My name is Hanzhang Yin, and I have developed this website as a resource to facilitate the review of key concepts in abstract algebra, and it is not fully completed. Feel free to email me at hanyin@ku.edu if there are any errors or suggestions for improvement.
1. Consider the iteration \[ x_{n+1} = \frac{1}{2} x_n + \frac{5}{2x_n}, \quad n = 0, 1, \ldots \]
Proof of 1. Let \(f(x) = \frac{1}{2}x + \frac{5}{2x}\). Hence, we can calculate the derivative of \(f(x)\) as \[ f'(x) = \frac{1}{2} - \frac{5}{2x^2} = \frac{1}{2}\left(1 - \frac{5}{x^2}\right) \] Since \(x_0 > 0\), we just assume that \(x>0\). When \( 0 \lt x \leq \sqrt{5} \), we have \( f'(x) \lt 0 \). When \( x \geq \sqrt{5} \), we have \( f'(x) \geq 0 \). Thus, we can know that when \( x = \sqrt{5} \), \( f(x) \) is at minimum for \( f(x) \) where \( x \geq 0 \). Then, \[ \min f(x) = f(\sqrt{5}) = \frac{1}{2} \sqrt{5} + \frac{5}{2\sqrt{5}} = \frac{1}{2} \sqrt{5} + \frac{1}{2} \sqrt{5} = \sqrt{5}. \] Therefore, we have \( f(x) \geq \sqrt{5} \). In other words, if \( x_0 > 0 \), then \( x_n \geq \sqrt{5} \). \[ \tag*{$\square$} \]
4. Let \( A = [a_{ij}] \in \mathbb{R}^{n \times n} \) be a real, square matrix which is strictly diagonally dominant, i.e. \[ a_{ii} > \sum_{j=1, j \neq i}^{n} |a_{ij}|, \quad i = 1, 2, \ldots, n. \]
(a) Prove that each eigenvalue of \( A \) has a positive real part.
(b) Show that \( \det A > 0 \).
(c) Prove that \( A \) has a unique LU factorization \( A = LU \), where \( L \) is unit lower triangular and \( U \) is upper triangular. Prove that all diagonal entries of \( U \) are strictly positive.
(d) Describe the main steps of a practical algorithm (based on LU factorization) for solving the linear system \( Ax = b \) for some vector \( b \in \mathbb{R}^n \).
Proof. According to \(\textbf{