Matrix Decomposition
Eigen Decomposition
\(\mathbf{A} \in \mathbb{R}^{n \times n}\), \(\mathbf{A}\) is diagonalizable, then
\(\mathbf{A}= \mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1}\), \(\mathbf{Q} = \begin{bmatrix} | & | & ... & |\\ \mathbf{v}_{1} & \mathbf{v}_{2} & ... & \mathbf{v}_{n} \\ | & | & ... & | \end{bmatrix}\), \(\mathbf{\Lambda} = \begin{bmatrix} \lambda_{1} & 0 & 0 & 0\\ 0 & \lambda_{2} & ... & 0\\ 0 & 0 & ... & \lambda_{n} \end{bmatrix}\)
\(\{\lambda_{i}\}^{n}_{i=0}\) are eigenvalues with corresponding eigenvectors \(\{\mathbf{v}_{i}\}^{n}_{i}\). This means \(\mathbf{A}\mathbf{v}_{i} = \lambda_{i}\mathbf{v}_{i}\) (eigenvalues equation)
If \(\mathbf{A}\) is symmetric (\(\mathbf{A}=\mathbf{A}^{T}\)), then we can always write \(\mathbf{A}= \mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1}\),
Application
Eigen Decomposition is very useful to simplify and reduce matrix product computation.
Problem 1. Let \(\mathbf{A} \in \mathbb{R}^{n \times n}\) be a diagonalizable matrix with eigenvalues \(\{\lambda_{i}\}^{n}_{i=0}\)
* Compute \(\mathbf{A}^{k}\) for a fixed \(k \in \mathbb{Z}_{+}\)
Singular Value Decomposition
Application
Reference
[1] Diagonalization