Basis Ortonormal - Proses Gram Schmidt

Basis Ortonormal - Proses Gram Schmidt (blogaritma.net)

Definisi:

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$

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

✔ 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

$v_{2}=\frac{u_2-proy_{w_1^{u_2}}}{||u_2-proy_{w_1^{u_2}}||}$   : $w_1$ 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

Baca Juga : Ortogonal dan Pembahasan Soal 
                     Vektor Eigen, Nilai Eigen dari Sebuah Matriks