ) dim(w 2 ), then dim(w 1 ) < dim(w 2 ) 43 PNU-MATH

47 1.6. ( ) By (a), v W 1 and v / W 2 By the exercise 20, dim(w 1 ) k and dim(w 2 ) k + 1 dim(w 1 ) dim(w 2 ) dim(w 1 ) < dim(w 2 ) 24. f(x) = k n x n + + k 1 x + k 0, k n 0, k i R Let a 0 f(x) + a 1 f (x) + + a n f (n) (x) = 0, for some scalars a 0,, a n R then (a 0 k n )x n + (a 0 k n 1 + a 1 k n )x n (a 0 k 1 + 2!a 1 k 2 + 3!a 2 k n!a n 1 k n )x + ( n m!a m k m ) = 0 m=0 By equating the coefficient of x k on both sides of this equation for k = 0, 1, 2,, n, we obtain a 0 = a 1 = = a n = 0 (since charr=0) It follows from (b) of corollary 2 (p.48) that {f(x), f (x),, f (n) (x)} is a basis for P n (R). g(x) P n (R), c 0,, c n R s.t. g(x) = c 0 f(x) + c 1 f (x) + + c n f (n) (x) 25. Z = {(v, w) v V and w W } = V W dim(z) = dim(z) dim(w ) = mn β = {v 1,, v m } a basis for V γ = {w 1,, w n } a basis for W then α = {(v 1, 0),, (v m, 0),, (0, w 1 ),, (0, w n )} a basis for V W 44 PNU-MATH

48 W = {f P n (R) f(a) = 0} is a subspace of V = P n (R) i.e. f W forms (x a)(a n 1 x n a 1 x + a 0 ) dim(w ) = n 27. dim(w 1 P n (F )) = n if n : even 2 n+1 if n : odd, dim(w 2 P n (F )) = 2 n + 1 if n : even 2 n+1 if n : odd dim V = 2n β = {v 1, v 2,, v n } a basis for V over C, dim V = n β = {v 1, v 2,, v n, v 1 i, v 2 i,, v n i} a basis for V over R a 1 v a n v n + b 1 v v i + + b n v n i = 0 (a 1 + ib 1 )v (a n + ib n )v n = Let β = {u 1,, u k, v 1,, v m, w 1,, w p } (i) β spans W 1 + W 2 v W 1 + W 2, v = ( k a i u i + m b i v i ) + ( k c i u i + p d i w i ) span(β), w 1 i=1 i=1 W 1, w 2 W 2 (ii) β is linearly independent Suppose that a 1 u 1 + +a k u k +b 1 v 1 + +b m v m +c 1 w 1 + +c p w p = 0 ( ) i=1, where a 1,, a k, b 1,, b m, c 1,, c p F Let v = k a i u i + m b i v i = p c i w i W 1 W 2 i=1 i=1 i=1 i=1 45 PNU-MATH

49 1.6. Since {u 1,, u k } is a basis for W 1 W 2, d i F s.t. v = k d i u i k i=1 d i u i + p c i w i = 0 c 1 = c 2 = = c p = 0 i=1 By ( ), a 1 = = a k = b 1 = = b m = 0 β is linearly independent β is a basis for W 1 + W 2 dim(w 1 + W 2 ) = dim(w 1 ) + dim(w 2 ) dim(w 1 W 2 ) (b) dim(w 1 + W 2 ) = 0 W 1 W 2 = i=1 30. (i) W 1 and W 2 are subspaces of V ( ) ( ) a b d e A =, B = W c a f d 1, α F ( ) αa+d αb+e αa + B = W αc+f αa+d 1 ( ) ( ) 0 a 0 c A =, B = W a b c d 2, α F ( ) 0 αa+c αa + B = W αa-c b+d 2 (ii) dim(w 1 ) = 3, dim(w 2 ) = 2, dim(w 1 W 2 ) = 1 and dim(w 1 + W 2 ) = (a) ( ) W 1 W 2 is a subspace of W 1, By the theorem 1.11, dim(w 1 W 2 ) dim(w 1 ) n dim(w 1 W 2 ) n 46 PNU-MATH

50 1.6. (b) dim(w 1 + W 2 ) = dim(w 1 ) + dim(w 2 ) dim(w 1 W 2 ) m + n dim(w 1 + W 2 ) dim(w 1 + W 2 ) m + n 32. (a) dim(w 1 W 2 ) = dim(w 2 ) W 1 = {(a 1, a 2, 0) a 1, a 2 F }, dim(w 1 )=2 W 2 = {(a 1, 0, 0) a 1 F }, dim(w 2 )=1 (b) dim(w 1 W 2 )=0 W 1 = {(a 1, 0, 0) a 1 F }, dim(w 1 )=1 W 2 = {(0, a 2, a 3 ) a 2, a 3 F }, dim(w 2 )=2 (c) W 1 = {(a 1, 0, a 3 ) a 1, a 3 F }, dim(w 1 )=2 W 2 = {(a 1, a 2, 0) a 1, a 2 F }, dim(w 2 )=2 33. V = W 1 W 2 β 1 β 2 =, β 1 β 2 : a basis for V ( ) (i) Suppose that a 1 u 1 + a 2 u b 1 w 1 + b 2 w 2 + = 0 a i, b j F Then a 1 u 1 + a 2 u 2 + = (b 1 w 1 + b 2 w 2 + ) W 1 W 2 = {0} a i = b j = 0 β 1 β 2 is linearly independent Let v = u + w W 1 + W 2 Since β 1, β 2 spans W 1, W 2, respectively 47 PNU-MATH

51 1.6. a i, b j F s.t. v = a 1 u 1 + a 2 u b 1 w 1 + b 2 w 2 + β 1 β 2 spans V (ii) If 0 u β 1 β 2, then W 1 W 2 {0} β 1 β 2 = ( ) (i) Since β 1 β 2 spans V, v span(β 1 β 2 ) v = a 1 u 1 + +a 2 u b 1 w 1 + b 2 w 2 + W 1 + W 2 V = W 1 + W 2 (ii) Since β 1 β 2 =, W 1 W 2 = {0} V = W 1 W (a) Let β 1 = {u 1, u 2,, u k } be a basis for W 1 and we can extend it to a basis for V, say β Let β = {u 1, u 2,, u k, u k+1,, u n } and β 2 = {u k+1, u k+2,, u n } By the exercise 33, V = W 1 W 2 (b) V = W 1 W 2, W 1 = {(a, 0) a R}, W 2 = {(0, b) b R} V = W 1 W 2, W 1 = {(a, 0) a R}, W 2 = {( a, b) a, b R} 35. (a) β = {u k+1 + W, u k+2 + W,, u n + W } is a basis for V/W. (p.23) 48 PNU-MATH

52 1.6. (i) α V/W = {v + W v V } α = (a 1 u 1 + a 2 u a n u n ) + W = (a 1 u 1 + W ) + (a 2 u 2 + W ) + + (a n u n + W ) = a 1 (u 1 + W ) + + a k (u k + W ) + a k+1 (u k+1 + W ) + + a n (u n + W ) = a k+1 (u k+1 + W ) + a k+2 (u k+2 + W ) + + a n (u n + W ) β spans V/W (ii) Suppose a k+1 (u k+1 + W ) + a k+2 (u k+2 + W ) + + a n (u n + W ) = W, a i F, i = k + 1,, n then a k+1 u k+1 + a k+2 u k a n u n = 0 Since {u k+1,, u n } is linearly independent, a k+1 = = a n = 0 β is a basis for V/W. (b) dim(v ) = k + n, dim(w ) = k, dim(v/w ) = n k dim(/w V ) = dim(v ) dim(w ) 49 PNU-MATH

53 Maximal Linearly Independent Subsets 1. Label the following statement as true or false. (a) Every family of sets contains a maximal element. (F) ( ) Let F be the family of all finite subsets of an infinite set S. then F has no maximal element. (b) Every chain contains a maximal element. (F) ( ) If A : a partial ordered set and every chain ( ) of A has an upper bound, then A has a maximal element. (ex) Z, Q, R (c) If a family of sets has a maximal element, then that maximal element is unique. (F) ( ) Let S = {a = x 3 2x 2 5x 3, b = 3x 3 5x 2 4x 9, c = 2x 3 2x 2 +12x 6} then {a, b}, {a, c}, {b, c} are maximal linearly independent subsets of S So maximal element need not be unique. (d) If a chain of sets has a maximal element, then that maximal element is unique. (T) (e) A basis for a vector space is a maximal linearly independent subset of that vector space. (T) (f) A maximal linearly independent subset of a vector space is a basis for that vector space. (T) 50 PNU-MATH

54 Show that the set of convergent sequences is an infinite-dimensional subspace of the vector space of all sequences of real numbers. ( See Exercise 21 in Section 1.3.) (i) By the exercise 21 of section 1.3, W is a suspace of V (ii) {1, 1, } is a basis for W 3. Let V be the set of real numbers regarded as a vector space over the field of rational numbers. Prove that V is infinite-dimensional. Let V be a finite-dimensional vector space β = {v 1,, v n } Since π V, π = α 1 v α n v n, α i Q, i = 1,, n But π is a transcendental number in Q It s a contrdiction Thus V is infinite dimensional 4. Let W be a subspace of a (not necessarily finite-dimensional) vector space V. Prove that any basis for W is a subset of a basis for V. ( ) Let β W be a basis for W, then β W V (i) By theorem 1.13, β : a maximal linearly independent subset of V that contains β W. (ii) We are going to show that β is a basis for V which contains β W. Since β is linearly independent, so it suffices to show that β spans V. 51 PNU-MATH

55 1.7. If v V and v is not contained in β (i.e. v is not in span(β) ), then β {v} is linearly independent. This contradicts to the maximality of β. v V, v span(β) V span(β) = V. β is a basis for V which contains β W. 5. ( ) ( ) Let β = {v 1, v 2, } is a basis for V. If v V, then v spanv Thus v is a linear combination of the vectors of β (that is v can be expressed by some finite vectors of β and scalars in F ) Suppose that v = a 1 v 1 + a 2 v a n v n and v = b 1 v 1 + b 2 v b n v n are two such representation of v. Then 0 = (a 1 b 1 )v 1 + (a 2 b 2 )v (a n b n )u n Since β is linearly independent, it follows that a i = b i, where 1 i n ( ) Suppose each v V can be uniquely expressed as a linear combination of vectors of β, then clearly β spans V. If 0 = a 1 u 1 + a 2 u a n u n And we also have 0 = 0 u u u n By hypothesis, the representation of zero as a linear combination of the u i is unique. Hence each a i = 0, and the u i are linearly independent. 52 PNU-MATH

56 (i) Since S 1 F, F = Let C be a chain in F and U = {A A C} Clearly S 1 U S 2 Let a 1 u a n u n = 0 s.t. u i U, a i F, i = 1,, n Since u i A, A i C s.t. u i A i Since C is a chain, A k a.t. A i A k Thus u i A k F a 1 = = a n = 0 So U is n upper bound of C By the maximal principle, F has a maximal element β i.e. β is a maximal linearly independent subset of V (ii) v S 2, if v β, then v span(β) If v / β, β {v} is linearly independent This contradicts to the maximality of β S 2 span(β) Then V = span(s 2 ) span(β) V span(β) = V Therefore β is a basis for V s.t. S 1 β S 2 53 PNU-MATH

57 Let S = {A β A S =, A S is linearly independent } S ( A = ) Let C be a nonempty chain in S and B = {A A C} If x B S, then x B and x S Since A B, x A for some A C thus x A S = It s a contradiction, so B S = If a 1 u a m u m + b 1 v b n v n = 0, a i, b j F, u i B, v j S Since B = {A A C}, A 1,, A m C s.t. u i A i So we may assume that u 1,, u m A m then u 1,, u m, v 1,, u n A m S is linearly independent a i = 0, b j = 0 B S is linearly independent i.e. B is an upper bound of C By the maximal principle, S has a maximal element, say S 1 Clearly S 1 β, S 1 S =, S 1 S 2 is linearly independent So we only need to show that either { S 1 S is a maximal linearly independent subset of V or V } or {β span(s 1 S)} v β, If v S 1 S, then v span(s 1 S) If v / S 1 S, then S 1 S {v} is linearly independent It s a contradiction, β span(s 1 S) Therefore S 1 S is a basis for V 54 PNU-MATH

58 Linear Transformations and Matrices 2.1. Linear Transformations, Null Spaces, and Ranges 1. (a) T (b) F ( ) If x, y V and c F, T(x + y) =T(x)+T(y) and T(cx) = ct(x), then T is a linear transformation (c) F ( ) T is linear and one-to-one if and only if (d) T ( ) T(0 V ) =T(0 V + 0 V ) =T(0 V )+T(0 V ) T(0 V ) = 0 W (e) F ( ) p.70 Theorem 2.3 (f) F ( ) T is linear and one to one, then (g) T ( ) p.73 Corollary to Theorem 2.6 (h) F ( ) p.72 Theorem (1) T ((a 1, a 2, a 3 ) + (b 1, b 2, b 3 )) = T (a 1 + b 1, a 2 + b 2, a 3 + b 3 ) = ((a 1 + b 1 ) (a 2 + b 2 ), 2(a 3 + b 3 )) = (a 1 a 2 ), 2a 3 ) + (b 1 b 2 ), 2b 3 ) = T (a 1, a 2, a 3 ) + T (b 1, b 2, b 3 ) T (c(a 1, a 2, a 3 )) = T (ca 1, ca 2, ca 3 ) = (ca 1 ca 2, 2ca 3 ) = c(a 1 a 2, 2a 3 ) = ct (a 1, a 2, a 3 ) Thus, T is linear. (2) N(T)=span{(1, 1, 0)} R(T)=span{(1, 0), (0, 1)} 55 PNU-MATH

59 2.1. (3) Dim(V)=nullity(T) + rank(t)= (4) T is not one-to-one ( N(T) {0} = {(0, 0, 0)}) T is onto ( 2=rank(T)=dim(W)=2) 3. (1) T is linear. (2) N(T)=span{(0, 0)} R(T)=span{(1, 0, 0), (0, 0, 1)} (3) Dim(V)=nullity(T) + rank(t)= (4) T is one-to-one ( N(T)= {0}) T is not onto ( 2=rank(T) dim(w)=3) 4. (1) T is linear. ( ) ( ) (2) N(T)=span,, ( 0 ) 0 ( 0 0 ) ( ) R(T)=span,,, (3) Dim(V)=nullity(T) + rank(t)= (4) T is not one-to-one ( N(T) {0}) T is not onto ( 2=rank(T) dim(w)=4) ( 0 0 ) (1) T is linear. 56 PNU-MATH

60 2.1. (2) N(T)=span{0} R(T)=span{1, x, x 2 } (3) Dim(V)=nullity(T) + rank(t)= (4) T is one-to-one ( N(T)= {0}) T is not onto ( rank(t) dim(w)) 6. (1) T is linear. (2) N(T)={A T (A) = tr(a) = 0}=span{E ij, E ij}, where E ij denote the matrix whose ij-entry is 1 and zero elsewhere and E ij denote the matrix of a 11 component is 1, a jj component is 1 and others are all zero. (2 j n) (*) β = {E ij i j} {E 11 E ii 2 j n} R(T)={T (A) A Mat n n }=span{1} (3) n 2 =Dim(V)=nullity(T) + rank(t)= (n 2 1) + 1 * dim N(T)=n 2 1 (p.56 exercise 15, sec 1.6) (4) T is not one-to-one but onto ( rank(t)= dim(w)) 7. (1) If T is linear, then T (0) = 0 T (0) = T (0 + 0) = T (0) + T (0) T (0) = 0 (2) 57 PNU-MATH

61 2.1. If T is linear, T (cx + y) = T (cx) + T (y) = ct (x) + T (y) Suppose T (cx + y) = ct (x) + T (y), then c = 1, T (x + y) = T (x) + T (y) j = 0, T (cx) = ct (x) (3) 0 = T (0) = T (x x) = T (x) + T ( x) T ( x) = T (x) x So T (x y) = T (x) + T ( y) = T (x) T (y) (4) ( ) If T is linear, then T ( n a i x i ) = n T (a i x i ) = n a i T (x i ) ( ) clear i=1 i=1 i=1 8. (a) (i) T θ ((a 1, a 2 ) + (b 1, b 2 )) = T θ (a 1 + b 1, a 2 + b 2 ) = ((a 1 + b 1 ) cos θ (a 2 + b 2 ) sin θ, (a 1 + b 1 ) sin θ + (a 2 + b 2 ) cos θ) = (a 1 cos θ a 2 sin θ, a 1 sin θ + a 2 cos θ) + (b 1 cos θ b 2 sin θ, b 1 sin θ + b 2 cos θ) = T θ (a 1, a 2 ) + T θ (b 1, b 2 ) (ii) T θ c((a 1, a 2 )) = T θ (ca 1, ca 2 ) = (ca 1 cos θ ca 2 sin θ, ca 1 sin θ + ca 2 cos θ)) = c(a 1 cos θ a 2 sin θ, a 1 sin θ + a 2 cos θ) = ct θ (a 1, a 2 ) (b) (i) T ((a 1, a 2 ) + (b 1, b 2 )) = T (a 1 + b 1, a 2 + b 2 ) = (a 1 + b 1, a 2 b 2 ) = (a 1 a 2 ) + (b 1 b 2 ) = T (a 1 + a 2 ) + T (b 1 + b 2 ) (ii) T (c(a 1, a 2 )) = (ca 1, ca 2 ) = c(a 1, a 2 ) = ct (a 1, a 2 ) 58 PNU-MATH

62 (a) T ((a 1 + a 2 ) + (b 1 + b 2 )) = (1, a 2 + b 2 ) T (a 1 + a 2 ) + T (b 1, b 2 ) = (2, a 2 + b 2 ) (b) T (c(a 1, a 2 )) = (ca 1, c 2 a 2 2) ct (a 1, a 2 ) = (ca 1, ca 2 1) (c) T ((a 1 + a 2 ) + (b 1 + b 2 )) = (sin(a 1 + b 1 ), 0) T (a 1 + a 2 ) + T (b 1, b 2 ) = (sin a 1 + sin b 1, 0) (d) T ((a 1 + a 2 ) + (b 1 + b 2 )) = ( a 1 + b 1, a 2 + b 2 ) T (a 1 + a 2 ) + T (b 1, b 2 ) = ( a 1 + b 1, a 2 + b 2 ) (e) T ((a 1 + a 2 ) + (b 1 + b 2 )) = (a 1 + b 1 + 1, a 2 + b 2 ) T (a 1 + a 2 ) + T (b 1, b 2 ) = (a 1 + b 1 + 2, a 2 + b 2 ) 10. (a) T (2, 3) = 1T (1, 0) + 3T (1, 1) = (5, 11) (b) T is one-to-one T (x, y) = xt (1, 0)+yT (0, 1) = xt (1, 0)+y(T (1, 1) T (0, 1)) = x(1, 4)+y((2, 5) (1, 4)) == (x + y, 4x + y) = (0, 0) (x, y) = (0, 0) 59 PNU-MATH

63 T (8, 11) = 2T (1, 1) + 3T (2, 3) = (5, 3, 16) 12. No (T (2, 0, 6) 2T (1, 0, 3)) 13. S={v 1, v 2,, v k } s.t T (v i ) = w i, i = 1, 2,, k Suppose k a i v i = 0 for some scalars a i F, i=1 then we need to show that all a i = 0 T ( k a i v i ) = k k a i T (v i ) = a i w i = T (0) = 0 i=1 i=1 i=1 Since w i s are linearly independent, so a i = 0 S is linearly independent. 14. (a) ( ) {v 1, v 2,, v n } is linearly independent subset of V If a i T (v i ) = 0, then T ( a i v i ) = 0 and a i v i N(T ) = {0} a i = 0 {T (v 1 ),, T (v n )} is linearly independent subset of W ( ) Let w 1 = a i T (v i ), a i, b i F If w 1 = w 2, then (a i b i )T (v i ) = 0 Since T (v i ) s are linearly independent, a i = b i 60 PNU-MATH

64 2.1. a i v i = b i v i T is one-to-one (b) ( ) By (a), T (S) is linearly independent ( ) Let T (S) = {T (v 1 ),, T (v n )} linearly independent If a i v i = 0, then a i T (v i ) = T (0) = 0 a i = 0 S = {v 1,, v n } is linearly independent (c) By (b), T (β) is linearly independent By the theorem 2.2, R(T ) = span(t (β)) Since T is onto, span(t (β)) = W T (β) is a basis for W 15. (a) (i) T (f(x)+g(x)) = x (f(t)+g(t))dt = x f(t)dt+ x 0 0 (ii) T (cf(x)) = x cf(t)dt = c x f(t)dt = ct (f(x)) 0 0 (b) If T (f(x)) = T (g(x)), then x 0 f(t)dt = x 0 g(t)dt i.e. x (f(t) g(t))dt = 0 0 f(x) = g(x) (c) {x, x 2, } is a basis for R(T ) Since spanr(t ) P (R), T is not onto 0 g(t)dt = T (f(x))+t (g(x)) 61 PNU-MATH

65 (a) Let f(x) = ax + b, g(x) = ax + d, b d (i.e. f(x) g(x)) But T (f(x)) = T (g(x)) (b) β = {1, x, x 2, } is a standard basis for P (R) It suffices to show that β R(T ) = span(t (β)) x n β, x n = T ( 1 n+1 xn ) R(T ) 17. (a) By the theorem 2.3, dim R(T ) dim V By the assumption, dim V < W dim R(T ) < W T can t be onto (b) If T is one-to-one i.e. nullity(t ) = 0, then dim V = rank(t ) > dim W It s a contradiction to dim R(T ) dim W 18. Let (a, b) R 2, T (a, b) = (0, a) Then N(T ) = {(0, b) b R}, R(T ) = {(0, a) a R} N(T ) = R(T ) 19. (a, b) R 2, Let T (a, b) = (0, a) and U(a, b) = (0, 2a) Then T U, N(T ) = {(0, b) b R}, N(U) = {(0, b) b R} R(T ) = {(0, a) a R}, R(U) = {(0, 2a) a R} 62 PNU-MATH

66 (a) Let w 1, w 2 T (v 1 ), a F then v 1, v 2 V 1 s.t. w i = T (v i ), i = 1, 2 So aw 1 + w 2 = at (v 1 ) + T (v 2 ) = T (av 1 + v 2 ) T (v 1 ) T (v 1 ) is a subspace of W (b) Let K = {x V T (x) W 1 }, a F Let x 1, x 2 ık s.t. T (x 1 ) = w 1, T (x 2 ) = w 2 W 1 T (ax 1 + x 2 ) = at (x 1 ) + T (x 2 ) = aw 1 + w 2 W 1 ax 1 + x 2 K, a F {x V T (x) W 1 } is a subspace of V 21. (a) (i) T (c(a 1, a 2, )+(b 1, b 2 )) = T (ca 1 +b 1, ca 2 +b 2, ) = (ca 2 +b 2, ca 3 +b 3, ) = c(a 2, a 3 ) + (b 2, b 3, ) = ct (a 1, a 2, ) + T (b 1, b 2, ) (ii) U(c(a 1, a 2, ) + (b 1, b 2 )) = U(ca 1 + b 1, ca 2 + b 2, ) = (0, ca 1 + b 1, ca 2 + b 2, ) = (0, ca 1, ca 2, ) + (0, b 1, b 2, ) = cu(a 1, a 2, ) + U(b 1, b 2, ) (b) (i) T is not one-to-one 0 (1, 0, 0, ) N(T ) (ii) T is onto R(T ) = {(a 2, a 3, ) a i F } = span{e 1, e 2, } = V 63 PNU-MATH

67 2.1. (c) (i) If U(a 1, a 2, ) = (0, a 1, a 2, ) = (0, 0, ), then a i = 0 N(U) = {0} (ii) U is not onto R(U) = {(0, a 1, a 2, ) a i F } = V 22. (i) T : R 3 R is linear Let β = {e 1 = (1, 0, 0), e 2 = (0, 1, 0), e 3 = (0, 0, 1)} a standard basis for R 3 Put T (e 1 ) = a, T (e 2 ) = b, T (e 3 ) = c then (x, y, z) R 3, T (x, y, z) = T (xe 1 + ye 2 + ze 3 ) = xt (e 1 ) + yt (e 2 ) + zt (e 3 ) = ax + bycz (ii) T : F n F is linear Let β = {e 1, e 2,, e n } a standard basis for F n Put T (e 1 ) = a 1, T (e 2 ) = a 2,, T (e n ) = a n then T (x 1, x 2,, x n ) = a 1 x a n x n (iii) T : F n F m For j (i j m), T (e j ) = m a ij w i i=1 T (x 1, x 2,, x n ) = T ( n x j e j ) = n x j T (e j ) = n x j ( m a ij w i ) = n m a ij x j w i = j=1 m ( n )a ij x j w i = ( n a 1j x j, i=1 j=1 j=1 j=1 n a 2j x j,, j=1 j=1 n a mj x j ) j=1 i=1 j=1 i=1 64 PNU-MATH

68 T : R 3 R linear transformation By the exercise 22, a, b, c R s.t. T (x, y, z) = ax + by + cz N(T ) = {(x, y, z) R 3 T (x, y, z) = 0} = {(x, y, z) R 3 ax + by + cz)0} = R 3 (where a = b = c = 0) a plane through 0 (where a 2 + b 2 + c 2 0) 24. (a) The projection on the y-axis along the x-axis ( ) 0 0 T (a, b) = (0, b), (a, b) R 2, A = 0 1 (b) T (a, b) = (a b, a b), (a, b) R 2 (a, b) (0, 0) if a = b (a b, a b) otherwise 25. (a) Let W 1 = {(a, b, 0) a, b R}, W 2 = {(0, 0, c) c R} Then R 3 = W 1 W 2 x = (a, b, c) R 3 s.t. x 1 = (a, b, 0) W 1 and x 2 = (0, 0, c) W 2 T (x) = x T is the projection on the W 1 along the W 2, A = PNU-MATH

69 2.1. (b) T (a, b, c) = (0, 0, c) (c) Let W 1 = {(a, b, 0) a, b R}, W 2 = L = {(a, 0, a) a R} then R 3 = W 1 + W 2 For all x = (a, b, c) R 3 s.t. x = x 1 + x 2, x 1 = (a c, b, 0) W 1, x 2 = (c, 0, c) W 2 T (a, b, c) = (a c, b, 0) W T : R 3 R 3 is the projection on W 1 along the W 2, A = x V s.t x = x 1 + x 2, x 1 W 1, x 2 W 2 and T(x) = x 1 (1) T is linear x, y V s.t. x = x 1 + x 2, y = y 1 + y 2, x 1, y 1 W 1, x 2, y 2 W 2 T (x + y) = T ((x 1 + y 1 ) + (x 2 + y 2 )) = x 1 + y 1 =T(x)+T(y) T (cx) = T (cx 1 + cx 2 ) = cx 1 = T (x) (2) W 1 = {x V T(x) = x} ( ) If x 1 W, then T (x 1 ) = x 1 x 1 {x V T (x) = x} ( ) If x V s.t T(x) = x and we have T (x) = x 1 ( ) x = x 1 W 1 (b) (1)W 1 = R(T ) 66 PNU-MATH

70 2.1. ( ) T (x) R(T ), T (x) = x 1 W 1 R(T ) W 1 ( ) x 1 W 1, x 1 = T (x 1 ) R(T ) W 1 R(T ) (2)W 2 = N(T ) x 2 W 2, T (x 2 ) = 0 x 2 N(T ) W 2 N(T ) ( ) x N(T ), T (x) = x 1 = 0 x = x 1 + x 2 = 0 + x 2 = x 2 W 2 ( ) N(T ) W 2 (c) Describe if W 1 = V T (x) = x, x W 1 = V T is the identity transformation(i V ) (d) Describe if W 1 = 0 T (x) = x x = 0 ( or (0) = W 1 = R(T ) ) T is the zero transformation(t 0 ) 27. Claim : W V, W V as subspace, T = W +W and T : V V, T (V ) = W (a) From the exercise 34 in Section 1.6(p.58), If W is any subspace of a finite dimensional vector space, W s.t V = W W so x V,!x 1 W x 2 W s.tx = x 1 + x 2 Define T : V V T (x) = x 1 is a desired linear transformation Then T is a projection on W along W 67 PNU-MATH

71 2.1. * Say dim V = n 2 {v 1, v 2,, v m } : a basis for W {v 1, v 2,, v m, v m+1,, v n } : a basis for V Let W = span{v m+1,, v n } and W = span{v m+1 v 1, v m+2 v 3,, v n v 5 } then V = W W = W W Clearly W W ( ) If W = W, then v m+1, v m+1 v 1 W v 1 W W = (0) It s a contradiction (b) Example (example 1) the projection on W along W 1 {(a, b)) = (0, b 1a) + (a, 1a)} 3 3 (example 2) the projection on W along W 2 (a, b) = (0, b) + (a, 0) 28. (1) {0} is T -invariant x {0}, T (x) = 0 {0} ( T is linear) (2) V is T -invariant (T(V) V) x V, T (x) V ( T : V V ) (3) R(T ) is T -invariant (T(T(V)) T(V)) 68 PNU-MATH