본 글은 주재걸교수님의 인공지능을 위한 선형대수 강의를 듣고 정리한 내용입니다.
Singular Value Decomposition
- 직사각행렬에 대한 decomposition
- 𝑈 ∈ ℝ𝑚×𝑚 , 𝑉 ∈ ℝ𝑛×𝑛 : matrices with orthonormal columns, providing an orthonormal basis of Col 𝐴 and Row 𝐴, respectively
- • Σ ∈ ℝ𝑚×𝑛 : a diagonal matrix whose entries are in a decreasing order, i.e., 𝜎1 ≥ 𝜎2 ≥ ⋯ ≥ 𝜎min (𝑚,n)
Another Perspective of SVD
- {𝐮1, … , 𝐮𝑛} for Col 𝐴 and {𝐯1, … , 𝐯𝑛} for Row 𝐴, by using, say, Gram–Schmidt orthogonalization
- 𝐴𝑉 = 𝑈Σ ⟺ [𝐴𝐯1 𝐴𝐯2 ⋯ 𝐴𝐯𝑛] = [𝜎1𝐮1 𝜎2𝐮2 ⋯ 𝜎𝑛𝐮𝑛]
- 𝑉 −1 = 𝑉^𝑇 since 𝑉 ∈ ℝ𝑛×𝑛 has orthonormal columns.
- Thus 𝐴𝑉 = 𝑈Σ ⟺ 𝐴 = 𝑈Σ𝑉^T
- Can we find the following? (Yes)
- Orthogonal eigenvector matrices 𝑈 and 𝑉
- Eigenvalues in Σ^2 that are all positive
- Eigenvalues in Σ^2 that are shared by 𝐴𝐴^𝑇 and 𝐴^𝑇𝐴
Diagonalization of Symmentric Matrices
In general, 𝐴 ∈ ℝ𝑛×𝑛 is diagonalizable if and only if 𝑛 linearly independent eigenvectors exist.
- symmetric matrix : diagonal을 기준으로 transpose 했을 때 변하지않는 행렬, always diagonalizable
- Orthogonally diagonalizable
- meaning that their eigenvectors are not only linearly independent, but also orthogonal to each other
- n개의 real eigenvalues를 가지고 있음(중복 허용, 실근)
- 각 eigenvalue에 대한 eigenspace의 dimension은 characteristic equation의 root의 multiplicity of 𝜆 와 같다
- The eigenspaces are mutually orthogonal. That is, eigenvectors corresponding to different eigenvalues are orthogonal.
Spectral Decomposition
Positive Definite Matrices
- 𝐴 ∈ ℝ𝑛×𝑛 is positive definite if 𝐱^𝑇𝐴𝐱 > 0, ∀𝐱 ≠ 𝟎.
- 𝐴 ∈ ℝ𝑛×𝑛 is positive semi-definite if 𝐱^𝑇𝐴𝐱 ≥ 0, ∀𝐱 ≠ 𝟎.
- 𝐴 ∈ ℝ𝑛×𝑛 is positive definite if and only if the eigenvalues of 𝐴 are all positive.
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