Vector Space5.1. Real Vector Spaces

5.2. Subspaces5.3. Linear Independence

5.4. Basis and Dimension5.5. Row Space, Column Space, Nullspace

5.6. Rank and Nullity

S = {v1, v2, v3, ., vr } is a non-empty set of vectors

Linear Independence:If the only solution of k1 v1+ k2 v2 + k3 v3. + kr vr = 0 is trivial (k1, k2, k3, ., kr = 0)

Linear Dependence:If the solution of k1 v1+ k2 v2 + k3 v3. + kr vr = 0 is non trivial

Diketahui : himpunan S = {v1, v2, v3, ., vr } Ditanyakan: apakah S linearly independent atau linearly dependent?

Jawab:1. Bentuk SPL Homogen 2. Tentukan solusinya 3. Jika solusinya trivial k1, k2, k3, ., kr = 0 maka S linearly independent k1 v1+ k2 v2 + k3 v3. + kr vr = 0

4. Jika solusinya non-trivial maka S linearly dependent

Example 4: determine whether u = (1, 2, 3), v = (5, 6, 1), w = (3, 2, 1) form a linear independent set. Solution : k1 (1, 2, 3) + k2 (5, 6, 1) + k3 (3, 2, 1) = (0, 0, 0)k1 + 5 k2 + 3 k3 = 0 2k1 + 6 k2 + 2 k3 = 0 3k1 k2 + k3 = 0 1 2 3 5 6 1 3 2 1 0 0 0

1 0 0

5 16 16

3 8 8

0 0 0

1 0 0

5 2 0

3 1 0

0 0 0

1 0 0

5 16 16

3 8 8

0 0 0

1 0 0

5 2 0

3 1 0

0 0 0

1 0 0 k1 = 0.5 k3

0 1 0

0.5 0.5 0

0 0 0

solusinya = { ( 0.5t, 0.5t, t ) }

k2 = 0.5 k3k3 = t

NON-TRIVIALmaka u, v, w tidak independen-linier ( linearly dependent )

Example 1: (page 241) If u = (2, 1, 0, 3), v = (1, 2, 5, 1), w = (7, 1, 5, 8), then the set S = {u, v, w} is linearly dependent, since 3u + v w = 0

linearly dependent (artinya: saling bergantung linier) u = (w v) / 3 v = w 3u w = 3u + v u depends linearly on v, w v depends linearly on u, w w depends linearly on u, v

Theorem 5.3.1 Theorem 5.3.2

Theorem 5.3.3:Let S = {v1, v2, v3, ., vr } be a set of vectors in Rn. If r n then S is linearly dependent The solution set of the following homogeneous system of linear equations is non trivial (Theorem 1.2.1) k1 v1+ k2 v2 + k3 v3. + kr vr = 0k1 v11 + k2 v21 + k3 v31 + .. + kr vr1 = 0 v12 v22 v32 vr2 v13 v23 v33 vr3 v1n v2n v3n vrnmore columns than rows

0 0 .. 0

Vector Space5.1. Real Vector Spaces

5.2. Subspaces5.3. Linear Independence

5.4. Basis and Dimension5.5. Row Space, Column Space, Nullspace

5.6. Rank and Nullity

Basis:V is any Vector Space S = { v1, v2, v3, , vn } where v1, v2, v3, , vn V then S is called a Basis for V if 1. S is linearly independent 2. S spans VA vector space can have more than one basisSee Figure 5.4.3

Cara menentukan apakah S basis untuk Ruang Vektor V:V adalah Ruang Vektor S = { v1, v2, v3, , vn } di mana v1, v2, v3, , vn V maka S disebut Basis dari V jika

1.

S linearly independentk1v1 + k2v2 + + knvn = 0 menghasilkan k1 = k2 = = kn = 0 (skalar)

2.

S merupakan rentang (span) dari Vuntuk sembarang vektor u V dipenuhi c1v1 + c2v2 + + cnvn = u artinya : ada solusi untuk c1, c2 , , cn (skalar)

DefinitionA non zero vector space V is called finite-dimensional if it contains a finite set of vectors { v1, v2, v3, ., vn } that forms a basis. If no such set exists, V is called infinite-dimensional. The dimension of the zero vector space { 0 = (0, 0, , 0) } is zero.

Contoh: (Example 1 page 253 & Example 3 page 254) Dalam contoh ini ditunjukkan dua basis untuk R3 B = {e1, e2, e3} dan S = {v1, v2, v3}

di mana e1 = (1, 0, 0); e2 = (0, 1, 0); e3 = (0, 0, 1) v1 = (1, 2, 1); v2 = (2, 9, 0); v3 = (3, 3, 4)

Cara menentukan apakah B / S basis untuk Ruang Vektor V:B disebut Basis dari V jika

1.

B linearly independent SPL Homogenk1v1 + k2v2 + + knvn = 0

menghasilkan k1 = k2 = = kn = 0 (skalar)

2.

B merupakan rentang (span) dari Vuntuk sembarang vektor u V dipenuhi c1v1 + c2v2 + + cnvn = u artinya : ada solusi untuk c1, c2 , , cn (skalar)

Contoh: (Example 1 page 253 & Example 3 page 254) Dalam contoh ini ditunjukkan dua basis untuk R3 B = {e1, e2, e3} dan S = {v1, v2, v3}

di mana e1 = (1, 0, 0); e2 = (0, 1, 0); e3 = (0, 0, 1) v1 = (1, 2, 1); v2 = (2, 9, 0); v3 = (3, 3, 4) Bukti bahwa B adalah basis untuk R3. B disebut basis standar untuk R3. B independen linier? B merentang R3?

k1e1 + k2e2 + k3e3 = 0

u = (x, y, z) R3

1 0 00 1 0 0 0 1

k1k2 k3

=

00 0

c1e1 + c2e2 + c3e3 = u1 0 0 0 1 0 c1 c2 = x y

B independen linier

0 0 1

c3

z

B merentang R3

Bukti bahwa S = {v1, v2, v3} juga basis R3 S = {v1, v2, v3} di mana v1 = (1, 2, 1); v2 = (2, 9, 0); v3 = (3, 3, 4)S independen linear?

k1v1 + k2v2 + k 3v3 = 0 1 2 3 2 9 3 1 0 4B2 2*B1 B3 B1

k1 k2 k3

=

0 0 0

1 2 12 5 -2 3 -3 1 0 0 0

2 9 0

3 3 4

0 0 0

1 0 0

Bukti bahwa S = {v1, v2, v3} juga basis R3 S = {v1, v2, v3} di mana v1 = (1, 2, 1); v2 = (2, 9, 0); v3 = (3, 3, 4)1 0 0 2 5 -2 3 -3 1 0 0 0

B3 + 2/5*B2

1 0 0

2 5 0

3 -3 -1/5

0 0 0

Solusi trivial: (0, 0, 0) S independen linier

Bukti bahwa S = {v1, v2, v3} juga basis R3S = {v1, v2, v3} di mana v1 = (1, 2, 1); v2 = (2, 9, 0); v3 = (3, 3, 4)u = (x, y, z) R3S merentang R3 ?

k1v1 + k2v2 + k 3v3 = u 1 2 3 k1 = x 1 2 3 x

2 9 31 0 4

k2k3

yz

21

90

34

yz

1 2 1B2 2*B1 B3 B1

2 9 01 0

3 3 42 5

x y z3 -3 x y 2x

0B3 + 2/5*B2 1 0 0

-22 5 0

13 -3 -1/5

zxx y 2x z + 2/5y 9/5x

1 0 0

2 5 0

3 -3 -1/5

x y 2x z + 2/5y 9/5x

k3 = 9x 2y 5z k2 = 1/5 (y 2x + 3k3) = 1/5 (25x 5y 15z) = 5x y 3z k1 = x 2k2 3k3 = x 10x + 2y 6z 27x + 6y + 15z = 36x + 8y + 9z S merentang R3. Maka S adalah (salah satu basis) R3

Dari contoh ini terlihat bahwa dimensi R3 = 3 karena B / S berisi 3 vektor B = { e1, e2, e3 } dan S = { v1, v2, v3 }

di mana e1 = (1, 0, 0); e2 = (0, 1, 0); e3 = (0, 0, 1) v1 = (1, 2, 1); v2 = (2, 9, 0); v3 = (3, 3, 4)

Theorem 5.4.3. All bases for a finite-dimensional vector space have the same number of vectors. Theorem 5.4.5. If V is a finite-dimensional vector space, and if S is a set in V with exactly n vectors, then S is a basis for V if either S spans V or S is linearly independent.

B = {e1, e2, e3}

dan

S = {v1, v2, v3}

di mana e1 = (1, 0, 0); e2 = (0, 1, 0); e3 = (0, 0, 1) v1 = (1, 2, 1); v2 = (2, 9, 0); v3 = (3, 3, 4)

Koordinat sebuah vektor akan berbeda jika dinyatakan berdasarkan basis yang berbeda(lihat Example 4 halaman 255)

(5, 1, 9)B ekivalen (1, 1, 2)S(11, 31, 7)B ekivalen (1, 3, 2)S(1, 1, 2)S = (1)*(1, 2, 1) + (1)*(2, 9, 0) + (2)*(3, 3, 4) = (1, 2, 1) + (2, 9, 0) + (6, 6, 8) = (5, 1, 9)

Contoh : Tentukan basis untuk ruang-solusi dari SPL berikut : x1 + 2x2 + 7x3 9x4 + 31x5 = 0 2x1 + 4x2 + 7x3 11x4 + 34x5 = 0 3x1 + 6x2 + 5x3 11x4 + 29x5 = 0 Penyelesaian : Jika ditulis dalam bentuk matriks augmented:12 3 1 0 0

24 6 2 0 0

77 5 7 -7 -16

-9-11 -11 -9 7 16

3134 29 31 -28 -64

00 0 0 0 0

B2 2*B1 B3 3*B1

B2 * (1/7) B3 * (1/16)

1 0 0

2 0 0

7 1 1

-9 -1 -1

31 4 4

0 0 0

B3 B2 B1 7*B2

1 0 0

2 0 0

0 1 0

-2 -1 0

3 4 0

0 0 0 2r + 2s 3t r = s 4t s t

x1 = 2x2 + 2x4 3x5 x2 bebas = r x3 = x4 4x5 x4 bebas = s x5 bebas = t

x1 x2 x3 x4 x5

x1 x2 x3

=

2r + 2s 3t r s 4t

=

2r + 2s 3t 1r + 0s + 0t 0r + 1s 4t

x4x5 = r 2 1 0 0 0

st

0r + 1s + 0t0r + 0s + 1t 3 0 4 0 1

+ s

2 0 1 1 0

+ t

v1

v2

v3

Basis Ruang Solusi = B = {v1, v2, v3 } dengan dimensi 3

Contoh : Tentukan basis untuk ruang-solusi dari SPL berikut : x1 + 2x2 + 7x3 9x4 + 31x5 = 0 2x1 + 4x2 + 7x3 11x4 + 34x5 = 0 3x1 + 6x2 + 5x3 11x4 + 29x5 = 0 Ruang Solusinya S = { ( 2r + 2s 3t, r, s 4t, s, t ) } = subset dari R5

Basis dari Ruang Solusi S = B = {v1, v2, v3 } Dimensi dari Ruang Solusi S = 3

P.R. u/ 2-12-2011 5.3. no. 1, 5

5.4. no. 10, 11, 13

Tes tgl. 3-12-2011 Sabtu, pk.8.00-9.00 Sifat : tertutup Materi: bab 5.1 s/d. 5.4