From 562ba9a9b6efd8cc27fc506f83b1125c2cfa4619 Mon Sep 17 00:00:00 2001
From: Elizabeth Hunt <elizabeth.hunt@simponic.xyz>
Date: Mon, 30 Oct 2023 19:07:43 -0600
Subject: hw 5

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+* Power Method for computing the largest eigenvalue of a square matrix
+
+An eigenvector, v \in R^n is a nonzero vector such that for some number, \lambda \in C, Av = \lambda v
+\Rightarrow || v || = 1
+
+
+Suppose we start with some vector v and assume, v = \alpha_0 v_0 + \alpha_1 v_1 + \cdots + \alpha_n v_n, where {v_1, \cdots, v_n}
+are the eigenvectors of A. Assume {v_1, \cdots, v_n} is a basis for R^n
+
+We can order the eigenvalues such that \lambda_1 \ge \lambda_2 \ge \lambda_3 \ge \cdots \ge \lambda_n
+
+Compute u = Av
+= A(\alpha_1 v_1 + \cdots + \alpha_n v_n)
+= \alpha_1 Av_1 + A(\cdots) + \alpha_n A v_n
+= \alpha_1 \lambda_1 v_1 + \alpha_2 \lambda_2 v_2 + \cdots + \alpha_n \lambda_n v_n
+
+w = A (Av)
+= \alpha_1 \lambda_1^2 v_1 + \alpha_2 \lambda_2^2 v_2 + \cdots + \alpha_n \lambda_n^2 v_n
+
+Thus,
+A^k v = \alpha_1 \lambda_1^k v_1 + \alpha_2 \lambda_2^k v_2 + \cdots + \alpha_n \lambda_n^k v_n
+= \lambda_1^k ( \alpha_1 v_1 + \alpha_2 \frac{\lambda_2^k}{\lambda_1^k} v_2 + \cdots + \alpha_n \frac{\lambda_3^k}{\lambda_1^k} v_n)
+
+As k \rightarrow \infty
+A^k v = \lambda_1^k (\alpha_1 v_1) + \text{negligble terms}
+
+Algorithm:
+v \ne 0 with v \in R^n
+y = Av = \alpha_1 v_1 + \cdots + \alpha_n v_n
+
+w = \frac{1}{||y||} \cdot y
+
+Rayleigh Quotient:
+If $v$ is an eigenvector of A with eigenvalue \lambda then \frac{v^T A v}{v^T v} = \lambda
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