Sebuah himpunan vektor pada ruang hasil kali dalam dinamakan himpunan ortogonal jika semua pasangaan vektor-vektor yang berbeda dalam himpunan tersebut ortogonal. sebuah himpunan ortogonal yang setiap vektornya mempunyai norma 1 dinamakan ortonormal.
v_{1}=\frac {u_1}{||u_1||}
v_{2}=\frac{u_2-proy_{w_1^{u_2}}}{||u_2-proy_{w_1^{u_2}}||} ; proy_{w_1^{v_2}} = < u_2, v_1 > v_1
v_{3}=\frac{u_3-proy_{w_2^{u_3}}}{||u_3-proy_{w_2^{u_3}}||} ; proy_{w_2^{v_3}} = < u_3, v_1 > v_1+< u_3, v_2 > v_2
v_{3}=\frac{u_3-proy_{w_2^{u_3}}}{||u_3-proy_{w_2^{u_3}}||} ; proy_{w_2^{v_3}} = < u_3, v_1 > v_1+< u_3, v_2 > v_2
Soal:
1. Diketahui S = { (2,1),(1,1)} adalah sebuah basis di R , Ubahlah basis tersebut menjadi basis ortonormal dengan menggunakan langkah-langkah proses Gram-Schmidt. Untuk perhitungannya menggunakan hasil kali dalam berikut:
< (x₁,y₁),(x₂,y₂) > = 2x₁y₁ + 3x₂y₂, ∀(x₁,y₁),(x₂,y₂)⋴R²
Penyelesaian:
S = { (2,1),(1,-1)}
Misal
u₁ = (2,1) S' = (v₁,v₂)
u₂ = (1,-1) Basis Ortonormal
u₁ = (2,1) S' = (v₁,v₂)
u₂ = (1,-1) Basis Ortonormal
✔ Langkah 1
v_1=\frac {u_1}{||u_1||} ||u_1||=<u_1,u_1>^{1/2}
=\frac{(2,1)}{\sqrt{11}} =<(2,1),(2,1)>^{1/2}
=(2.2.2 + 3.1.1)^{1/2}
=\sqrt{11}
\therefore \frac {u_1}{||u_1||}=\left ( \frac{2}{\sqrt{11}},\frac{1}{\sqrt{11}}\right )
✔ Langkah 2
v_1=\frac {u_1}{||u_1||} ||u_1||=<u_1,u_1>^{1/2}
=\frac{(2,1)}{\sqrt{11}} =<(2,1),(2,1)>^{1/2}
=(2.2.2 + 3.1.1)^{1/2}
=\sqrt{11}
\therefore \frac {u_1}{||u_1||}=\left ( \frac{2}{\sqrt{11}},\frac{1}{\sqrt{11}}\right )
✔ Langkah 2
v_{2}=\frac{u_2-proy_{w_1^{u_2}}}{||u_2-proy_{w_1^{u_2}}||} : w_1 ruang yg direntang oleh v_1
⚫ proy_{w_1^{u_2}} = < u_2, v_1 > v_1
=<(1,-1),\left ( \frac{2}{\sqrt{11}},\frac{1}{\sqrt{11}}\right )>\left ( \frac{2}{\sqrt{11}},\frac{1}{\sqrt{11}}\right )
=\left(2.1. \frac{2}{\sqrt{11}} + 3.(-1). \frac{1}{\sqrt{11}}\right).\left(\frac{2}{\sqrt{11}},\frac{1}{\sqrt{11}}\right)
=\left(\frac{4}{\sqrt{11}}-\frac{3}{\sqrt{11}}\right)\left(\frac{2}{\sqrt{11}},\frac{1}{\sqrt{11}}\right)
=\frac{1}{\sqrt{11}}\left(\frac{2}{\sqrt{11}},\frac{1}{\sqrt{11}}\right )
=\left(\frac{2}{11},\frac{1}{11}\right)
⚫u_2-proy_{w_1^{u_2}}=(1,-1)-\left(\frac{2}{11},\frac{1}{11}\right)
=\left(1-\frac{2}{11},-1-\frac{1}{11}\right)
=\left(\frac{9}{11},\frac{-12}{11}\right)
⚫ ||u_2-proy_{w_1^{u_2}}||=<\left(\frac{9}{11},\frac{-12}{11}\right)\left(\frac{9}{11},\frac{-12}{11}\right)>^{1/2}
=\left(2.\frac{9}{11}.\frac{9}{11}+3.\frac{-12}{11}.\frac{-12}{11}\right)
=\left(\frac{162}{121}+\frac{432}{121}\right)^{1/2}
=\sqrt{\frac{594}{121}}
=\sqrt{\frac{9.66}{121}}
=\frac{3.\sqrt{66}}{11}
Sehingga :
v_{2}=\frac{u_2-proy_{w_1^{u_2}}}{||u_2-proy_{w_1^{u_2}}||}
=\left(\frac{\frac{9}{11},\frac{-12}{11}}{\frac{3.\sqrt{66}}{11}}\right)
=\left(\frac{9}{11}.\frac{11}{3\sqrt{66}},\frac{-12}{11}.\frac{11}{3\sqrt{66}}\right)
=\left(\frac{3}{\sqrt{66}},\frac{-4}{\sqrt{66}}\right)
S'=\left \{ \left(\frac{2}{11},\frac{1}{11}\right),\left(\frac{3}{\sqrt{66}},\frac{-4}{\sqrt{66}}\right) \right \}
atau
S'=\left \{ \left(\frac{2\sqrt{11}}{11},\frac{\sqrt{11}}{11}\right),\left(\frac{\sqrt{66}}{22},\frac{-2\sqrt{66}}{33}\right) \right \}
Bagaimana kita bisa tahu bahwa jawaban kita benar?
1. Ortoghonal S' harus = 0
2. ||S'|| harus = 1
Pembuktian:
1. Ortoghonal S' = 0
S'=\left \{ \left(\frac{2\sqrt{11}}{11},\frac{\sqrt{11}}{11}\right),\left(\frac{\sqrt{66}}{22},\frac{-2\sqrt{66}}{33}\right) \right \}
=2.\left(\frac{2\sqrt{11}}{11}\right).\left(\frac{\sqrt{66}}{22}\right)+3.\left(\frac{\sqrt{11}}{11}\right).\left(\frac{-2\sqrt{66}}{33}\right)
=\frac{2.\sqrt{726}}{121}-\frac{2.\sqrt{726}}{121}=0 (terbukti)
2. ||S'|| = 1
< v₁,v₁ > = < \left(\frac{2\sqrt{11}}{11},\frac{\sqrt{11}}{11}\right), \left(\frac{2\sqrt{11}}{11},\frac{\sqrt{11}}{11}\right)>^{1/2}
=2.\left(\frac{2\sqrt{11}}{11}\right).\left(\frac{2\sqrt{11}}{11}\right)+3.\left(\frac{\sqrt{11}}{11}\right).\left(\frac{\sqrt{11}}{11}\right)
=\frac{2.44}{121}+\frac{33}{121}
=2.\frac{4}{11}+\frac{3}{11}
=\frac{8}{11}+\frac{3}{11}=\frac{11}{11}=1^{1/2} = 1 (terbukti)
< v₂,v₂ > = <\left(\frac{\sqrt{66}}{22},\frac{-2\sqrt{66}}{33}\right),\left(\frac{\sqrt{66}}{22},\frac{-2\sqrt{66}}{33}\right)>^{1/2}
=2.\left(\frac{\sqrt{66}}{22}\right).\left(\frac{\sqrt{66}}{22}\right)+3.\left(\frac{-2\sqrt{66}}{33}\right).\left(\frac{-2\sqrt{66}}{33}\right)
=\frac{66}{242}+\frac{264}{363}
=\frac{23.958+63.888}{87.846}=\frac{87846}{87846}=1^{1/2}=1 (terbukti)
∴ S' merupakan basis ortonormal
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