MATH 218 Problems Solution Manual Chapter 5.1, Fall 2011
American University of Beirut · Elementary Linear Algebra with Applications · 2 pages · with solution
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5.1 Eigenvalues and Higenvectors 119
@ The matrix can have three distinct real eigenvalues.and three linearly independent eigenvectors,
@ The matrix can have two distinct real eigenvalues such that one has algebraic multiplicity 2 and the
other algebraic multiplicity 1, But the matrix can still have three linearly independent eigenvectors,
two from the eigenvalue of multiplicity 2 and one form the other.
Solutions to Exercises
1. Since the matrix equation ; ° | , = , is satisfied if and only if A = 3, the eigenvalue’
corresponding to the eigenvector 1 isA =3,
« oo fat ot] ft i ;
2, Since the matrix equation 0 -2 1 = \ 1 {38 satisfied if and only if \ = —2, the eigenvalue
corresponding to the eigenvector 7 is A= —2.
3. The corresponding eigenvalue is \ = 0. 4. The corresponding eigenvalue is \ = —2,
5, The corresponding eigenvalue is A = 1. 6. The corresponding eigenvalue is.) = —1.
7. a. The characteristic equation is det(A ~ AZ) = 0, that is,
-2-% 2 *
aet(A— M1) =| 3 3d |=(-2- 9-8-9) 6 =P H5r=0.
b. Since the eigenvalues are the solutions to the characteristic equation \? + 5A = A(A +5) = 0, the |
eigenvalues are \; = 0 and Ay = —5. c, The corresponding eigenvectors are found by solving, respectively,
—2 2 _ —2 2 _ or ance the elec stare ay fl _ | 72
3-3 |v = 0 and 3 3 ly = —5v. Hence the eigenvectors are vy; = 1 and vo = 3 ;
respectively, :
d. . .
_f-2 2]f1])_fo]_.fa _f[-2 10 ]_ ~2
a= [3 a] [t]-[o]-o[t] meae=[% as ]=-9[ 3
8 a M4443 =0b A =-lLy=-3e v=
LA} S]f-0f)
8a =I =0b a=tew=| 5] a | j male
10. a 7 +3A4¢2=0b. A= lag =-2e vi
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120 Chapter _5 Higenvalues and Bigenvectors
11. a, The characteristic equation det(A — AZ) = 0, is
-1-A 0 1
0 1-2 0 =0.
0 2 -1-A
Expanding down column one, we have that
-1-\ 0 1 *
o=| 0 1-A 0 = (21-277) y= 0+ya-9.
0 2 -1—2 “
1 1
b. Ay =—-l,Ag=le vi =] 0 |, vo= |} 2
0 2
d.
-1 0 1 1 —1 1 -l 0 12 1 1 1
0 o|[o]=|o |-co[a] oe | 0 o|[2]=[2]-c] 2)
Q0 2 -1 0 0 0 0 2 -1 2 2 2
1
)
~2 2
12, a, AA+1)(A—1) =0b. Ay =0, Ap = —1,A3 = le. ak Jw 1 w= H
0 0 2
1 —3
18, a, (\—2)(A~1)? =0b. 1 = 2, w= le ne [o|w- | a |
0 1
2 1 2 1 2 1 2 1 2 —3 =3 —3
d.}0 2 -1]/}0}]=/0]=(@)|0] and }o 2 -1 1 f=]; 1]=q@]1
01 0 0) 0 0 0 1 0 1 1
0 1 :
14, a. (A-1)8 =0b.A=1av1.=] -1 |,ve= | 0
I 0
15. a. (A+ 1)(A ~ 2)(A + 2)(A— 4) = 0b. Ay = 1, Ag = 2, Ag = -2, =
1 0 0 .
0 1 0 0 . eyes
G@vi= |] 9 [> v2= | 9 |r va= |] q fo va= 0 d. For the eigenvalue \ = —1, the verification is
0 0 0 1
-1 0 0 1 —1 1
, ; 3 , t = ; = (-1) ; . The other cases are similar.
0 0 0 4 0 0 0
1
16, a (A+1)(A—1)(A~-2)A-3) = 0b. A = WL Wp = Lrg =H 2M BO ME 4 ,
~2
1 -7 1
wall —{ 2 _ | 0
2 0 V3 = 1 1v4 = 0
0 0 0