**Unformatted text preview: **451(4)in If
Β» 2 0 O
A: β-6 l ~β4 ,
β3 0 β1
then
pat) = ββdet(A ββ AI)
2β)Β» 0 0
=β- β6 1~A β4 ~3 0 β-1βA Expanding across the ο¬rst row, 170Β») = β(2 β 1) 1β11 β-4 0 β-1β)t
= (A -2)(1 ~ A)(ββl β~ A)
= (A ββ 2)()Β» ~ 1)(}t +1). Thus, the eigenvalues are 2, 1, and βl, respectively.
Because A β 21 reduces, 0 O 0 1 0 1
A~β21= β6 β1 - ββ> 0 1 β2 ,
β-3 0 β3 0 O 0 it is easily seen that the nullspace of A β 21 is generβ
ated by the eigenvector v1 = (β1,2, 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 β1 0 0
det 2 1 2 : β1,
1 O 1 the eigenvectors are independent. . If 1 O 0
A = 3 β-Z 1 ,
5 β5 2
then β
pot) 2: β-det(A β M)
1 β A 0 0
= β 3 ββ2 β it 1 .
5 β5 2 -β A
Expanding across the ο¬rst row, V
β2 β A 1 : (A β 1)((~2 β m2 ~ A) + 5)
= (A ~ 1)(A2 +1) Thus, the eigenvalues are 1, i, and βi, respectively.
Because A ~ 1'] reduces, β - lβi 0 0
Aβil: 3 β2ββi 1
5 ~5 Zβi 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 1ββ-> 1
5 . Because the eigenvectors are independent. ...

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