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x
1
x
2
x
3
x
4
x
5
=
1
0
2
0
1
+ λ
1
1
1
0
0
1
+ λ
2
4
1
1
−2
−1
π ∩ π
0
π
0
{u
1
, u
2
, u
3
}
−2u
1
− 2u
2
− u
3
= 1 π ∩ π
0
π
0
π
m
∩ π
0
k
= ∅
π
m
= {M
0
, L
m
} π
0
k
= {M
0
0
, L
0
k
}
m 6 k
π
m
π
0
k
(π
m
kπ
0
k
) π
m
∩ π
0
k
= ∅ L
m
⊂ L
0
k
π
m
π
0
k
π
m
∩ π
0
k
= ∅
L
0
k
∩ L
m
= {0}
1) π
m
= {M
0
, L
m
}
π
0
k
= {M
0
0
, L
0
k
}
m 6 k π
m
⊂ π
00
k
π
00
k
kπ
0
k
k π
00
k
π
00
k
= {M
0
, L
0
k
}
2) π
m
π
0
k
(k + m) π
1
k+m
= {M
0
, L
0
k
⊕L
m
} π
2
k+m
= {M
0
0
, L
0
k
⊕L
m
}
M
0
0
π
0
k
M
0
π
m
π
00
k
π
2
k+m
M
0
0
π
0
k
π
1
k+m
M
0
π
m
π
m
π
0
k
r
0
0
− r
0
/∈
L
m
⊕L
0
k
dim(L
m
⊕L
0
k
) = m+k 1+m+k 6 n
n − 1
7/D5 x1 1 1 4 x2 0 1 1 x3 = 2 + λ1 0 + λ2 1 x4 0 0 −2 x5 1 1 −1 -/1-6>12 .6-/1-/7> π ∩ π0 35//:5730B5*:58 151 .-D:,-*/7B- B 5 0,,-: .3-/735,/7B* π0 / 1--3D0,575:0 {u1, u2, u3} E5D5*7/8 -D,0: 235B ,*,0*: −2u1 − 2u2 − u3 = 1 7- π ∩ π0 F 90.*3.6-/1-/7> B π0 < µ^ π ∩ π 0 = ∅ < VWXYZY[Y\]Y^ }hi njbc nkm eoghvghia πm = {M0, Lm} a πk0 = {M00 , L0k } m k aw not gefmnmombbghiaw m 6 k oghvghia πm a πk0 bjlckj{iht ejfjoomobcda k hifgxgd hdchom hogp kj (πmkπk0 ) w mhoa πm ∩ πk0 = ∅ a Lm ⊂ L0k oghvghia πm a πk0 bjlckj{iht hvfmqakj{qadahtw mhoa πm ∩ πk0 = ∅ a L0k ∩ Lm = {0} :**7 :*/7- /6*D2** -)*B0D,-* XYZ[¡Y\]Y^ 1) Æhoa eoghvghi π = {M , L } ejfjoomobj eoghvgp hia πk0 = {M00 , L0k } w m 6 k w ig πm ⊂ πk00 w xnm πk00kπk0 Ðij k peoghvghi πk00 m 0 m admmi homn}{qau kan πk00 = {M0, L0k } v v 2) Æhoa eogh ghia πm a πk0 h fmqakj{ihtw ig gba omyji k ejfjoomop bcr (k + m) peoghvghitr πk+m 1 = {M0 , L0k ⊕ Lm } a πk+m 2 = {M00 , L0k ⊕ Lm } 1 πk00 πk+m πm πk0 M0 M00 M0 πm M00 πk0 2 πk+m E 2/6-B08 ¶ /6*D2*7 )7- */60 π 0 π0 /13*0B57/8 7- r0 − r ∈/ Lm ⊕Lk0 < 51 151 7- m k dim(Lm ⊕Lk ) = m+k 1+m+k 6 n < 0 6*D-B57*6>,- 0 /2: 0 :5 35E:*3,-/7* /13*0B50C/8 .6-/1-/7* ,* :-*7 .3*BA57> n − 1 < Ñ
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