03_004_Eigenvector_Eigenvalue.html ========================================================= matrix A (square matrix) vector x $$$\neq$$$ 0 $$$x \in R^{n}$$$ (n-D real number vector) if there is vector x which satisfies following relation : (matrix A)(vector x)=(scalar value)(vector x) $$$Ax=\lambda x$$$ then, you can call vector x as eigenvector (corresponding to $$$\lambda$$$) "of matrix A" $$$\lambda$$$ : scalar value, eigenvalue of "matrix A" ========================================================= example $$$\begin{bmatrix} 2&3\\2&1 \end{bmatrix} \begin{bmatrix} 3\\2 \end{bmatrix} = \begin{bmatrix} 12\\8 \end{bmatrix} = 4 \begin{bmatrix} 3\\2 \end{bmatrix} $$$ $$$\begin{bmatrix} 2&3\\2&1 \end{bmatrix}$$$ : A $$$\begin{bmatrix} 3\\2 \end{bmatrix}$$$ : x 4 : $$$\lambda$$$ ========================================================= induce steps $$$Ax=\lambda x$$$ $$$Ax-\lambda x=0$$$ $$$(A-\lambda I)x=0$$$ then, $$$x=0$$$ or $$$(A-\lambda I)=0$$$ you choose $$$(A-\lambda I)=0$$$ Solving $$$(A-\lambda I)=0$$$ is equivalent to solving $$$|A-\lambda I|=0$$$ Therefore, you can call $$$|A-\lambda I|=0$$$ as characteristic equation of matrix A Since $$$A-\lambda I$$$ is also matrix : you can find its determinant $$$(A-\lambda I)=0 \Rightarrow |A-\lambda I|=0 \Rightarrow \lambda^{N} + a_{1}\lambda^{N-1} + ... + a_{N-1}\lambda + a_{N}=0 $$$ A : nxn matrix $$$\lambda$$$ : eigenvalue of A set composed of all eigenvectors corresponding to eigenvalue $$$\lambda$$$ forms subspace of $$$R^{n}$$$ along with 0 vector This subspace is called eigenspace of A ========================================================= characteristic of eigenvalue and eigenvector ========================================================= eigenvalue of diagonal matrix is its diagonal elements ========================================================= eigenvalue of tri-matrix is its diagonal elements ========================================================= If vector x is eigenvector of matrix A, kx is also eigenvector of matrix A ========================================================= eigenvalue of A == eigenvalue of $$$A^{T}$$$ ========================================================= eigenvalue of $$$A = \lambda$$$ eigenvalue of $$$A^{-1} = \frac{1}{\lambda}$$$ ========================================================= eigenvalues of A : $$$\lambda_{1}, \lambda_{2}, ..., \lambda_{n}$$$ $$$\lambda_{1} \times \lambda_{2} \times ... \times \lambda_{n}$$$ = determinant of A ========================================================= eigenvectors of each eigenvalue are linearly independent diagonal matrix $$$A = \begin{bmatrix} 1&0&0 \\ 0&2&0 \\ 0&0&3 \end{bmatrix}$$$ eigenvalues of A : $$$\lambda_{1} = 1$$$ $$$\lambda_{2} = 2$$$ $$$\lambda_{3} = 3$$$ eigenvector of A : $$$u_{\lambda_{1}} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} $$$ $$$u_{\lambda_{2}} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $$$ $$$u_{\lambda_{3}} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$$ $$$u_{\lambda_{1}}, u_{\lambda_{2}}, u_{\lambda_{3}}$$$ are linearly independent ========================================================= eigenvectors of real number symmetric matrix are orthogonal to each others ========================================================= eigenvalues of real number symmetric matrix are real number ========================================================= if A is positive-definite matrix, all eigenvalues of A are positive