21 215 show that the n n identity matrix i is

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Unformatted text preview: of eigenvectors for A even when the roots are repeated. We first consider the eigenspace W1 corresponding to the eigenvalue λ1 = 1, which consists of the solutions to the linear system 5−1 4 0 b1 4 5−1 0 b2 = 0, 0 0 1−1 b3 or 4b1 4b 1 +4b2 +4b2 0 = 0, = 0, = 0. 2 There are many excellent linear algebra texts that prove this theorem in detail; one good reference is Bill Jacob, Linear algebra , W. H. Freeman, New York, 1990; see Chapter 5. 33 The coefficient matrix of this linear system is 440 4 4 0 . 000 Applying the elementary row operations 440 11 4 4 0 → 4 4 000 00 to this matrix 0 1 0 → 0 0 0 yields 1 0 0 0 0 . 0 Thus the linear system is equivalent to b1 + b2 0 0 = 0, = 0, = 0. Thus W1 is a plane with equation b1 + b2 = 0. We need to extract two unit length eigenvectors from W1 which are perpendicular to each other. Note that since the equation for W1 is b1 + b2 = 0, the unit length vector √ 1 n= 2 1 √ 2 0 is perpendicular to W1 . Since √ 1 0 2 −1 b1 = 0 ∈ W1 , we find that b2 = n × b1 = √2 ∈ W1 . 1 0 The vectors b1 and b2 are unit...
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This document was uploaded on 01/12/2014.

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