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# Let x = (x1, . . . , xn)T be an eigenvector of A belonging to . Show that if |xi| = _x_ ISBN: 9780136009290 436

## Solution for problem 7 Chapter 7.6

Linear Algebra with Applications | 8th Edition

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Problem 7

Let x = (x1, . . . , xn)T be an eigenvector of A belonging to . Show that if |xi| = _x_, then (a) _n j=1 ai j x j = xi (b) |aii| _n j=1 j _=i |ai j | (Gerschgorins theorem)

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Math 1314 Lesson 1: Prerequisites 1. Simplifying Algebraic Expressions 1 2 1 Example 1: Simplify 15x (2x 3  x ) 3x (x  x ) 5 . 3 2. Identifying Polynomials To begin with, let’s review the definition of a polynomial function. A polynomial is the...

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##### ISBN: 9780136009290

This full solution covers the following key subjects: . This expansive textbook survival guide covers 47 chapters, and 921 solutions. The answer to “Let x = (x1, . . . , xn)T be an eigenvector of A belonging to . Show that if |xi| = _x_, then (a) _n j=1 ai j x j = xi (b) |aii| _n j=1 j _=i |ai j | (Gerschgorins theorem)” is broken down into a number of easy to follow steps, and 44 words. The full step-by-step solution to problem: 7 from chapter: 7.6 was answered by , our top Math solution expert on 03/15/18, 05:24PM. Linear Algebra with Applications was written by and is associated to the ISBN: 9780136009290. Since the solution to 7 from 7.6 chapter was answered, more than 207 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Linear Algebra with Applications, edition: 8.

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