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Unformatted text preview: and λ 2 =2 (with algebraic multiplicity 1). λ 1 = 0 Aλ 1 I = A = 1 0 1 3 03 1 01 ∼ 1 01 0 0 0 0 x y z = z y z = y 1 + z 1 1 Therefore 1 , 1 1 is a basis of the eigenspace corresponding to λ 1 = 0. λ 2 =2 Aλ 2 I = A +2 I = 1 0 1 3 23 1 0 1 ∼ 1 0 1 0 26 0 0 ∼ 1 0 1 0 13 0 0 x y z = z 3 z z = z 1 3 1 Therefore 1 3 1 is a basis of the eigenspace corresponding to λ 2 =2....
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This note was uploaded on 03/22/2010 for the course MATH 1104 taught by Professor Unknown during the Winter '10 term at Carleton.
 Winter '10
 Unknown
 Math, Linear Algebra, Algebra

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