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Math 21

# Math 21 - 451(4)in If Β 2 0 O A β-6 l ~β4 β3 0 β1...

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Unformatted text preview: 451(4)in If Β» 2 0 O A: β-6 l ~β4 , β3 0 β1 then pat) = ββdet(A ββ AI) 2β)Β» 0 0 =β- β6 1~A β4 ~3 0 β-1βA Expanding across the ο¬rst row, 170Β») = β(2 β 1) 1β11 β-4 0 β-1β)t = (A -2)(1 ~ A)(ββl β~ A) = (A ββ 2)()Β» ~ 1)(}t +1). Thus, the eigenvalues are 2, 1, and βl, respectively. Because A β 21 reduces, 0 O 0 1 0 1 A~β21= β6 β1 - ββ> 0 1 β2 , β-3 0 β3 0 O 0 it is easily seen that the nullspace of A β 21 is generβ ated by the eigenvector v1 = (β1,2, 1)T. In a sim- ilar manner, we arrive at the following eigenvalue- eigenvector pairs. 0 0 1 β> 1 ,, and ~ 1 β-> 2 0 1 Because β1 0 0 det 2 1 2 : β1, 1 O 1 the eigenvectors are independent. . If 1 O 0 A = 3 β-Z 1 , 5 β5 2 then β pot) 2: β-det(A β M) 1 β A 0 0 = β 3 ββ2 β it 1 . 5 β5 2 -β A Expanding across the ο¬rst row, V β2 β A 1 : (A β 1)((~2 β m2 ~ A) + 5) = (A ~ 1)(A2 +1) Thus, the eigenvalues are 1, i, and βi, respectively. Because A ~ 1'] reduces, β - lβi 0 0 Aβil: 3 β2ββi 1 5 ~5 Zβi l 0 0 β> 0 1 ~2/5+1/5i , 0 0 0 it is easily seen that the nullspaces of A ~ i I is gener- ated by the eigenvector v z (0, 2 ~ 1', 5)T. In a simβ ilar manner, we arrive at the following eigenvalueβ eigemvector pairs. /0 1 ο¬i_+5\2+i} and 1ββ-> 1 5 . Because the eigenvectors are independent. ...
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