Apa itu SVD
Singular Value Decomposition (SVD) adalah suatu teknik dalam aljabar linier yang memfaktorkan sebuah matriks \(A\) berukuran \(m \times n\) menjadi tiga buah matriks:
\[
A = U \Sigma V^T
\]
\(U \in \mathbb{R}^{m \times m}\) adalah matriks ortogonal, berisi vektor-vektor eigen dari \(A A^T\).
\(\Sigma \in \mathbb{R}^{m \times n}\) adalah matriks diagonal, dengan elemen-elemen diagonalnya disebut sebagai \textbf{nilai singular}
\(V \in \mathbb{R}^{n \times n}\) adalah matriks ortogonal, berisi vektor-vektor eigen dari \(A^T A\).
Kegunaan SVD
Contoh Soal 1
Matriks Awal
\[\begin{split}
A = \begin{bmatrix}
3 & 1 \\
1 & 3
\end{bmatrix}
\end{split}\]
Mencari Transpose A
Transpose = baris jadi kolom:
\[\begin{split}
A^T = \begin{bmatrix}
3 & 1 \\
1 & 3
\end{bmatrix}
\end{split}\]
Karena \(A\) simetris, maka \(A^T = A\).
Hitung \(A^T A\)
\[\begin{split}A^T A = A A =
\begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix}
\begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix}
=\begin{bmatrix}
10 & 6 \\
6 & 10
\end{bmatrix}\end{split}\]
Cari Nilai Eigen \(A^T A\)
\[
\det(A^T A - \lambda I) = 0
\]
\[\begin{split}
\det \begin{bmatrix}
10 - \lambda & 6 \\
6 & 10 - \lambda
\end{bmatrix} = 0
\end{split}\]
\[
(10 - \lambda)^2 - 36 = 0
\]
\[
(10 - \lambda)^2 = 36
\]
\[
10 - \lambda = \pm 6
\]
\[
\lambda = 10 \mp 6
\]
\[
\Rightarrow \lambda_1 = 16, \quad \lambda_2 = 4
\]
Singular Value
\[
\sigma_1 = \sqrt{16} = 4, \quad \sigma_2 = \sqrt{4} = 2
\]
Cari Vektor Eigen untuk \(V\)
Untuk \(\lambda = 16\):
\[\begin{split}
\begin{bmatrix} -6 & 6 \\ 6 & -6 \end{bmatrix}
\Rightarrow v_1 = v_2
\end{split}\]
Normalisasi:
\[
v_1 = \frac{1}{\sqrt{2}}, \quad v_2 = \frac{1}{\sqrt{2}}
\]
Untuk \(\lambda = 4\):
\[\begin{split}
\begin{bmatrix} 6 & 6 \\ 6 & 6 \end{bmatrix}
\Rightarrow v_1 = -v_2
\end{split}\]
Normalisasi:
\[
v_1 = \frac{1}{\sqrt{2}}, \quad v_2 = -\frac{1}{\sqrt{2}}
\]
Matriks V
\[\begin{split}
V = \begin{bmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{bmatrix}
\end{split}\]
Cari U
\[
U = A V \Sigma^{-1}
\]
\[\begin{split}
A V =
\begin{bmatrix}
2 \sqrt{2} & \sqrt{2} \\
2 \sqrt{2} & -\sqrt{2}
\end{bmatrix}
\end{split}\]
Normalisasi:
\[\begin{split}
U = \begin{bmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{bmatrix}
\end{split}\]
Hasil Akhir SVD
\[
A = U \Sigma V^T
\]
dengan:
\[\begin{split}
U = \begin{bmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{bmatrix}, \quad
\Sigma = \begin{bmatrix}
4 & 0 \\
0 & 2
\end{bmatrix}, \quad
V = \begin{bmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{bmatrix}
\end{split}\]
Pembuktian \(A\)
\(A = \begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix}\)
\(U = \frac{1}{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}\)
\(\Sigma =
\begin{bmatrix}
4 & 0 \\
0 & 2
\end{bmatrix}\)
\(V = \frac{1}{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}\)
Karena \(V\) ortogonal, \(V^T = V\)
Jadi:
\[
V^T = V
\]
Hitung \(\Sigma V^T)\)
\[\begin{split}
\Sigma V^T =
\Sigma V =
\begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}
\frac{1}{\sqrt{2}}
\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}
\end{split}\]
Hasil:
\[\begin{split}
= \frac{1}{\sqrt{2}}
\begin{bmatrix} 4 \cdot 1 + 0 \cdot 1 & 4 \cdot 1 + 0 \cdot (-1) \\
0 \cdot 1 + 2 \cdot 1 & 0 \cdot 1 + 2 \cdot (-1)
\end{bmatrix}
= \frac{1}{\sqrt{2}}
\begin{bmatrix} 4 & 4 \\ 2 & -2 \end{bmatrix}
\end{split}\]
Hitung \(U \Sigma V^T\)
\[\begin{split}
U \Sigma V^T =
\frac{1}{\sqrt{2}}
\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}
\frac{1}{\sqrt{2}}
\begin{bmatrix} 4 & 4 \\ 2 & -2 \end{bmatrix}
\end{split}\]
Karena
$\(
\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{2} = \frac{1}{2}
\begin{bmatrix}
1 \cdot 4 + 1 \cdot 2 & 1 \cdot 4 + 1 \cdot (-2) \\
1 \cdot 4 + (-1) \cdot 2 & 1 \cdot 4 + (-1) \cdot (-2)
\end{bmatrix}
= \frac{1}{2}
\begin{bmatrix}
6 & 2 \\
2 & 6
\end{bmatrix}
= \begin{bmatrix}
3 & 1 \\
1 & 3
\end{bmatrix} = A\)$
Kesimpulan
terbukti bahwa :
\[
A = U \Sigma V^T
\]
dengan hasil sesuai matriks aslinya.
Contoh Soal 2
Perhitungan Singular Value Decomposition (SVD)
Diketahui Matriks
\[\begin{split}
A = \begin{bmatrix}
3 & 7 \\
2 & 5 \\
4 & 3 \\
1 & 1
\end{bmatrix}
\end{split}\]
Langkah 1: Hitung (A^T A)
\[\begin{split}A^T =\begin{bmatrix}
3 & 2 & 4 & 1 \\
7 & 5 & 3 & 1
\end{bmatrix}\end{split}\]
\[\begin{split}A^T A =\begin{bmatrix}
3 & 2 & 4 & 1 \\
7 & 5 & 3 & 1\end{bmatrix}
\begin{bmatrix}
3 & 7 \\
2 & 5 \\
4 & 3 \\
1 & 1
\end{bmatrix}=
\begin{bmatrix}
30 & 44 \\
44 & 84
\end{bmatrix}\end{split}\]
Langkah 2: Hitung Eigenvalue dari (A^T A)
Misal \(\lambda \text{ adalah eigenvalue dari } A^T A\)
\[\begin{split}
\left|
\begin{array}{cc}
30 - \lambda & 44 \\
44 & 84 - \lambda
\end{array}
\right|
= 0
\Rightarrow
(30 - \lambda)(84 - \lambda) - 44^2 = 0
\end{split}\]
\[
\lambda^2 - 114 \lambda + 2520 - 1936 = 0
\Rightarrow
\lambda^2 - 114 \lambda + 584 = 0
\]
\[
\lambda = \frac{114 \pm \sqrt{114^2 - 4 \cdot 584}}{2}
= \frac{114 \pm \sqrt{10660}}{2}
\Rightarrow
\lambda_1 \approx 108.62, \quad \lambda_2 \approx 5.38
\]
Langkah 3: Hitung Singular Value
\[
\sigma_1 = \sqrt{108.62} \approx 10.42, \quad
\sigma_2 = \sqrt{5.38} \approx 2.32
\]
\[\begin{split}
\Sigma =
\begin{bmatrix}
10.42 & 0 \\
0 & 2.32 \\
0 & 0 \\
0 & 0
\end{bmatrix}
\end{split}\]
Langkah 4: Cari Eigenvector dari (A^T A)
Hasil eigenvector ter-normalisasi (dari komputasi numerik):
\[\begin{split}
v_1 =
\begin{bmatrix}
0.4884 \\
0.8726
\end{bmatrix}, \quad
v_2 =
\begin{bmatrix}
-0.8726 \\
0.4884
\end{bmatrix}
\end{split}\]
\[\begin{split}
V =
\begin{bmatrix}
0.4884 & -0.8726 \\
0.8726 & 0.4884
\end{bmatrix}
\end{split}\]
Langkah 5: Hitung Matriks (U)
\[
u_1 = \frac{1}{10.42} A v_1, \quad
u_2 = \frac{1}{2.32} A v_2
\]
\[\begin{split}
u_1 \approx
\begin{bmatrix}
0.6817 \\
0.4992 \\
0.4725 \\
0.1668
\end{bmatrix}, \quad
u_2 \approx
\begin{bmatrix}
0.3241 \\
0.3882 \\
-0.3541 \\
-0.7987
\end{bmatrix}
\end{split}\]
Lengkapi (U) menjadi matriks ortogonal (4 \times 4) (dengan Gram-Schmidt atau QR decomposition):
\[
U =
\begin{bmatrix}
\vec{u}_1 & \vec{u}_2 & \vec{u}_3 & \vec{u}_4
\end{bmatrix}
\]
Langkah 6: Bentuk SVD
\[
A = U \Sigma V^T
\]
Dengan:
\[\begin{split}
\Sigma =
\begin{bmatrix}
10.42 & 0 \\
0 & 2.32 \\
0 & 0 \\
0 & 0
\end{bmatrix}, \quad
V^T =
\begin{bmatrix}
0.4884 & 0.8726 \\
-0.8726 & 0.4884
\end{bmatrix}
\end{split}\]
Pembuktian
\[
U \Sigma V^T = A
\]
Perkalian kembali ketiga matriks akan menghasilkan matriks semula (atau sangat mendekati secara numerik karena pembulatan).
