cu’a c´ac vecto.. riˆeng kh´ac khˆong nˆen c´ac vecto... riˆeng kh´ac khˆong nˆen c´ac vecto.. riˆeng cu’a ph´ep bdtt d˜a cho.. Ph´ep biˆe´n dˆo’i L l`a ph´ep biˆe´n dˆo’i dˆ` ng nhˆa´t
Trang 1Nhu vˆa.y L(x) = −x v`a do d´o x l`a vecto riˆeng ´u.ng v´o.i gi´a tri riˆeng
5 − λ 4
8 9 − λ
= 0 ⇔ λ
2+ Ca’ hai gi´a tri λ = 1 v`a λ = 13 dˆe`u l`a c´ac gi´a tri riˆeng
3+ Dˆe’ t`ım to.a dˆo cu’a c´ac vecto riˆeng ta c´o hai hˆe phu.o.ng tr`ınh
i) V`ı λ1 = 1 nˆen hˆe (I) c´o da.ng
4ξ1+ 4ξ2 = 0, 8ξ1+ 8ξ2 = 0.
T`u d´o suy ra ξ2 = −ξ1, do d´o nghiˆe.m cu’a hˆe n`ay c´o da.ng ξ1 = α1,
ξ2 = −α1, trong d´o α1 l`a da.i lu.o ng t`uy ´y V`ı vecto riˆeng kh´ac khˆong
nˆen c´ac vecto ´u.ng v´o.i gi´a tri riˆeng λ1 = 1 l`a c´ac vecto u(α1, −α1),
trong d´o α1 6= 0 l`a t`uy ´y
ii) Tu.o.ng tu khi λ2 = 13 hˆe (II) tro.’ th`anh
−8ξ1 + 4ξ2 = 0, 8ξ1 − 4ξ2 = 0,
Trang 2t´u.c l`a ξ2 = 2ξ1 D˘a.t ξ1 = β ⇒ ξ2 = 2β Vˆa.y hˆe (II) c´o nghiˆe.m l`a
ξ1 = β, ξ2 = 2β V`ı vecto riˆeng kh´ac khˆong nˆen c´ac vecto riˆeng ´u.ngv´o.i gi´a tri λ2 = 13 l`a c´ac vecto v(β, 2β) N
V´ ı du 6 T`ım gi´a tri riˆeng v`a vecto riˆeng cu’a ph´ep biˆe´n dˆo’i tuyˆe´n
1 − λ5 4 − λ2
= −λ
Gia’ su.’ x = (ξ1, ξ2, ξ3) 6= 0 l`a vecto riˆeng ´u.ng v´o.i gi´a tri riˆeng λ.
Khi d´o x l`a nghiˆe.m cu’a hˆe thuˆa` n nhˆa´t
(1 − λ)ξ1+ ξ2+ 4ξ3 = 0, 2ξ1− λξ2− 4ξ3 = 0,
−ξ1+ ξ2+ 4ξ3 = 0.
⇒ nghiˆ e.m tˆo’ng qu´at l`a (0, −4α, α), α 6= 0 t`uy ´y.
Vˆa.y v´o.i gi´a tri riˆeng λ1 = 1 ta c´o c´ac vecto riˆeng ´u.ng v´o.i n´o l`a
(0, −4α, α), α ∈ R, α 6= 0.
2+ Khi λ = 2 ta c´o
(∗) ⇒
−ξ1+ ξ2 + 4ξ3 = 0, 2ξ1− 2ξ2− 4ξ3 = 0,
Trang 4Khi λ = 3, thu c hiˆe.n tu.o.ng tu nhu o.’ 1+ v`a 2+
ta thu du.o cvecto riˆeng tu.o.ng ´u.ng (2γ, 0, γ), γ 6= 0 t`uy ´y N
V´ ı du 8 T`ım gi´a tri riˆeng v`a vecto riˆeng cu’a ph´ep bdtt v´o.i ma trˆa.n
10 −19 − λ 10
... )ξ − 12 η + 6ζ = 0, 10 ξ − ( 19 + λ i )η + 10 ζ = 0, 12 ξ − 24η + (13 − λ i )ζ = 0; i = 1, 2.
1< small>+ Khi λ = ta c´o
6ξ − 12 η +... v´o.i ma trˆa.n
10 ? ? 19 − λ 10
=
c´o nghiˆe.m λ1< /small> = λ2 = 1, λ1< /small> = ? ?1 C´ac vecto d˘a.c... data-page="5">
2+ Khi λ2 = ? ?1 ta c´o
8ξ − 12 η + 6ζ = 0, 10 ξ − 18 η + 10 ζ = 0, 12 ξ − 24η + 14 ζ = 0.
Ha.ng cu’a ma trˆa.n (A−λ3E)