Algebre
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]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﺒﻨﻴﺔ ﺍﻝﻔﻀﺎﺀ ﺍﻝﺸﻌﺎﻋﻲ: ﻝﻴﻜﻥ ) • ( K , +,ﺤﻘل ﺘﺒﺩﻴﻠﻲ ،ﻭﻝﺘﻜﻥ Eﻤﺠﻤﻭﻋﺔ ﻏﻴﺭ ﺨﺎﻝﻴﺔ .
ﻤﺜﺎل :1
ﺘﻌﺭﻴﻑ :ﻨﻘﻭل ﺃﻥ Eﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ﻋﻠﻰ ﺍﻝﺤﻘل ، Kﺇﺩﺍ ﺯﻭﺩﻨﺎ E
ﻜل ﺤﻘل Kﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ﻋﻠﻰ ﻨﻔﺴﻪ ( K , +, • ) ،ﺤﻘل.
ﺒﻌﻤﻠﻴﺘﻴﻥ (1ﻋﻤﻠﻴﺔ ﺩﺍﺨﻠﻴﺔ ) : ( +ﺘﺠﻌل ﻤﻥ Eﺯﻤﺭﺓ ﺘﺒﺩﻴﻠﻴﺔ
E =K " "+ﺍﻝﺩﺍﺨﻠﻴﺔ ﻋﻠﻰ Eﻫﻲ " "+ﺍﻝﻌﻤﻠﻴﺔ ﺍﻝﺩﺍﺨﻠﻴﺔ ﻋﻠﻰ . K
(2ﺍﻝﻌﻤﻠﻴﺔ ﺍﻝﺨﺎﺭﺠﻴﺔ ) • ( :ﻤﻌﺭﻓﺔ ﻜﻤﺎ ﻴﻠﻲ : K ×E → E
" • " ﺍﻝﺨﺎﺭﺠﻴﺔ ﻋﻠﻰ Eﻫﻲ " • " ﺍﻝﻌﻤﻠﻴﺔ ﺍﻝﺨﺎﺭﺠﻴﺔ ﻋﻠﻰ . K
(α , x ) → α • x
ﺤﺎﻝﺔ ﺨﺎﺼﺔ ℝ :ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ﻋﻠﻰ . ℝ
ﺒﺤﻴﺙ : 1) ∀α , β ∈ K : ∀u ∈ E : (α + β ) u = α u + β u
ﻤﺜﺎل :2
2) ∀α ∈ K : ∀u ,v ∈ E : α (u + v ) = α u + α v
) : ℑ ( ℝ, ℝﻤﺞ ﺍﻝﺘﻁﺒﻴﻘﺎﺕ ﻤﻥ ℝﻓﻲ . ℝ
) 3) ∀α , β ∈ K : ∀u ∈ E : (αβ ) u = α ( β u
fﺘﻁﺒﻴﻕ }ℑ ( ℝ, ℝ ) = { f : ℝ → ℝ
4) 1K • u = u
ﻋﻠﻰ Eﻨﻌﺭﻑ ﺍﻝﻌﻤﻠﻴﺘﻴﻥ E ×E → E
( f .g ) → f + g ) ∀x ∈ ℝ : ( f + g )( x ) = f ( x ) + g ( x
ﺍﺼﻁﻼﺤﺎﺕ: •
ﻋﻨﺎﺼﺭ Eﺘﺴﻤﻰ ﺃﺸﻌﺔ.
•
ﻋﻨﺎﺼﺭ ) Kﺍﻝﺤﻘل( ﺘﺴﻤﻰ ﺴﻠﻤﻴﺎﺕ .
•
ﺍﻝﻌﻨﺼﺭ ﺍﻝﺤﻴﺎﺩﻱ ﻝـ " "+ﻓﻲ Eﻨﺭﻤﺯ ﻝﻪ ﺒـ. O E :
•
ﻨﺭﻤﺯ ﻝﻠﻔﻀﺎﺀ ﺍﻝﺸﻌﺎﻋﻲ ﺒـ ) • ( K , +,ﻭﻨﻘﻭل
ℝ×E → E
( λ.f ) → λ f ) )( x ) = λ f ( x
∀x ∈ ℝ : ( λ f
ﺇﺫﻥ ℝﻑ.ﺵ
ﺃﻥ E Kﻑ.ﺵ.
ﻤﺜﺎل :3ﻤﺠﻤﻭﻋﺔ ﻜﺜﻴﺭﺍﺕ ﺍﻝﺤﺩﻭﺩ ﺒﻤﻌﺎﻤﻼﺕ ﻤﻥ ﺍﻝﺤﻘل : K
ﻗﻭﺍﻋﺩ ﺍﻝﺤﺴﺎﺏ ﻓﻲ ﻑ.ﺵ :
] = ℝ ∨ ℂ) , E = K [x
ﻝﻴﻜﻥ K , Eﻑ .ﺵ α , β ∈ K , u ,v ∈ E ، ) 1) ∀α ∈ K , ∀u ∈ E :αu = O E ⇒ (α = O K ) ∨ (u = O E
ﻨﺯﻭﺩ Eﺒـ :
) α −1 (α .α −1 = 1Kد ⇒ ) (α u = O E ) ∧ (α ≠ O K ⇒ α u = O E ⇒ α −1 (α u ) = α −1O E
)
] + : K [x ]× K [x ] → K [x
(
⇒ αα −1 u = O E ⇒ 1K u = O E ⇒ u = O E 2) α (u − v ) = α (u + ( −v ) ) = α u − α v
3) (α − β ) u = α u − β u
www.math.3arabiyate.net
] E = K [ xﻫﻭ ﻑ.ﺵ
1
(K ( p ,q ) → p + q ) ( p + q )( x ) = p ( x ) + q ( x ] K × K [x ] → K [x (λ, p ) → λ p ) ( λ p )( x ) = λ p ( x
â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
:4 ﻤﺜﺎل
؟u1 , u 2 ﻫﻭ ﻋﺒﺎﺭﺓ ﺨﻁﻴﺔ ﻓﻲv (1, −1)
.ﺵ. ﻑE K ﻤﺞ ﻭX ≠ φ
∃α , β ∈ ℝ : v = α u 1 + β u 2
H = ℑ ( X , E ) = {f : X → E , f
(1, −1) = α ( 5, 2 ) + β (1, 0 ) = ( 5α + β , 2α )
f , g ∈ H , ( f + g )( x ) = f ( x ) + g ( x )
1 α =− 5 + = 1 α β 2 ⇒ 2α = −1 β = 7 2
λ ∈ K , f ∈ H , ( λ f ( x ) ) = λ if ( x ) ﺵ. ﻑKH
}
ﺇﺫﻥ
ﺵ. ﻑℝ ﻫﻭℑ ([a , b ] , ℝ )
1 7 v = u 1 + u 2 ﻭﻤﻨﻪ 2 2
:5ﻤﺜﺎل ﺠﺩﺍﺀ ﻑ ﺸﻌﺎﻋﻴﺔ ﻋﻠﻰ ﻨﻔﺱ ﺍﻝﺤﻘل
:ﺍﻝﻔﻀﺎﺀ ﺍﻝﺸﻌﺎﻋﻲ ﺍﻝﺠﺯﺌﻲ
.ﺵ. ﻑK
E n ,............, E 3 , E 2 , E 1
E ⊇ F ≠ φ ﺵ ﻭﻝﻨﻜﻥ. ﻑE , K ﻝﻴﻜﻥ
E = E 1 × E 2 × E 3 × ..................E n
ﺝ ﻤﻥ ﺍﻝﻔﻀﺎﺀ ﺍﻝﺸﻌﺎﻋﻲ.ﺵ. ﻑF ﻨﻘﻭل ﺃﻥ:ﺘﻌﺭﻴﻑ
X ∈ E ⇔ X = ( x 1 , x 2 ,..........., x n ) / x i ∈ E i , i =1....n
: ( ﺇﺫﺍ ﻜﺎﻥK , +, • )
X ,Y ∈ E , X +Y = ( x 1 , x 2 ,..., x n ) + ( y 1 , y 2 ,..., y n )
F ≠φ ♦
= ( x 1 + y 1 , x 2 + y 2 , ... , x n + y n )
K ﺵ ﻋﻠﻰ ﺍﻝﺤﻘل. ( ﺒﺩﻭﺭﻫﺎ ﻑF , +, • ) ♦
λ ∈ K , X ∈ E : λ X = ( λ x 1 , λ x 2 ,........... λ x n )
(
O E = O E1 ,O E 2 ,................,O E n
:ﺝ.ﺵ.ﺘﻤﻴﻴﺯ ﺍﻝﻑ E ⊇F ﻭ
)
n
ﺵ. ﻑK ﻫﻭE = ΠE i
ﺵ. ﻑK ( E , + , • )
i =1
.ﺵ. ﻑℝ ﻫﻭE = ℝ ← E i = ℝ n
iF ≠ φ ( I ) i∀u ,v ∈ F :u + v ∈ F ⇔ (ﺝ.ﺵ. ﻑF ) i∀α ∈ K , ∀u ∈ F , α u ∈ F
.ﺵ. ﻑℝ ﻫﻭℝ 2 , n = 2 .ﺵ. ﻑℝ ﻫﻭℝ 3 , n = 3
iF ≠ φ ⇔ i∀u ,v ∈ F , ∀ α , β ∈ K , α u + β v ∈ F
:ﺍﻝﻌﺒﺎﺭﺓ ﺍﻝﺨﻁﻴﺔ ﻝﻤﺠﻤﻭﻋﺔ ﺃﺸﻌﺔ
( II )
ﺵ. ﻑE , K ﻝﻴﻜﻥ
α1 , α 2 ∈ K , u1 , u 2 ∈ E ﻭﻝﻴﻜﻥ
iF ≠ φ ( III ) ⇔ i∀u ,v ∈ F , ∀α ∈ K , α u + v ∈ F
ﻴﺴﻤﻰ ﻋﺒﺎﺭﺓ ﺨﻁﻴﺔ ﻓﻲ ﺍﻝﺸﻌﺎﻋﻴﻥv = α1 u 1 + α 2 u 2 ∈ E ﺍﻝﺸﻌﺎﻉ u 2 , u1
:ﻭﻨﻌﻤﻡ ﺫﻝﻙ ﻜﻤﺎ ﻴﻠﻲ
O E ∈ F ﺘﺄﻜﺩ ﺃﻥF ≠ φ ﻹﺜﺒﺎﺕ:ﻤﻼﺤﻅﺔ
E ( ﻤﺠﻤﻭﻋﺔ ﺃﺸﻌﺔ ﻤﻥu i )i ∈I ﻝﺘﻜﻥ
: ( ﻜل ﻋﺒﺎﺭﺓ ﻤﻥ ﺍﻝﺸﻜلu i ) ﻨﺴﻤﻲ ﻋﺒﺎﺭﺓ ﺨﻁﻴﺔ ﻓﻲ ﺍﻷﺸﻌﺔ
∑α u i
i
, αi ∈ K
i ∈I
2
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â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
E ﺝ ﻤﻥ.ﺵ. ﻑF2 ، F1 :ﺘﻌﺭﻴﻑ
{( x , y ) ∈ ℝ
u ∈ F1 ⇔ u = ( −2 y , y ) , y ∈ ℝ
(ﺵ. )ﻑE ﺝ ﻤﻥ.ﺵ. ﻑF1 + F2 ﺇﻥ:ﻤﻼﺤﻅﺔ
v ∈ F2 ⇔ u = ( −2 y ′, y ′ ) , y ′ ∈ ℝ
α u + v = ( −2α y , α y ) + ( −2 y ′, y ′ )
E ﺝ ﻤﻥ.ﺵ. ﻑF1 , F2 ,..........., Fp ، ﺵ. ﻑK , E :ﺘﻌﻤﻴﻡ
= ( −2α y − 2 y ′, α y + y ′ )
ﻭﺍﻝﻤﻌﺭﻑ ﺒـH = F1 + F2 + ........ + Fp ﺝ.ﺵ.ﻨﺴﻤﻲ ﺍﻝﻑ
= ( −2 (α y + y ′ ) , α y + y ′ ) ∈ F1
H = {w ∈ E / w = u 1 + u 2 + ......... + u p , u i ∈ Fi } F1 , F2 ,..........., Fp ﺒﻤﺠﻤﻭﻉ
F2 = {( x , y , z ) ∈ ℝ3 / x + 2 y + z = 3}
φ ≠ X ⊂ E ، ﺵ. ﻑK E
K = ℝ , E = ℝ [x ]
V ect ( x ) ﺒـX ﻨﺴﻤﻲ ﻤﺠﻤﻭﻋﺔ ﻜل ﺍﻝﻌﺒﺎﺭﺍﺕ ﺍﻝﺨﻁﻴﺔ ﻝﻌﻨﺎﺼﺭ OE ∈ F ⇒ F ≠ φ ?
p , q ∈ F ,α ∈ ℝ ⇒α p + q ∈ F
v ,w ∈V ect ( x ) , α ∈ K , (α .v + w ) ∈V ect ( x ) ♦
i ∈I
)
= α p ′ (0) + q ′ ( 0)
♦
α U +V = ∑ (α λi ) u i + ∑ β j v j
= (α p + q )′ ( 0 ) = A ′ ( 0 )
j ∈J
= ∑ λ ′u i + ∑ β j v j i ∈I
(A = α p + q )
p ( 0 ) = p ′ ( 0 ) q ′ ( 0 ) = q ′ ( 0 ) A ( 0 ) = (α p + q )( 0 ) = α p ( 0 ) + q ( 0 )
. X ﺍﻝﻤﻭﻝﺩ ﺒـ
j ∈J
(3
F = { p ∈ E / p ( 0 ) = p ′ ( 0 )}
V ect ( x ) = ∑ λi u i , λi ∈ K , u i ∈ X i ∈I ﻴﺴﻤﻰ ﺍﻝﻔﻀﺎﺀ ﺍﻝﺠﺯﺌﻲE ﺝ ﻤﻥ.ﺵ. ﻑV ect ( x ) ♦ ﺇﻥ
V = ∑ β j v j (v j ∈ X
(2
O ℝ3 = ( 0, 0, 0 ) ∉ F2 ⇒ ﺝ.ﺵ. ﻝﻴﺱ ﻑF2
E ﺍﻝﻔﻀﺎﺀ ﺍﻝﺸﻌﺎﻋﻲ ﺍﻝﺠﺯﺌﻲ ﺍﻝﻤﻭﻝﺩ ﺒﺠﺯﺀ ﻤﻥ
)
(1
u ,v ∈ F1 , α ∈ ℝ ⇒ α u + v ∈ F1
F1 + F2 = { u + v / u ∈ F1 ,v ∈ F2 }
i ∈I
}
/ x + 2y = 0
O E = ( 0, 0 ) ∈ F1 ⇒ F1 ≠ φ
: ﺍﻝﻤﺠﻤﻭﻋﺔ ﺍﻝﺘﺎﻝﻴﺔF2 ﻭF1 ﻨﺴﻤﻲ ﻤﺠﻤﻭﻉ
U = ∑ λi u i (u i ∈ X
2
E = ℝ 2 :ﺃﻤﺜﻠﺔ
j ∈J
V ect ( F ) = F ، ﺝ.ﺵ. ﻑF :ﺤﺎﻝﺔ ﺨﺎﺼﺔ
:ﺨﻭﺍﺹ
∑α u
ﺵ. ﻑE , K :ﻤﺜﺎل
i
X = { u1 ,u 2 } ♦
p ∈ℕ*
∈ F ﺝ ﻓﺈﻥ.ﺵ. ﻑF ﻭu i ∈ F (1
E ﺝ ﻤﻥ.ﺵ. ﻑG ﻭG ﺝ ﻤﻥ.ﺵ. ﻑF (2
V ect ( X ) = { α1 u1 + α 2 u 2 , α1 , α 2 ∈ K } X = { u 1 , u 2 ,........., u p }
i
i ∈I
E ﺝ ﻤﻥ.ﺵ. ﻑF ﻓﺈﻥ
♦
ﻫﻭ ﺃﻴﻀﺎF1 ∩ F2 ﻓﺈﻥE ﺝ ﻤﻥ.ﺵ. ﻑF2 ﻭF1 (3
p V ect ( X ) = ∑ α i u i , α i ∈ K i =1 V ect ( X ) ﻴﺭﻤﺯ ﻓﻲ ﻫﺫﻩ ﺍﻝﺤﺎﻝﺔ ﻝـ
ﺝ.ﺵ.ﻑ .ﺝ.ﺵ. ﻋﻠﻰ ﻋﺩﺩ ﻜﻴﻔﻲ ﻤﻥ ﻑ3 ﺘﻌﻤﻡ ﺍﻝﻨﺘﻴﺠﺔ E ﺝ ﻤﻥ ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ.ﺵ. ﻑFi ، i ∈ I
V ect ( X ) = u 1 , u 2 ,............, u p ﺒـ
.ﺝ.ﺵ.ﻫﻭ ﺃﻴﻀﺎ ﻑ
∩F
i
ﻓﺈﻥ
i ∈I
3
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â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
E = ℝ 2 , u = (1, 2, 0 ) , v ( 0, 0,1) , X = {u ,v } ♦ V ect ( X ) = u ,v
u = u 1 + u 2 ⇔ ( x , y ) = ( x 1 , − x 1 ) + ( x 2 , 2x 2 )
= { α u + β v , α , β ∈ ℝ}
1 x 1 = ( 2x − y ) + = x x x 1 3 2 ⇔ ⇔ − x 1 + 2x 2 = y x = 1 ( x + y ) 2 3 ﻤﻭﺠﻭﺩﺍﻥ ﺩﺍﺌﻤﺎx 1 , x 2
= { (α , 2α , 0 ) + ( 0, 0, β ) , α , β ∈ ℝ} = { (α , 2α , β ) , α , β ∈ ℝ}
. ﻤﻭﺠﻭﺩﺍﻥ ﺩﺍﺌﻤﺎu 2 ( x 2 , 2x 2 ) , u1 ( x 1 , − x 1 ) ⇐
:ﺍﻝﻤﺠﻤﻭﻉ ﺍﻝﻤﺒﺎﺸﺭ ﻝﻔﻀﺎﺀﺍﺕ ﺸﻌﺎﻋﻴﺔ ﺠﺯﺌﻴﺔ E ﺝ ﻤﻥ.ﺵ. ﻑF2 ﻭF1 ، ﺵ. ﻑK , E ﻝﻴﻜﻥ
:ﺍﻝﻔﻀﺎﺀﺍﺕ ﺍﻝﺸﻌﺎﻋﻴﺔ ﺍﻝﺠﺯﺌﻴﺔ ﺍﻹﻀﺎﻓﻴﺔ
ﺝ.ﺵ. ﻑH ﻨﻌﻠﻡ ﺃﻥ، H = F1 + F2 ﻨﻀﻊ
ﺝ.ﺵ. ﻑF ,G ، ﺵ. ﻑE ﻝﻴﻜﻥ
ﻭﻨﻜﺘﺏF2 ﻭF1 ﻤﺠﻤﻭﻉ ﻤﺒﺎﺸﺭ ﻝـH ﻨﻘﻭل ﺃﻥ:ﺘﻌﺭﻴﻑ
F ≡ )ﺃﻭF ﺝ ﺇﻀﺎﻓﻲ ﻝـ.ﺵ. ﻑG ﻨﻘﻭل ﺃﻥ:ﺘﻌﺭﻴﻑ
ﻴﻜﺘﺏ ﻋﻠﻰ ﺸﻜل ﻭﺤﻴﺩH ﺇﺫﺍ ﻜﺎﻥ ﻜل ﻋﻨﺼﺭ ﻤﻥ، H = F1 ⊕ F2
E = F ⊕ G : ﺇﺫﺍ ﺘﺤﻘﻕ، ( G ﺝ ﺇﻀﺎﻓﻲ ﻝـ.ﺵ.ﻑ
: ﺤﻴﺙw = u + v :
(w ∈ H , u ∈ F1 , v ∈ F2 )
:ﺘﻤﻴﻴﺯ ﺍﻝﻤﺠﻤﻭﻉ ﺍﻝﻤﺒﺎﺸﺭ
E = ℝ 2 :ﻤﺜﺎل F = {( x , y , z ) / x + y = z } ,G = {( x , y , z ) / x = y = z }
H = F1 + F2 ﻭ، E ﺝ ﻤﻥ.ﺵ. ﻑF2 , F1 ﻝﻴﻜﻥ
H = F1 + F2 H = F1 ⊕ F2 ⇔ F1 ∩ F2 = { O E }
ℝ = F ⊕ G ⇔ ﺇﻀﺎﻓﻴﺎﻥG ﻭF 3
F ∩ G = { O E } 3 ℝ ⊂ F + G
E = ℝ 2 :ﻤﺜﺎل
♦
u ( x , y , z ) ∈ F ∩ G ⇒ u ∈ F ∧ u ∈G ⇒x +y =z ∧ x = y =z ⇒ 2x = x ⇒ x = 0 ⇒ y = z = 0
( x , y , z ) = ( 0, 0, 0 ) = O ℝ
{( x , y ) ∈ ℝ = {( x , y ) ∈ ℝ
}
F1 =
2
/x + y =0
F2
2
/ 2x − y = 0
}
؟E = F1 ⊕ F2 ﻫل
3
F ∩ G = {O E } ﺃﻱ
i) E = F1 + F2 E = F1 ⊕ F2 ⇔ ii) F1 ∩ F2 = {O E }
w ( x , y , z ) ∈ ℝ ♦ ﻝﻴﻜﻥ 3
∃u ∈ F ,v ∈G : w = u + v
?
ii) u ( x , y ) ∈ F1 ∩ F2 ⇒ u = O ℝ2
u ( x 1 , y 1 , x 1 + y 1 ) ∈ F , v ( x 2 , x 2 , x 2 ) ∈G u +v = w ⇔ ( x 1 + x 2 , y 1 + x 2 , x 1 + y 1 + x 2 )
u ∈ F1 ∩ F2 ⇒ (u ∈ F1 ) ∧ (u ∈ F2 )
x 1 + x 2 = x .................. (1) y 1 + x 2 = y ................. ( 2 ) x 1 + y 1 + x 2 = z .......... ( 3) x 1 = z − y ( 3) − ( 2 ) ⇒ x 2 = x + y − z y = z − x 1
x + y = 0 ⇒ x = y = 0 ⇒ u = O ℝ2 2x − y = 0 ii) ℝ 2 = F1 + F2 i ℝ 2 ⊃ F1 + F2 ﻭﺍﻀﺢ ?
i F1 + F2 ⊃ ℝ 2
∀u ( x , y ) ∈ ℝ 2 , ∃u1 ∈ F1 , u 2 ∈ F2 : u = u1 + u 2 4
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]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﻭﻤﻨﻪ uﻭ vﻤﺴﺘﻘﻼﻥ ﺨﻁﻴﺎ
u ( z − y , z − x , 2z − x − y ) ∈ F v ( x + y − z , x + y − z , x + y − z ) ∈G
ℝ 2 [ x ] .3ﻑ.ﻙ.ﺡ ﺍﻝﺘﻲ ﺩﺭﺠﺘﻬﺎ ≥ 2
u +v = ( x , y , z ) = w
u = x 2 , v = x , w =1 α u + β v + δ w = Oℝ ⇒ α = β = δ = 0
ﻤﻼﺤﻅﺔ :ﻓﻲ ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ، Eﻜل ﻑ.ﺵ.ﺝ ﻴﻤﻠﻙ ﺇﻀﺎﻓﻴﺎ.
α x + β x + δ .1 = O ℝ , ∀x ∈ ℝ 2
x =0:δ =0
x =1 : α + β +δ = 0 ⇒α = β = δ = 0 x = −1 : α − β + δ = 0 ﻭﻤﻨﻪ w , v , uﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ
ﺍﻝﻌﺎﺌﻠﺔ ﺍﻝﺤﺭﺓ – ﺍﻝﻌﺎﺌﻠﺔ ﺍﻝﻤﻭﻝﺩﺓ – ﺍﻷﺴﺱ ﻓﻲ ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ: ﺍﻝﻌﺎﺌﻠﺔ ﺍﻝﺤﺭﺓ: ﻝﻴﻜﻥ K , Eﻑ.ﺵ ،
*p ∈ℕ
} S = {u 1 , u 2 ,........, u p
ﻤﻼﺤﻅﺎﺕ ﻋﺎﻤﺔ:
ﺘﻌﺭﻴﻑ :ﻨﻘﻭل ﺃﻥ ﺍﻝﻌﺎﺌﻠﺔ Sﺤﺭﺓ ،ﺃﻭ ﺃﻥ ﺍﻷﺸﻌﺔ u1 , u 2 ,......., u p
، S = { u } (1
ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ ﺇﺫﺍ ﺘﺤﻘﻕ
∀α1 , α 2 ,..., α p ∈ K ,
)ﺍﻝﺸﻌﺎﻉ uﻤﺴﺘﻘل ﺨﻁﻴﺎ ⇔ ( u ≠ φ
α1u1 + α 2u 2 + ... + α pu p = 0 ⇒ α1 = α 2 = ... = α p = 0
S = { u ,v } (2
ﺃﻱ ﻻ ﺘﻘﺒل ﺍﻝﻤﻌﺎﺩﻝﺔ ﺇﻻ ﺍﻝﺤل ﺍﻝﻤﻌﺩﻭﻡ
Sﻤﻘﻴﺩﺓ ⇔ u ,vﻤﺭﺘﺒﻁﺎﻥ ﺨﻁﻴﺎ
) (α ,α ,..., α ) = ( 0, 0,..., 0 p
2
⇔ ) ∃ λ ∈ K * : (u = λ v ) ∨ (v = λ u
1
ﻭﺇﺫﺍ ﻭﺠﺩ α iﻏﻴﺭ ﻤﻌﺩﻭﻡ ﺒﺤﻴﺙ = O E
Sﺤﺭﺓ ⇔ u ≠ O E
i =p
i
∑α u i
ﻤﺜﺎل :ﻓﻲ ، ℝ3
ﻨﻘﻭل ﺃﻥ ﺍﻝﻌﺎﺌﻠﺔ
) u (1, 2 ) , v ( −1, −2
u ,vﻤﺭﺘﺒﻁﺎﻥ ﺨﻁﻴﺎ ﻷﻥ u + v = O ℝ2 ⇔ v = −u
i =1
Sﻤﺭﺘﺒﻁﺔ ،ﺃﻭ ﺃﻥ ﺍﻷﺸﻌﺔ u1 , u 2 ,..., u pﻤﺭﺘﺒﻁﺔ ﺨﻁﻴﺎ.
(3
*
p ∈ℕ
} S = {u 1 , u 2 ,........, u pﻤﺭﺘﺒﻁﺔ ﺨﻁﻴﺎ
⇔ ﺃﺤﺩ ﻋﻨﺎﺼﺭ ﺍﻝﻌﺎﺌﻠﺔ ﻴﻜﺘﺏ ﻋﻠﻰ ﺸﻜل ﻋﺒﺎﺭﺓ ﺨﻁﻴﺔ ﻓﻲ ﺒﺎﻗﻲ
ﺃﻤﺜﻠﺔ: .1
2
ﺍﻝﻌﻨﺎﺼﺭ .
u = (1,3) , v = ( 2, −1) , E = ℝ ?
⇒α = β = 0
2
ﻤﺜﺎل:
α u + β v = Oℝ
♦
) (α ,3α ) + ( 2β , − β ) = ( 0, 0
.w = u + 2v
♦ ] w = 3x 2 − 2x , v = x , u = x 2 , E = ℝ 2 [ x ﻭﺍﻀﺢ ﺃﻥ w = 3u − 2vﻭﻤﻨﻪ } { u ,v ,wﻤﺭﺘﺒﻁﺔ ﺨﻁﻴﺎ .
v = e x , u = sin x , E = ℑ ( ℝ, ℝ ) .2 ?
α u + β v = 0 ⇒α = β = 0 α sin x + β e x = 0 , ∀x ∈ ℝ x = 0 ⇒ β = 0 π x = ⇒ α = 0
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w = ( 5, 6, −2 ) , v ( 2, 2, −1) , u (1, 2, 0 ) , E = ℝ
} { u ,v ,wﻤﻘﻴﺩﺓ ﺃﻱ u , v , wﻤﺭﺘﺒﻁﺔ ﺨﻁﻴﺎ ﻷﻥ
α + 2 β = 0 ⇒ 3α − β = 0 ⇒ α = β = 0 ﻭﻤﻨﻪ uﻭ vﻤﺴﺘﻘﻼﻥ ﺨﻁﻴﺎ ،ﺃﻭ } S = { u ,vﺤﺭﺓ.
2
3
5
(4
ﻜل ﻋﺎﺌﻠﺔ ﺠﺯﺌﻴﺔ ﻤﻥ ﻋﺎﺌﻠﺔ ﺤﺭﺓ ﻫﻲ ﺒﺩﻭﺭﻫﺎ ﻋﺎﺌﻠﺔ ﺤﺭﺓ .
(5
ﻜل ﻋﺎﺌﻠﺔ ﺘﺸﻤل ﻋﺎﺌﻠﺔ ﻤﻘﻴﺩﺓ ﻫﻲ ﺒﺩﻭﺭﻫﺎ ﻤﻘﻴﺩﺓ .
(6
ﻜل ﻋﺎﺌﻠﺔ ﺘﺸﻤل O Eﻫﻲ ﺒﺎﻝﻀﺭﻭﺭﺓ ﻤﻘﻴﺩﺓ .
]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﺍﻝﻌﺎﺌﻠﺔ ﺍﻝﻤﻭﻝﺩﺓ: ﻝﻴﻜﻥ K , Eﻑ.ﺵ ،ﻭ
*p ∈ℕ
(α , −α ) + ( 0, 3 β ) = ( 0, 0 ) ⇒
} S = {u 1 , u 2 ,........, u pﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ
ﻤﻥ ، Eﻨﻘﻭل ﺃﻥ ﺍﻝﻌﺎﺌﻠﺔ Sﺘﻭﻝﺩ ﺍﻝﻔﻀﺎﺀ Eﺇﺫﺍ ﻜﺎﻥ ﻜل ﺸﻌﺎﻉ ﻤﻥ
{ u ,vﺘﻭﻝﺩ ℝ 2
• } = ( x , y ) ∈ ℝ2
Eﻴﻜﺘﺏ ﻋﻠﻰ ﺸﻜل ﻋﺒﺎﺭﺓ ﺨﻁﻴﺔ ﻓﻲ ﺃﺸﻌﺔ ، S i =p
∀v ∈ E : ∃α1 , α 2 ,..., α p ∈ K / v = ∑ α i u i
ﺃﻱ
wﻜﻴﻔﻲ ∃α , β ∈ ℝ , w = α u + β v ،
) ( x , y ) = (α , −α ) + ( 0,3β
i =1
ﺃﻤﺜﻠﺔ: v (0, −1) , u (1, 2) , E = ℝ 2 (1
ﻭﻤﻨﻪ ﻓﻌﻼ
ﻝﻴﻜﻥ w = ( x , y ) ∈ ℝ 2ﻜﻴﻔﻲ ∃α , β ∈ ℝ : w = α u + β v
}
α = x α = x ⇒ y +x α + 3 β = y β = 3 { u ,vﺃﺴﺎﺱ ﻝـ . ℝ 2
ﺘﻌﺭﻴﻑ ﻤﻜﺎﻓﺊ ﻷﺴﺎﺱ ﻓﻲ ﻑ.ﺵ:
α = x α = x ⇔ 2 α − β = y β = 2x − y )w = x (1, 2 ) + ( 2x − y )( 0, −1
ﻝﻴﻜﻥ K , Eﻑ.ﺵ ،
*p ∈ℕ
} S = {u 1 , u 2 ,........, u pﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ
ﻤﻥ E
u ,vﻴﻭﻝﺩﺍﻥ . ℝ 2
Sﺃﺴﺎﺱ ﻝـ Eﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ﻜل ﺸﻌﺎﻉ ﻤﻥ Eﻴﻜﺘﺏ ﻋﻠﻰ ﺸﻜل ﻋﺒﺎﺭﺓ ﺨﻁﻴﺔ ﻓﻲ ﻋﻨﺎﺼﺭ Sﻭﻫﺫﻩ ﺍﻝﻜﺘﺎﺒﺔ ﻭﺤﻴﺩﺓ k =n
u = x 2 v = x ، E = ℝ 2 [ x ] (2 w = 1
∀v ∈ E , ∃! α1 , α 2 ,..., α n ∈ K : v = ∑ α k u k k =1
ﺘﺭﻤﻴﺯ :ﺘﺴﻤﻰ α1 , α 2 ,..., α nﻤﺭﻜﺒﺎﺕ vﻓﻲ ﺍﻷﺴﺎﺱ S
ﺇﻥ } { u ,v ,wﺘﻭﻝﺩ E
ﺃﻤﺜﻠﺔ:
∀p ∈ E : p = a x 2 + b x + c
E = ℝ n (1 ♦ E = ℝ2 ← n = 2 )v = ( x , y ) = ( x ,0 ) + ( 0, y ) = x (1, 0 ) + y ( 0,1
ﻤﻼﺤﻅﺔ :ﻜل ﻋﺎﺌﻠﺔ ﺘﺸﻤل ﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓ ﻫﻲ ﺒﺩﻭﺭﻫﺎ ﻤﻭﻝﺩﺓ.
ﺇﺫﻥ ) e 2 ( 0,1) , e1 (1, 0ﻴﻭﻝﺩﺍﻥ ℝ 2
ﺍﻷﺴﺎﺱ:
( x , y ) = x e1 + y e 2 ⇒ (α , β ) = ( 0, 0 ) ⇒ α = β = 0 ﻭﻤﻨﻪ } { e1 , e 2ﺃﺴﺎﺱ ﻝـ ℝ 2
ﻭ e 2 , e1ﻤﺴﺘﻘﻼﻥ ﺨﻁﻴﺎ.
ﺘﻌﺭﻴﻑ :ﻨﻘﻭل ﻋﻥ ﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ ﺃﻨﻬﺎ ﺃﺴﺎﺱ ﻓﻲ ﺍﻝﻔﻀﺎﺀ Eﺇﺫﺍ ﻜﺎﻨﺕ ﺤﺭﺓ ﻭﻤﻭﻝﺩﺓ ﻓﻲ ﺁﻥ ﻭﺍﺤﺩ.
ﻴﺴﻤﻰ ﻫﺫﺍ ﺍﻷﺴﺎﺱ ﺍﻝﻁﺒﻴﻌﻲ )ﺃﻭ ﺍﻝﻘﺎﻨﻭﻨﻲ( ﻝﻠﻔﻀﺎﺀ . ℝ 2
ﻤﻭﻀﻭﻋﺔ :ﻓﻲ ﻜل ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ﻏﻴﺭ ﻤﻌﺩﻭﻡ ﺘﻭﺠﺩ ﺃﺴﺱ.
♦ E = ℝ3 ← n = 3 ) u ( x , y , z ) = ( x , 0, 0 ) + ( 0, y , 0 ) + ( 0, 0, z
ﻤﺜﺎلV ( 0,3) , u (1, −1) , E = ℝ 2 :
}
)= x (1, 0, 0 ) + y ( 0,1, 0 ) + z ( 0, 0,1
{ u ,vﺃﺴﺎﺱ ﻝـ ℝ 2 •
}
ﻭﻤﻨﻪ ) e 3 ( 0, 0,1) , e 2 ( 0,1, 0 ) , e1 (1, 0, 0ﺘﻭﻝﺩ ℝ3
{ u ,vﺤﺭﺓ
ﻭﻫﻲ ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ ﻭﻤﻨﻪ } { e1 , e 2 , e 3ﺃﺴﺎﺱ ﻝـ ℝ3
α u + β v = Oℝ ⇒ α = β = 0
ﻴﺴﻤﻰ ﺍﻝﻘﺎﻨﻭﻥ ﺍﻝﻁﺒﻴﻌﻲ ﻝـ . ℝ3
2
www.math.3arabiyate.net
2
α e1 + β e 2 = O ℝ
6
]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﺘﻌﺭﻴﻑ ﺒﻌﺩ ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ): (Dimension n
v ∈ℝ
ﻝﻴﻜﻥ K , Eﻑ.ﺵ ﻤﻨﺘﻬﻲ ﺍﻝﺒﻌﺩ
)v = x 1 (1, 0,..., 0 ) + x 2 ( 0,1, 0,..., 0 ) + ... + x n ( 0, 0,..., 0,1
ﻤﺒﺭﻫﻨﺔ ﻭﺘﻌﺭﻴﻑ :ﻜل ﺍﻷﺴﺱ ﻝﻬﺎ ﻨﻔﺱ ﺍﻝﻌﺩﺩ ﻤﻥ ﺍﻝﻌﻨﺎﺼﺭ ) ، (Card
ﺍﻷﺸﻌﺔ:
ﻨﺴﻤﻲ ﺒﻌﺩ Eﻋﺩﺩ ﻋﻨﺎﺼﺭ ﺃﺴﺎﺱ ﻜﻴﻔﻲ ﻤﺎ ،ﻭﻨﻜﺘﺏ
) e1 (1, 0,...,0 ) + e 2 ( 0,1, 0,..., 0 ) + ... + e n ( 0, 0,..., 0,1ﺘﻭﻝﺩ
) ، dim K E = card ( Bﺤﻴﺙ : Bﺃﺴﺎﺱ ﻤﺎ .
ℝ nﻭﻫﻲ ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ ،ﻓﻬﻲ ﺇﺫﻥ ﺃﺴﺎﺱ ﻝـ ، ℝ nﺘﺴﻤﻰ ﺍﻷﺴﺎﺱ
ﻤﻼﺤﻅﺔ :ﺇﻥ ﺒﻌﺩ ﻑ.ﺵ ﻴﺘﻌﻠﻕ ﺒﺎﻝﺤﻘل ﺍﻝﻤﺴﺘﺨﺩﻡ
ﺍﻝﻘﺎﻨﻭﻨﻲ .
♦ ℂﻑ.ﺵ ﻋﻠﻰ ) ℂﻜل ﺤﻘل ﻑ.ﺵ ﻋﻠﻰ ﻨﻔﺴﻪ( ﺒﻌﺩﻩ ، 1 dim ℂ ℂ = 1
E = ℝ n [ x ] (2 p = a0 + a1x + a2 x 2 + ... + an x n , ai ∈ ℝ
)( z ∈ ℂ
p ∈ ℝ n [x ] ,
z = z ⋅1
u = 1∈ ℂ = E ,
♦ E = ℂﻫﻭ ﻑ.ﺵ ﻋﻠﻰ ﺍﻝﺤﻘل K = ℝ z = α .1 + β . i , α , β ∈ ℝ , z = α + β . i , z ∈ ℂ = E
ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ 1, x , x 2 ,..., x nﺘﻭﻝﺩ ] ℝ n [ xﻭﻫﻲ ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ ﻭﻤﻨﻪ ﺍﻝﻌﺎﺌﻠﺔ } { 1, x , x 2 ,..., x nﺃﺴﺎﺱ ﻝـ ] . ℝ n [ x
} { u1 ,u 2 } = { 1, iﺃﺴﺎﺱ ﻝﻠﻔﻀﺎﺀ ℂﻋﻠﻰ ﺍﻝﺤﻘل ، ℝ dim ℝ ℂ = 2 ♦ ∞dim ℚ ℂ = +
ﺍﻝﻔﻀﺎﺀ ﺍﻝﺸﻌﺎﻋﻲ ﺍﻝﻤﻨﺘﻪ ﺍﻝﺒﻌﺩ: ﺘﻌﺭﻴﻑ :ﻨﻘﻭل ﻋﻥ ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ﺃﻨﻪ ﻤﻨﺘﻬﻲ ﺍﻝﺒﻌﺩ ﺇﺫﺍ ﻗﺒل ﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓ
ﺃﻤﺜﻠﺔ:
ﻤﻨﺘﻬﻴﺔ . ﻤﻭﻀﻭﻋﺔ K , E :ﻑ.ﺵ ،
*p ∈ℕ
n ≥ 1 ، E = ℝ n (1ﻫﻲ ﻑ.ﺵ ﻤﻨﺘﻪ ﺍﻝﺒﻌﺩ ﻤﻊ . dim ℝ n = n
} S = {u 1 , u 2 ,........, u pﻋﺎﺌﻠﺔ
E = ℝ n [ x ] (2ﺒﻌﺩﻩ ﻤﻨﺘﻬﻲ ﻤﻊ dim ℝ n [ x ] = n + 1
ﺃﺸﻌﺔ ﻤﻥ E
ﺃﺴﺎﺴﻪ } { 1, x , x ,..., x ﻤﻥ ﺃﺠل ] { 1, x , x } : ℝ [ xﺍﻷﺴﺎﺱ ﺍﻝﻘﺎﻨﻭﻨﻲ } {2,3x + 1, x − 2ﺃﺴﺎﺱ ﺁﺨﺭ. n
ﺃ( ﺇﺫﺍ ﻜﺎﻨﺕ Sﺘﻭﻝﺩ Eﻓﺈﻥ ﻜل ﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ ﺒﻬﺎ ﺃﻜﺜﺭ ﻤﻥ p
2
ﻋﻨﺼﺭ ﻓﻬﻲ ﺒﺎﻝﻀﺭﻭﺭﺓ ﻤﻘﻴﺩﺓ.
S ′ﻤﻘﻴﺩﺓ ⇒ Card ( S ′ ) ≥ p + 1
2
2
2
S′⊆E
ﺏ( ﺇﺫﺍ ﻜﺎﻨﺕ Sﺤﺭﺓ ﻓﺈﻥ ﻜل ﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ ﺒﻬﺎ ﺃﻗل ﻤﻥ pﺸﻌﺎﻉ
ﻨﺘﻴﺠﺔ ﻫﺎﻤﺔ:
ﻻ ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﻭﻥ ﻤﻭﻝﺩﺓ .
ﻝﻴﻜﻥ K , Eﻑ.ﺵ ﺒﻌﺩﻩ ﻤﻌﻠﻭﻡ ) ∞= n ≺ +
( dim E
ﻭﻝﺘﻜﻥ Bﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ ﻤﻥ Eﻤﻊ Card B = n
ﻨﻅﺭﻴﺔ :ﻝﻴﻜﻥ K , Eﻑ.ﺵ ﻤﻨﺘﻬﻲ ﺍﻝﺒﻌﺩ )ﻴﻤﻠﻙ ﻤﺠﻤﻭﻋﺔ ﻤﻭﻝﺩﺓ ﻤﻨﺘﻬﻴﺔ( } S = { e1 , e 2 ,...., e mﻤﻭﻝﺩﺓ B = {u 1 , u 2 ,........, u p } ،ﺤﺭﺓ
Bﺃﺴﺎﺱ ⇐ Bﺤﺭﺓ ⇐ Bﻤﻭﻝﺩﺓ .
ﻭﻝﻴﺴﺕ ﻤﻭﻝﺩﺓ ،ﻋﻨﺩﺌﺫ ﻴﻤﻜﻥ ﺇﻜﻤﺎل ﺍﻝﻌﺎﺌﻠﺔ Bﺒﻌﻨﺎﺼﺭ ﻤﻥ S
ﻤﺜﻼB = {(1, 2 ) , ( 3, −1)} , dim E = 2 , E = ℝ 2 :
ﻝﻠﺤﺼﻭل ﻋﻠﻰ ﺃﺴﺎﺱ ﻝـ . E
ﺒﻤﺎ ﺃﻥ Card B = dim Eﻴﻜﻔﻲ ﻹﺜﺒﺎﺕ ﺃﻥ Bﺃﺴﺎﺱ ﺃﻥ ﻨﺩﺭﺱ
ﺍﻻﺴﺘﻘﻼل ﺍﻝﺨﻁﻲ ﻝﻠﺸﻌﺎﻋﻴﻥ )u (1, 2 ) , v ( 3, −1
α u + β v = OE ⇔ α = β = 0
) (α , 2α ) + ( 3β , − β ) = ( 0, 0 α + 3 β = 0 ⇒ ⇒α =β =0 2α − β = 0 www.math.3arabiyate.net
7
]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﻤﻼﺤﻅﺎﺕ: (1ﻝﻴﻜﻥ K , Eﻑ.ﺵ ﺒﻌﺩﻩ nﻤﻨﺘﻬﻲ ،
ﻝﻴﻜﻥ ﺍﻝﻔﻀﺎﺀ ﺍﻹﻀﺎﻓﻲ G
ﺇﻥ ﻜل ﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ ﻤﻜﻭﻨﺔ ﺃﻜﺜﺭ ﻤﻥ nﺸﻌﺎﻉ ﻫﻲ ﺒﺎﻝﻀﺭﻭﺭﺓ
Gﺇﻀﺎﻓﻲ ﻝـ E = F ⊕ G ⇔ F dim E = dim F + dim G ⇒ dim G = dim E − dim F = 2 − 1 = 1 ﻭﻤﻨﻪ dim G = 1
ﻤﺭﺘﺒﻁﺔ ﺨﻁﻴﺎ . ﻭﻤﻨﻪ ﺍﻷﺴﺎﺱ ﻋﺒﺎﺭﺓ ﻋﻥ ﻋﺎﺌﻠﺔ ﺤﺭﺓ ﺃﻋﻅﻤﻴﺔ. (2ﻜل ﻋﺎﺌﻠﺔ ﺒﻬﺎ ﺃﻗل ﻤﻥ nﺸﻌﺎﻉ ﻻ ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﻭﻥ ﻤﻭﻝﺩﺓ ،
ﺇﺫﻥ Gﻤﻭﻝﺩ ﺒﺸﻌﺎﻉ ﻭﺍﺤﺩ
ﻓﺈﻥ ﻜل ﺃﺴﺎﺱ ﻓﻬﻭ ﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓ ﻭﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓ ﺃﺼﻐﺭﻴﺔ ،
ﺒﺤﻴﺙ wﻤﺴﺘﻘل ﺨﻁﻴﺎ ﻤﻊ . u
ﺒﻤﻌﻨﻰ ﺃﻥ ﺃﻗل ﻋﺩﺩ ﻤﻥ ﺍﻷﺸﻌﺔ ﺍﻝﻤﻭﻝﺩﺓ ﻫﻭ . n
ﻝﻨﺤﺼل ﻋﻠﻰ } { u ,wﺃﺴﺎﺱ ﻝـ E = ℝ 2
) w ( 0,1) , u (1, 2
ﺒﻌﺩ ﻑ.ﺵ.ﺝ: ﻝﻴﻜﻥ K , Eﻑ.ﺵ ﺒﻌﺩﻩ nﻤﻨﻬﻲ
) , F (1, 2
) ∞( dim E ≺ +
F ⊂ Eﻑ.ﺵ.ﺝ ﻤﻥ E
1) dim F ≤ dim E 2) dim F = dim E ⇔ E = F G , Fﻑ.ﺵ.ﺝ ﻤﻥ E
ﻨﻌﻠﻡ ﺃﻥ F + Gﻫﻭ ﺃﻴﻀﺎ ﻑ.ﺵ.ﺝ
) dim ( F + G ) = dim F + dim G − dim ( F ∩ G ﺘﻁﺒﻴﻕ K , E :ﻑ.ﺵ ﺒﻌﺩﻩ ﻤﻨﺘﻪ G , Fﻑ.ﺵ.ﺝ ﻤﻨﻪ ﺤﻴﺙ . E = F ⊕ G } 1) F ∩ G = { O E E = F ⊕G ⇔ 2) E = F + G } F ∩ G = {O E
) ⇔ dim E = dim F + dim G − dim ( F ∩ G ⇔ dim E = dim F + dim G ﻤﺜﺎلF = {( x , y ) / 2x − y = 0} :
♦ ﺘﻌﻴﻴﻥ ﺃﺴﺎﺱ ﻝـ : F u = ( x , y ) = ( x , 2x ) , x ∈ ℝ = x (1, 2 ) , x ∈ ℝ
ﺇﺫﻥ ) u1 = (1, 2ﻴﻭﻝﺩ ، Fﻭ ﻤﻨﻪ } { uﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓ ﻭﻫﻲ ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ
≠ O ℝ2 ) G = w
) (u ≠ O ℝ2
ﻭﻤﻨﻪ } { uﺃﺴﺎﺱ ﻝـ Fﺇﺫﻥ . dim F = 1
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8
)( 0,1
.G
(w
â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
( ( 0, 0, 0 ) ) = ( 0,1) ≠ ( 0, 0 ) = O
ﺍﻝﺘﻁﺒﻴﻘﺎﺕ ﺍﻝﺨﻁﻴﺔ
ℝ2
. ﻝﻴﺱ ﺨﻁﻲf ﺇﺫﻥ
ϕ : ℝ [x ] → ℝ [x ]
. ﺘﻁﺒﻴﻕf : E → F ﻭﻝﻴﻜﻥ، ﺵ. ﻑK F , E ﻝﻴﻜﻥ
(3
:ﺘﻌﺭﻴﻑ
p → ϕ ( p ) = p′
⇔ ﺨﻁﻲf
ϕ ( p + q ) = ( p + q )′ = p ′ + q ′
f ∀u ,v ∈ E , ∀α ∈ K : f
ϕ (α p ) = (α p )′ = α p ′
( u + v ) = f (u ) + f (v ) ( α u ) = α f (u )
:ﺘﻌﺭﻴﻑ ﻤﻜﺎﻓﺊ ∀u ,v ∈ E , ∀α , β ∈ K : f (α u + β v ) = α f (u ) + β f (v )
E = ℓ ( [ a , b ] , ℝ ) , F = ℝ (4
ϕ :E → ℝ f → ϕ (f
:ﺘﻌﺭﻴﻑ ﻤﻜﺎﻓﺊ
b
) = ∫a f (t ) dt
∀u ,v ∈ E , ∀α ∈ K : f (α u +v ) = α f (u ) + f (v )
ﺨﻁﻲϕ
( f (O ) = O ) ⇐ ( ﺨﻁﻲf ) :ﺨﺎﺼﻴﺔ . ( ﻝﻴﺱ ﺨﻁﻲf ) ⇐ ( f (O ) ≠ O ) :ﻋﻜﺱ ﺍﻝﻨﻘﻴﺽ E
b
ϕ (α f + g ) = ∫ α f (t ) + g (t ) dt a
b
b
a
a
F
E
= α ∫ f (t ) dt + ∫ g (t ) dt
F
:ﺘﺭﻤﻴﺯ F ﻭE ( ﻨﺭﻤﺯ ﻝﻤﺠﻤﻭﻋﺔ ﺍﻝﺘﻁﺒﻴﻘﺎﺕ ﺍﻝﺨﻁﻴﺔ ﺒﻴﻥ ﺍﻝﻔﻀﺎﺌﻴﻥ1
ℓ ( E , F ) ﺒـ
. ﺨﻁﻲf : E → F ﻝﻴﻜﻥ:ﺨﻭﺍﺹ . F ﺝ ﻤﻥ.ﺵ. ﻑf ( E 1 ) ⇐ E ﺝ ﻤﻥ.ﺵ. ﻑE 1 . E ﺝ ﻤﻥ.ﺵ. ﻑf
−1
( F1 ) ⇐ F
ﺝ ﻤﻥ.ﺵ. ﻑF1
ﻤﺞ ﺍﻝﺘﻤﺎﺘﻼﺕℓ ( E , E ) = ℓ ( E ) : E = F ( ﺇﺫﺍ ﻜﺎﻥ2
• •
.ﺍﻝﺩﺍﺨﻠﻴﺔ E ﻤﺞ ﺍﻷﺸﻜﺎل ﺍﻝﺨﻁﻴﺔ ﻋﻠﻰℓ ( E , K ) = E ′ : F = K
. ﻤﺞ ﺍﻝﺘﺸﺎﻜﻼﺕ ﺍﻝﺩﺍﺨﻠﻴﺔ: GL ( E ) (4
:ﻨﻭﺍﺓ ﻭﺼﻭﺭﺓ ﺘﻁﺒﻴﻕ ﺨﻁﻲ
. ﻤﺘﻘﺎﺒل+ ﺍﻝﺘﺸﺎﻜل ≡ ﺨﻁﻲ
ﺨﻁﻲf : E → F
:ﺃﻤﺜﻠﺔ
ﺍﻝﻤﻌﺭﻑ ﺒـE ﺍﻝﺠﺯﺀ ﻤﻥf ﻨﺴﻤﻲ ﻨﻭﺍﺓ: f ﺘﻌﺭﻴﻑ ﻨﻭﺍﺓ ker f = f
−1
({O }) = {x ∈ E F
/ f (x ) = OF }
. E ﺝ ﻤﻥ.ﺵ.ﺍﻝﻨﻭﺍﺓ ﻫﻲ ﻑ
•
. ker f ≠ φ ⇐ f (O E ) = O F
•
، f 1 : ℝ2 → ℝ2
. ﺨﻁﻲf
(1
(x , y ) → (x + y , y − x ) u ( x , y ) , y ( x ′, y ′ ) , α ∈ ℝ f (α u + v ) = f
( (α x + x ′, α y + y ′) )
= (α x + x ′ + α y + y ′, α y + y ′ − α x − x ′ )
ﺍﻝﻤﻌﺭﻑ ﺒـF ﺍﻝﺠﺯﺀ ﻤﻥf ﻨﺴﻤﻲ ﺼﻭﺭﺓ: f ﺘﻌﺭﻴﻑ ﺼﻭﺭﺓ
= (α x + α y , α y − α x ) + ( x ′ + y ′, y ′ − x ′ )
Im f = f ( E ) = {f ( x ) , x ∈ E }
= α ( x + y , y − x ) + ( x ′ + y ′, y ′ − x ′ )
= {y ∈ F / ∃ x ∈ E :f ( x ) = y } . F ﺝ ﻤﻥ.ﺵ. ﻫﻲ ﻑIm f
(3
= α f (u ) + f (v )
•
f : ℝ3 → ℝ 2
(2
( x , y , z ) → ( x + y − z ,1) 9
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â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
f : ℝ 2 → ℝ 2 :ﻤﺜﺎل
( x , y ) → ( 0, x ) f ∈L
( ﺨﻁﻲf ) f ∈ L
(E , F )
2
2
ﻝﻴﻜﻥ
2
ker f = {O E } ⇔ ﻤﺘﺒﺎﻴﻥf -1
= {( 0, y ) , y ∈ ℝ}
Im f = F ⇔ ﻏﺎﻤﺭf -2
= { y ( 0,1) , y ∈ ℝ}
F ﺘﺒﻘﻰ ﺤﺭﺓ ﻓﻲE ﻤﺘﺒﺎﻴﻥ ⇔ ﺼﻭﺭﺓ ﻜل ﻋﺎﺌﻠﺔ ﺤﺭﺓ ﻤﻥf -3
= ( 0,1)
ﻫﻲ ﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓE ﻏﺎﻤﺭ ⇔ ﺼﻭﺭﺓ ﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓ ﻝـf -4
{ = {( u ,v ) ∈ ℝ = {( u ,v ) ∈ ℝ = {( u ,v ) ∈ ℝ
}
Im f = y = ( u ,v ) ∈ ℝ 2 / ∃ ( x , y ) ∈ ℝ 2 : ( 0, x ) = ( u ,v )
Im f ﻝـ
ﻓﺈﻥE { ⇔ ﺘﻭﻝﺩu1 ,u 2 ,........,u n } Im f = f (u 1 ) , f (u 2 ) ,...., f (u n )
:ﻨﻅﺭﻴﺔ ﺍﻝﺒﻌﺩ
/ ∃ ( x , y ) ∈ ℝ 2 : ( 0, x ) = ( u ,v )
2
/ ∃( x , y ) ∈ ℝ 2
2
/ u = 0 v ∈ℝ
}
= ( 0,1)
: ﻝﺩﻴﻨﺎ، f ∈ L ( E , F ) dim E = dim ker f + dim Im f
} :u = 0 ∧v = x }
2
= {( 0,v ) , v ∈ ℝ }
ﻭﻝﻴﻜﻥ، ﻤﻨﺘﻪdim E = n ﺵ ﻤﻊ. ﻑF ﻭE ﻝﻴﻜﻥ
ker f ∩ Im f = ( 0,1) f : ℝ3 → ℝ 2
ﻤﻨﺘﻬﻲdim E = dim F ﻤﻊf ∈ L ( E , F ) :ﻨﺘﻴﺠﺔ ( ﺘﻘﺎﺒﻠﻲ )ﺘﺸﺎﻜلf ⇔ ﻏﺎﻤﺭf ⇔ ﻤﺘﺒﺎﻴﻥf ker f
dim E = n . F ﻭE ⇐ ﻻ ﻴﻭﺠﺩ ﺃﻱ ﺘﺸﺎﻜل ﺒﻴﻥdim F = m n ≠ m
. ﺍﻝﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎS ﺃﻜﺒﺭ ﻋﺩﺩ ﻤﻥ ﺃﺸﻌﺔS ﻨﺴﻤﻲ ﺭﺘﺒﺔ
ker f = (1, −1, 0 ) dim ker f = 1 ﺃﺴﺎﺱ ﻝﻠﻨﻭﺍﺓ ﻭ
) = dim ( Im f )
10
{(1, −1, 0 )}
Im f = {(u ,v ) ∈ ℝ 2 / ∃ ( x , y , z ) ∈ ℝ 3 , f ( x , y , z ) = (u ,v )}
:( ﺭﺘﺒﺔ ﺘﻁﺒﻴﻕ ﺨﻁﻲ2 ﺒﻌﺩ ﺼﻭﺭﺘﻪf : E → F ﻨﺴﻤﻲ ﺭﺘﺒﺔ ﺘﻁﺒﻴﻕ ﺨﻁﻲ
(x , y , z ) → (x + y − z , x + y + z ) = {( x , y , z ) / x + y + z = 0 ∧ x + y − z = 0}
= x (1, −1, 0 ) , x ∈ ℝ
:( ﺭﺘﺒﺔ ﺠﻤﻠﺔ ﺃﺸﻌﺔ1 ، E ﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ ﻤﻥS = { u1 ,u 2 ,........,u n } ، ﺵ. ﻑE
:ﻤﺜﺎل
y = −x ⇒ 2(x + y ) = 0 ⇒ x + y = 0 ⇒ z = 0 u ( x , −x , 0 ) ∈ ker f , x ∈ ℝ
: Rang ﺍﻝﺭﺘﺒﺔ
rg ( f
2
{( x , y ) ∈ ℝ / f ( x , y ) = ( 0, 0 )} = {( x , y ) ∈ ℝ / ( 0, x ) = ( 0, 0 )} = {( x , y ) ∈ ℝ / x = 0 , ∀y }
ker f =
:ﺨﻭﺍﺹ ﺍﻝﺼﻭﺭﺓ ﻭﺍﻝﻨﻭﺍﺓ ﻭﻋﻼﻗﺘﻬﺎ ﺒﺎﻝﺘﻁﺒﻴﻕ
(ℝ )
x z
x + y − z = u = (u ,v ) ∈ ℝ 2 / ∃ ( x , y , z ) ∈ ℝ 3 , x + y + z = v u +v u v +y = x = ,y = 2 2 2 v −u v −u = z = 2 2
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â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
:ﻤﺜﺎل
⇒ f (u ) = u − g (u )
ϕ : ℝ 3 [x ] → ℝ
( g (u ) = 0 ) ﻷﻥ
p → p ( 2)
⇒ u ∈ Im f
K = ℝ ﻭE ﺨﻁﻲ ﺒﻴﻥϕ ⇔ ϕ ∈ ( ℝ 3 [ x ])
*
Im f ⊂ ker g
•
. ﺨﻁﻲp ﺇﺫﻥ
∃u ∈ E : f (u ) = v ⇐ v ∈ Im f ﻝﻴﻜﻥ ⇒ g ( f (u ) ) = g (v )
⇒ (g f
: ker ϕ ♦ ﺘﻌﻴﻴﻥ
ker ϕ = { p ∈ E / ϕ ( p ) = 0}
)(u ) = g (v )
= { p ∈ E / ϕ ( 2 ) = 0}
g f = 0 ﺒﻤﺎ ﺃﻥ
p (x ) ∈ E ⇔ p (x ) = a x 3 +b x + c x + d
⇒ g (v ) = 0 v ∈ ker g
p ( 2) = 8 a + 4b + 2c + d
ﺃﻱ
p ∈ ker ϕ ⇔ p ( 2 ) = 0 ⇔ 8 a + 4 b + 2 c + d = 0 ⇔ d = −8 a − 4 b − 2 c
. ker g = Im f
:ﻭﻤﻨﻪ
p ( x ) = a x 3 + b x 2 + c x + ( −8 a − 4 b − 2 c )
(
) (
)
= a x 3 − 8 + b x 2 − 4 + c ( x − 2) , a,b ,c ∈ ℝ
: rg ( f ) + rg ( g ) = n ﺘﺒﻴﻴﻥ ﺃﻥ-2
ker ϕ = x 3 − 8, x 2 − 4, x − 2
:ﻝﺩﻴﻨﺎ
(ﻭﻫﺫﻩ ﺍﻷﺸﻌﺔ ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ )ﺒﺩﺭﺠﺎﺕ ﻤﺨﺘﻠﻔﺔ ﻤﺜﻨﻰ ﻤﺜﻨﻰ
n = dim E = dim ker g + dim Im g
dim ker ϕ = 3
= dim Im f + dim Im g
: Im ϕ ♦ ﺘﻌﻴﻴﻥ
= rg ( f ) + rg ( g )
dim ( Im ϕ ) = dim E − dim ( ker ϕ ) = 4 − 3 = 1 dim ( Im ϕ ) = dim ℝ = 1 ﻭ، ﺝ.ﺵ. ﻑIm ϕ ⊆ ℝ ﺒﻤﺎ ﺃﻥ Im ϕ = ℝ : ﻓﺈﻥ
:ﺘﻤﺭﻴﻥ
( dim E
= n ) ﺵ ﺒﻌﺩﻩ ﻤﻨﺘﻬﻲ. ﻑE ﻝﻴﻜﻥ
. g f = 0 , f + g = id E : ﺒﺤﻴﺙf , g ∈ L
(E )
ﻭﻝﻴﻜﻥ
. ker g = Im f : ﺒﺭﻫﻥ ﺃﻥ-1 . rg ( f ) + rg ( g ) = n : ﺒﺭﻫﻥ ﺃﻥ-2
:ﺍﻝﺤل : ker g = Im f ﺘﺒﻴﻴﻥ ﺃﻥ-1 ker g ⊂ Im f •
g (u ) = 0 ⇐ u ∈ ker g ﻝﻴﻜﻥ id (u ) = u ﻝﻜﻥ ⇒ ( f + g )(u ) = u
11
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]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﺍﻝﻤﺼﻔﻭﻓﺎﺕ ﻝﻴﻜﻥ Kﺤﻘﻼ ﺘﺒﺩﻴﻠﻲ ( K = ℂ ∨ K = ℝ ) ، ﺘﻌﺭﻴﻑ :ﻨﺴﻤﻲ ﻤﺼﻔﻭﻓﺔ ﻤﻥ ﺍﻝﻨﻭﻉ ) ( m , nﺃﻭ ) ( m × n ﺘﻁﺒﻴﻕ ﻤﻥ } E = {1, 2,3,...., m } × {1, 2,3,...., nﻨﺤﻭ
ﻨﺭﻤﺯ ﻝﻤﺠﻤﻭﻋﺔ ﺍﻝﻤﺼﻔﻭﻓﺎﺕ ﻤﻥ ﺍﻝﻨﻭﻉ m × nﺒﻤﻌﺎﻤﻼﺕ ﻤﻥ K
ﻜل . K
ﺒﺎﻝﺭﻤﺯ : ﻤﺜﻼ:
1 2 A • ) ∈ Μ 2×2 ( ℝ 3 6 2 5 3 7 B • ) ∈ Μ 2×4 ( ℝ − 1 2 0 − 1 3 0 i • ) C i + 1 j j 2 ∈ Μ 3×3 ( ℂ 1 −3 0 2π i 2π 2π = e 3 = cos + i sin j 3 =1 3 3
ﻨﺭﻤﺯ ﻝﻠﻤﺼﻔﻭﻓﺔ ﺒـ . M , C , B , A A : E →K
( i , j ) → ai j ﻨﻜﺘﺏ ﻋﻨﺎﺼﺭ ﺍﻝﻤﺼﻔﻭﻓﺔ ﻋﻠﻰ ﺸﻜل ﺠﺩﻭل
a11 a12 a13 .......... a1n a21 a22 a23 .......... a2 n A ⋮ am 1 am 2 am 3 .......... amn ai jﺍﻝﻌﻨﺼﺭ ﺍﻝﻤﻭﺠﻭﺩ ﻓﻲ ﺍﻝﻤﺼﻔﻭﻓﺔ Aﻓﻲ ﺍﻝﻤﻭﻀﻊ : ﺘﻘﺎﻁﻊ ﺍﻝﺴﻁﺭ iﻤﻊ ﺍﻝﻌﻤﻭﺩ . j : iﺭﻗﻡ ﺍﻝﺴﻁﺭ ،
: jﺭﻗﻡ ﺍﻝﻌﻤﻭﺩ .
ﻋﺩﺩ ﺍﻷﺴﻁﺭ ﻫﻭ ، mﻋﺩﺩ ﺍﻷﻋﻤﺩﺓ ﻫﻭ . n
• •
• •
1 A ﻤﻥ ﺍﻝﻨﻭﻉ . 2 × 2 2 1 3 B ﻤﻥ ﺍﻝﻨﻭﻉ . 2 × 3 0 −1
3 − 1 2 3 1 C 2 ﻤﻥ ﺍﻝﻨﻭﻉ . 3 × 1 0 ) D (1, 2, −5ﻤﻥ ﺍﻝﻨﻭﻉ . 1× 3
•
H = 5ﻤﻥ ﺍﻝﻨﻭﻉ . 1× 1
•
1 i 0 M iﻤﻥ ﺍﻝﻨﻭﻉ . 3 × 3 i + 1 −1 1 1 i 2
ﻨﺴﻤﻲ ﺍﻝﻤﺼﻔﻭﻓﺔ ﺍﻝﻤﻌﺩﻭﻤﺔ ﻓﻲ ) Μ m ×n ( Kﺍﻝﻤﺼﻔﻭﻓﺔ
)
(
O α i jﺒﺤﻴﺙ ∀i , j : α i j = O k
0 ﻤﺜﺎل = : 0
O Μ 2×1
0 0 = 0 0 0 0
,
O Μ3×2
ﻤﻨﻘﻭل ﻤﺼﻔﻭﻓﺔ: ﻝﺘﻜﻥ ) ∈ Μ m ×n ( K
)
1≤i ≤ m 1≤ j ≤ n
(
A αi j
ﻨﺴﻤﻲ ﻤﻨﻘﻭل Aﺍﻝﻤﺼﻔﻭﻓﺔ t Aﻭﺍﻝﻤﻌﺭﻓﺔ ﻜﻤﺎ ﻴﻠﻲ: ﻤﺼﻔﻭﻓﺔ ﻤﻥ ﺍﻝﻨﻭﻉ n × mﻭﺍﻝﻨﺎﺘﺠﺔ ﺒﺘﺒﺩﻴل ﺃﺴﻁﺭ ﻭﺃﻋﻤﺩﺓ A
ﻤﺜﺎل: 3 ♦ 7
7 ♦ 0
−1 1 ←A 7 −1 2 1 3 5 ← B 2 5 0
1 ♦ C (1, 2 ) ← C 2
t
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j
ﺍﻝﻤﺼﻔﻭﻓﺔ ﺍﻝﻤﻌﺩﻭﻤﺔ ﻓﻲ ) Μ m ×n ( K
1≤ i ≤ m 1≤ j ≤ n
ﺃﻤﺜﻠﺔ:
) Μ m ×n ( K
12
1 A 3 1 t B 3 7 t
â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
−4 8 2 7
:ﺨﻭﺍﺹ t
(A + B ) = t A + tB t (λ A ) = λ . t A t
: ﺍﻝﻀﺭﺏ ﻓﻲ ﺴﻠﻤﻴﺔ-2
( )
A ai j
1≤ i ≤ m 1≤ j ≤ n
t
( A .B ) = t B .t A
.4
ﻤﺭﺒﻌﺔ ﺇﺫﺍ ﻜﺎﻥ ﻋﺩﺩ ﺃﺴﻁﺭﻫﺎ ﻴﺴﺎﻭﻱ ﻋﺩﺩA ﻨﻘﻭل ﻋﻥ ﻤﺼﻔﻭﻓﺔ
:ﻤﺜﺎل
. ( n = m ) ﺃﻋﻤﺩﺘﻬﺎ
9 −3 3 −1 3A ← A , λ =3 6 15 2 5
(ﺃﺜﺭ ﻤﺼﻔﻭﻓﺔ )ﺨﺎﺹ ﺒﺎﻝﻤﺼﻔﻭﻓﺔ ﺍﻝﻤﺭﺒﻌﺔ
:ﺨﻭﺍﺹ
n × n ﻤﺭﺒﻌﺔ ﻤﻥ ﺍﻝﻨﻭﻉA ∈ Μ n ( K ) ﻝﺘﻜﻥ
ﺍﻝﻤﺯﻭﺩ ﺒﺎﻝﺠﻤﻊ ﻭﺍﻝﻀﺭﺏ ﻓﻲ ﺴﻠﻤﻴﺔ ﻝﻬﺎ ﺒﻨﻴﺔΜ m ×n ( K ) ﺇﻥ
( )
A = ai j
: ﻭﺃﺴﺎﺴﻪ ﺍﻝﻘﺎﻨﻭﻨﻲ، m × n ﺒﻌﺩﻩ ﻫﻭK ﺵ ﻋﻠﻰ.ﻑ 1 0 ⋯ 0 0 1 ⋯ 0 0 0 ⋯ 0 0 0 ⋯ 0 0 0 ⋯ 0 0 0 ⋯ 0 E1 , E2 , ⋯ , E mn ⋮ ⋮ ⋮ 0 0 0 0 0 0 0 0 0 0 0 1
a b a, b , c , d ∈ ℝ = Μ 2×2 ( ℝ ) :ﻤﺜﺎل c d dim ( Μ 2×2 ( ℝ ) ) = 4 1 0 0 1 0 0 0 0 E1 , E2 , E3 , E4 0 0 0 0 1 0 0 1
n
Tr ( A ) = a11 + a22 + a33 + ......... + ann = ∑ akk
: ﺍﻝﺠﻤﻊ-1 : ﺤﻴﺙA , B ∈ Μ m ×n ( K ) ﻝﺘﻜﻥ
ﻭﺍﻝﻤﻌﺭﻓﺔC = A iB ﺍﻝﻤﺼﻔﻭﻓﺔB ﻭA ﻨﺴﻤﻲ ﺠﺩﺍﺀ:ﺘﻌﺭﻴﻑ
( )
. ﻤﺠﻤﻭﻋﺔ ﻋﻨﺎﺼﺭ ﻗﻁﺭﻫﺎA ﻨﺴﻤﻲ ﺃﺜﺭ ﺍﻝﻤﺼﻔﻭﻓﺔ
ﻋﻤﻠﻴﺎﺕ ﻋﻠﻰ ﺍﻝﻤﺼﻔﻭﻓﺎﺕ
A ∈ Μ k ×l ( K ) , B ∈ Μ l ×n ( K ) ﻝﺘﻜﻥ k × n ﻤﻥ ﺍﻝﻨﻭﻉC c i j
1≤i , j ≤ n
k =1
: ﺠﺩﺍﺀ ﻤﺼﻔﻭﻓﺘﻴﻥ-3
)
t ( A ) = A .3
:ﺍﻝﻤﺼﻔﻭﻓﺔ ﺍﻝﻤﺭﺒﻌﺔ
1≤ j ≤ n
(
.2
∈ Μ m ×n ( K ) , λ ∈ K
λ A = λ ( ai j )1≤i ≤ m ﻨﻌﺭﻑ:ﺘﻌﺭﻴﻑ
C ∈ Μk n (K )
.1
( )
A ai j
: ﺒـ
1≤ i ≤ m 1≤ j ≤ n
( )
, B bi j
1≤ i ≤ m 1≤ j ≤ n
ﺍﻝﻤﻌﺭﻓﺔ ﻜﻤﺎC ﺍﻝﻤﺼﻔﻭﻓﺔB ﻭA ﻨﺴﻤﻲ ﻤﺠﻤﻭﻉ ﺍﻝﻤﺼﻔﻭﻓﺘﻴﻥ
l
C i j = ∑ ai k i b k j
( )
∀i , j : c i j = ai j + bi j ﺤﻴﺙC c i j
k =1
j ﻓﻲ ﺍﻝﻌﻤﻭﺩA ﻤﻥ ﺍﻝﻤﺼﻔﻭﻓﺔi ﻫﻭ ﻨﺎﺘﺞ ﻀﺭﺏ ﺍﻝﺴﻁﺭc i j
1≤i ≤ m 1≤ j ≤ n
: ﻴﻠﻲ
:ﻤﺜﺎل
. ﻤﻌﺎﻤﻼ ﻤﻌﺎﻤﻼB ﻤﻥ
1 −2 5 −2 1 5 0 −2 5 2 A ,B ,D ,F 3 7 0 1 3 2 7 −1 1 0
1 7 0 7 1 3 C −2 6 , B , A :ﻤﺜﺎل −1 3 2 4 5 1
13
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]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﻜل ﻤﻥ B iAﻭ A iBﻭ C iAﻤﻌﺭﻓﺔ ،
1 −1 0 A 2 6 1 3 5 7
A iCﻏﻴﺭ ﻤﻌﺭﻓﺔ .
0 7 3 −1 1 3 1× 0 + 3 × ( −1) 1× 7 + 3 × 3 A iB = 2 4 2 × 0 + 4 × ( −0 ) 2 × 7 + 4 × 2 −3 10 = −4 26
1 = ( 42 − 3 + 0 ) − ( 0 + 5 − 14 ) = 48 7
6 5
det ( A ) = 2 3
ﺤﺴﺎﺏ ﻤﻘﻠﻭﺏ ﻤﺼﻔﻭﻓﺔ:
ﺨﺎﺼﻴﺔ: ♦
1 −1 0
A i( B iC ) = A iB + A iC
ﺘﻌﺭﻴﻑ :ﻝﺘﻜﻥ ) A ∈ Μ n ( K
( B iC )iA = B iA + C iA
ﻨﻘﻭل ﻋﻥ Aﺃﻨﻬﺎ ﻗﺎﺒﻠﺔ ﻝﻠﻘﻠﺏ ﺇﺫﺍ ﻭﺠﺩﺕ A ′ ∈ Μ nﺒﺤﻴﺙ:
♦ ﺤﺎﻝﺔ ﺍﻝﻤﺼﻔﻭﻓﺔ Aﻤﺭﺒﻌﺔ ﻤﻥ ﺍﻝﻨﻭﻉ n × n A n = A n −1 iA ,............, A 3 = A 2 iA , A 2 = A iA
A iA ′ = A ′iA = I n
ﺤﻴﺙ I nﻫﻲ ﻤﺼﻔﻭﻓﺔ ﺍﻝﻭﺤﺩﺓ ﻓﻲ ﺍﻝﻔﻀﺎﺀ Μ n ×nﻭﻫﻲ ﻤﻌﺭﻓﺔ ﺒـ: 1 0 ⋯ 0 0 1 0 In ⋮ ⋱ 0 ⋯ ⋯ 1
ﺤﺴﺎﺏ ﺍﻝﻤﺤﺩﺩﺍﺕ ﻤﻥ ﺭﺘﺒﺔ ﺩﻨﻴﺎ : ﺤﺎﻝﺔ ﺍﻝﻤﺼﻔﻭﻓﺎﺕ ﻤﻥ ﺍﻝﻨﻭﻉ 2x2 a b A ﻝﺘﻜﻥ ) ∈ Μ 2×2 ( K c d ﻨﺴﻤﻲ ﻤﺤﺩﺩ Aﺍﻝﻌﺩﺩ = ab − bc
ﻤﺜﺎل: a b c d
= ) det ( A
1 −5 3 5 A = , B ﻤﺜﺎل : 3 7 −2 13 1 −5 = ) det ( A = 22 , det ( B ) = 49 3 7
ﺤﺎﻝﺔ ﺍﻝﻤﺼﻔﻭﻓﺎﺕ ﻤﻥ ﺍﻝﻨﻭﻉ 3x3 a11 a12 a13 ﻝﺘﻜﻥ ) A a21 a22 a23 ∈ Μ 3×3 ( K a 31 a32 a33 ﻨﺴﻤﻲ ) det ( Aﺍﻝﻌﺩﺩ ﺍﻝﻤﻌﺭﻑ ﺒـ:
1 0 0 1 0 ، I 2 ﻓﻲ I 3 0 1 0 : Μ 3×3 ﻓﻲ : Μ 2×2 0 1 0 0 1 ﻨﺭﻤﺯ ﻝﻠﻤﻘﻠﻭﺏ ﺇﻥ ﻭﺠﺩ ﺒـ A −1 ﺸﺭﻁ ﻻﺯﻡ ﻭﻜﺎﻓﻲ ﻝﻭﺠﻭﺩ A −1
) A −1ﻤﻭﺠﻭﺩﺓ(
⇔ ) ( det ( A ) ≠ 0
ﻗﺎﻋﺩﺓ ﺤﺴﺎﺏ ﻤﻘﻠﻭﺏ ﻤﺼﻔﻭﻓﺔ: -1ﺍﻝﺘﺄﻜﺩ ﻤﻥ ﺃﻥ . det ( A ) ≠ 0 -2ﺘﻌﻴﻴﻥ ﻤﺼﻔﻭﻓﺔ ﺍﻝﻤﻜﻤﻤﺎﺕ ﺍﻝﺠﺒﺭﻴﺔ ﺍﻝﻤﻘﺎﺒﻠﺔ ﻨﺭﻤﺯ ﻝﻬﺎ ﺒـ . Aɶ 1 t ɶ = A −1 A -3 ) det ( A
) det ( A ) = ( a11 a22 a33 + a12 a23 a31 + a13 a21 a32 ) − ( a31 a22 a13 + a32 a23 a11 + a33 a21 a12
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)= ℂ ∨ ℝ
(K
14
]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﺘﻌﻤﻴﻡ ﺤﺴﺎﺏ ﺍﻝﻤﺤﺩﺩﺍﺕ:
ﺃ -ﺤﺎﻝﺔ ﻤﺼﻔﻭﻓﺔ A ∈ Μ 2×2 a b ɶ d −c → t Aɶ = d −b A = →A = c d −b a −c a d −b 1 d −b 1 = A −1 = det ( A ) −c a a d − b c −c a
ﺤﺴﺎﺏ ﻤﺤﺩﺩ ﻤﻥ ﺍﻝﺭﺘﺒﺔ
a21 a22 a23 .......... a2 n ⋮ ai 1 ai 2 ai 3 .......... ain an 1 an 2 an 3 .......... an n
ﺤﺴﺎﺏ ) det ( Aﻴﺘﻡ ﻭﻓﻕ ﻁﺭﻴﻘﺘﻴﻥ:
ﺃ -ﻨﺸﺭ ﺍﻝﻤﺤﺩﺩ ﻭﻓﻕ ﺴﻁﺭ: ﻨﺸﺭ ﻤﺤﺩﺩ Aﻭﻓﻕ ﺍﻝﺴﻁﺭ i 0ﻴﻌﻁﻰ ﺒﺎﻝﻘﺎﻋﺩﺓ:
ai 0 j ∆i 0 j
) (
: aɶi , j ، Aɶ = aɶi jﻴﺴﻤﻰ ﺍﻝﻤﻜﻤﻡ ﺍﻝﺠﺒﺭﻱ ﻝـ ai j
∆i j
= ) det ( A
⋮
a11 a12 a13 ﻝﺘﻜﻥ A a21 a22 a23 a 31 a32 a33 .1ﺍﻝﺘﺄﻜﺩ ﻤﻥ ﺃﻥ . det ( A ) ≠ 0 .2ﻨﺤﺴﺏ ﻤﺼﻔﻭﻓﺔ ﺍﻝﻤﻜﻤﻤﺎﺕ Aɶ 1≤i , j ≤3
n
a11 a12 a13 .......... a1n
ﺏ -ﺤﺎﻝﺔ ﻤﺼﻔﻭﻓﺔ A ∈ Μ 3×3
i+j
) ( A ∈ Μ n ×n
)aɶi j = ( −1
n
det ( A ) = ∑ ( −1) 0
i +j
j =1
ﺤﻴﺙ : ∆ i 0 jﺍﻝﻤﺤﺩﺩ ﺍﻝﻨﺎﺘﺞ ﻤﻥ ﻤﺤﺩﺩ Aﺒﺤﺫﻑ ﺍﻝﺴﻁﺭ i 0ﻭﺍﻝﻌﻤﻭﺩ j
ﺤﻴﺙ : ∆ i jﺍﻝﻤﺤﺩﺩ ﺍﻝﻨﺎﺘﺞ ﻤﻥ ﻤﺤﺩﺩ Aﺒﺤﺫﻑ ﺍﻝﺴﻁﺭ iﻭﺍﻝﻌﻤﻭﺩ j
ﺏ -ﻨﺸﺭ ﺍﻝﻤﺤﺩﺩ ﻭﻓﻕ ﻋﻤﻭﺩ:
ﻤﺜﻼ:
ﻨﺸﺭ ﻤﺤﺩﺩ Aﻭﻓﻕ ﺍﻝﻌﻤﻭﺩ j 0ﻴﻌﻁﻰ ﺒﺎﻝﻘﺎﻋﺩﺓ: a23
a22
a33
a32
a12
a11
a32
a32
= ∆11 = ∆ 23
ai j 0 ∆i j 0
)aɶ11 = ( −1
1+1
2 +3
)aɶ23 = ( −1
1 t ɶ -1ﻨﻁﺒﻕ ﺍﻝﻘﺎﻋﺩﺓ A : ) det ( A
= A −1
i + j0
n
)det ( A ) = ∑ ( −1 i =1
ﺤﻴﺙ : ∆ i j 0ﺍﻝﻤﺤﺩﺩ ﺍﻝﻨﺎﺘﺞ ﻤﻥ ﻤﺤﺩﺩ Aﺒﺤﺫﻑ ﺍﻝﺴﻁﺭ iﻭﺍﻝﻌﻤﻭﺩ j 0
ﻤﺜﺎل:
1 2 3 1 2 3 A 3 0 1 , A = 3 0 1 −1 1 3 −1 1 3 ﻨﻨﺸﺭﻩ ﻭﻓﻕ ﺍﻝﺴﻁﺭ : i 0 = 2 a2 j ∆ 2 j
2+ j
3
)det ( A ) = ∑ ( −1 j =1
3
1
−1 3
a22
2+ 2
)+ ( −1 2
2 3 1 3 1
−1 1
a23
a21 2+3
2 +1
)= ( −1
)+ ( −1
= −9 + 0 − 3 = −12
www.math.3arabiyate.net
15
â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
g : ℝ 2 → ℝ3
:ﻤﺼﻔﻭﻓﺔ ﺘﻁﺒﻴﻕ ﺨﻁﻲ . ﻋﻠﻰ ﺍﻝﺘﺭﺘﻴﺏm ﻭn ﺒﻌﺩﺍﻫﻤﺎ، ﻓﻀﺎﺌﻴﻥ ﺸﻌﺎﻋﻴﻥF ﻭE ﺘﻁﺒﻴﻕ ﺨﻁﻲf : E → F
(x , y ) → (x + y , 2 x − y , x − 3 y ) B E = {(1, 0 ) , ( 0,1)} B F = (1, 0, 0 ), ( 0,1, 0 ), ( 0, 0,1) v2 v3 v 1 f f
2
1
ﺍﻝﺘﻲ ﺃﻋﻤﺩﺘﻬﺎ ﻫﻲ ﺼﻭﺭ ﺃﺴﺎﺱ ﺍﻝﻤﻨﻁﻠﻕ ﻤﻜﺘﻭﺒﺔ ﻓﻲ ﺃﺴﺎﺱ ﺍﻝﻭﺼﻭل f (u 1 ) = α11v 1 + α 21v 2 + α 31v 3 + .......... + α m 1v m
3
2
f (u 2 ) = α12v 1 + α 22v 2 + α 32v 3 + .......... + α m 2v m
3
1 1 ⇒ M = 2 −1 1 −3
⋮
f (u j ) = α1 jv 1 + α 2 jv 2 + α 3 jv 3 + .......... + α mjv m ⋮
h : ℝ 3 [x ] → ℝ 3 [x ] p → p′
♦
B E = {1, x , x 2 , x 3 } , B F = {1, x , x 2 , x 3 }
↓ α11 α 21 M (f , B E , B F ) = ⋮ αm1
h (1) = 0 = 0i1 + 0ix + 0ix 2 + 0ix 3 h ( x ) = 1 = 1i1 + 0ix + 0ix 2 + 0ix 3
( ) = 2x = 0i1 + 2ix + 0ix h ( x ) = 3x = 0i1 + 0ix + 3ix 2
3
2
2
0 0 M (h ) = 0 0
+ 0i x
2
f (u n ) = α1nv 1 + α 2 nv 2 + α 3nv 3 + .......... + α mnv m f (u1 ) f (u 2 ) ⋯ f (u j ) ⋯ f (u n )
dim E = dim F = 4 ﻷﻥ4 × 4 ﻤﻥ ﺍﻝﻨﻭﻉh ﺇﻥ ﻤﺼﻔﻭﻓﺔ
h x
F ﺃﺴﺎﺱ ﻝـB F = { v 1 ,v 2 ,⋯ ,v m }
ﺍﻝﻤﺼﻔﻭﻓﺔB F ﻭB E ﺒﺎﻝﻨﺴﺒﺔ ﻝﻸﺴﺎﺴﻴﻥf ﻨﺴﻤﻲ ﻤﺼﻔﻭﻓﺔ:ﺘﻌﺭﻴﻑ
( (1, 0 ) ) = (1, 2,1) = v + 2v + v ( ( 0,1) ) = (1, −1, −3) = v −v + 3v 1
E ﺃﺴﺎﺱ ﻝـB E = { u1 , u 2 ,⋯ , u n }
3
↓
⋯
↓
⋯
α12 α 22
⋯
α1 j α2 j
⋯
⋮
⋯ ⋮
αm 2
⋯
⋮
⋯ ⋮
α mj
⋯
↓
α1n v 1 α 2 n v 2
⋮ ⋮ α m n v m
+ 0i x 3
m × n : M ﻨﻭﻉ ﺍﻝﻤﺼﻔﻭﻓﺔ
1 0 0 0 2 0 :ﻭﻤﻨﻪ 0 0 3 0 0 0
k : ℝ → ℝ3 t → ( t , 2t , 6t )
E ﺒﻌﺩ: M ﻋﺩﺩ ﺃﻋﻤﺩﺓ، F ﺒﻌﺩ: M ﻋﺩﺩ ﺃﺴﻁﺭ
. 3 × 2 ﻤﺼﻔﻭﻓﺔ ﻤﻥ ﺍﻝﻨﻭﻉ، ﺨﻁﻲf : ℝ 2 → ℝ 3 :ﻤﺜﻼ
♦
3 × 1 ﻤﻥ ﺍﻝﻨﻭﻉk ﻤﺼﻔﻭﻓﺔ B E = { 1} , B F = {(1, 0, 0 ) , ( 0,1, 0 ) , ( 0, 0,1)} f (1) = (1, 2, 6 ) = 1e1 + 2 e 2 + 6 e 3
1 M ( k , B E , B F ) = 2 :ﻭﻤﻨﻪ 6
16
dim E = n ﺘﻘﺎﺒﻠﻪ f ∈ L ( E , F ) ﻜل dim F = m . Μ m ×n ﻤﺼﻔﻭﻓﺔ ﻤﻥ ﺍﻝﻨﻭﻉ
ﻭﻁﻠﺏ ﺍﻝﺘﻁﺒﻴﻕ ﺍﻝﺨﻁﻲΜ m ×n ﺇﺫﺍ ﻋﻠﻤﺕ ﻤﺼﻔﻭﻓﺔ ﻤﻥ ﺍﻝﻨﻭﻉ
♦
♦
:ﺍﻝﻤﻘﺎﺒل ﻝﻬﺎ ﻓﺈﻥ ﻋﺒﺎﺭﺘﻪ ﺘﻌﻁﻰ ﺒﺎﻝﻘﺎﻋﺩﺓ
f : E ( dim E = n ) → F ( dim F = m ) f (X ) = A t X a11 a12 a13 .......... a1n x 1 a a a .......... a2 n x 2 f ( X ) = 21 22 23 ⋮ ⋮ am 1 am 2 am 3 .......... amn x n www.math.3arabiyate.net
]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل )ﺘﻐﻴﻴﺭ ﺍﻝﻘﺎﻋﺩﺓ( :
1 0 −1 A 2 0 5 3 1 6 ﻝﻨﺒﺤﺙ ﻋﻥ ﺍﻝﺘﻁﺒﻴﻕ ﺍﻝﺨﻁﻲ ﺍﻝﻤﻘﺎﺒل ﻝﻬﺎ.
Eﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ،ﺒﻌﺩﻩ . n
} B 2 = { v 1 ,v 2 ,⋯ ,v n } ، B 1 = { u1 , u 2 ,⋯ , u n ﺃﺴﺎﺴﻴﻥ ﻝـ . E ﺘﻌﺭﻴﻑ :ﻨﺴﻤﻲ ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل ﻤﻥ ﺍﻷﺴﺎﺱ B 1ﻨﺤﻭ ﺍﻷﺴﺎﺱ : B 2
ﺇﻨﻬﺎ ﻤﻥ ﺍﻝﻨﻭﻉ 3 × 3
ﺍﻝﻤﺼﻔﻭﻓﺔ Pﺍﻝﺘﻲ ﺃﻋﻤﺩﺘﻬﺎ ﻫﻲ ﻋﻨﺎﺼﺭ ﺍﻷﺴﺎﺱ B 2ﻤﻜﺘﻭﺒﺔ ﻓﻲ
ﻝﻨﻔﺭﺽ ﺃﻥ Aﻤﺭﻓﻘﺔ ﺒﺎﻝﺘﻁﺒﻴﻕ: 3
, X ( x , y , z ) ∈ ℝ3
)
f :ℝ → ℝ 3
ﺍﻷﺴﺎﺱ . B 1 v 1 = α11u1 + α 21u 2 + α 31u 3 + ⋯ + α n 1u n
(
f (x , y , z ) = t A. t X
v 2 = α12u1 + α 22u 2 + α 32u 3 + ⋯ + α n 2 u n
x −z 1 0 −1 x 2 0 5 y = 2 x + 5 z 3 1 6 z 3 x + y + 6 z ) f (x , y , z ) = (x − z , 2 x + 5 z , 3x + y + 6 z
⋮ v i = α1i u 1 + α 2i u 2 + α 3i u 3 + ⋯ + α n i u n ⋮ v n = α1nu1 + α 2 nu 2 + α 3nu 3 + ⋯ + α n n u n
1 2 ، M = ﻭﻝﻴﻜﻥ fﺍﻝﺘﻁﺒﻴﻕ ﺍﻝﺨﻁﻲ ﺤﻴﺙ: ♦ ﻝﺘﻜﻥ 3 4 ) f : M 2×2 ( ℝ ) → M 2×2 ( ℝ X → f ( X ) = MX
2 1 0 1 = 4 0 0 3 2 0 1 0 = 4 0 0 0 2 0 0 2 = 4 1 0 4
0 = 1B1 + 0B 2 + 3B3 + 0B 4 0 1 = 0B1 + 1B 2 + 0B3 + 3B 4 3 0 = 2B1 + 0B 2 + 4B3 + 0B 4 0 1 2 0 0 0 2 f ( B 4 ) = MB 4 = = = 0B1 + 2B 2 + 0B3 + 4B 4 3 4 0 1 0 4
0 2 0 1 0 2 ﺇﺫﻥ: 0 4 0 3 0 4
1 0 M (f , B , B ) = 3 0
1 f ( B1 ) = MB1 = 3 1 f ( B 2 ) = MB 2 = 3 1 f ( B3 ) = MB3 = 3
↓
⋯
⋮ ⋮ α n n u n
ﺇﻴﺠﺎﺩ ﻤﺼﻔﻭﻓﺔ fﺒﺎﻝﻨﺴﺒﺔ ﻝﻠﻘﺎﻋﺩﺓ ﺍﻝﻤﻌﺘﺎﺩﺓ 1 0 0 1 0 0 0 0 B = B1 = , B2 = , B3 = , B4 = 0 0 0 0 1 0 0 1 ﻝﺩﻴﻨﺎ:
vn
⋯
α1n u1 α 2 n u 2
⋯
v2
v1
↓
⋯
↓
↓
⋯
α1i α 2i
⋯ ⋯
α12 α 22
⋮
⋮
⋮
⋮
⋯
α ni
⋯
αn 2
⋯
vi
α11 α 21 P = ⋮ α n1
ﻨﻭﻉ ﺍﻝﻤﺼﻔﻭﻓﺔ n × n : M
ﺘﻌﺭﻴﻑ :2ﻨﺴﻤﻲ ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل ﻤﻥ ﺍﻷﺴﺎﺱ B 1ﻨﺤﻭ ﺍﻷﺴﺎﺱ : B 2 ﻤﺼﻔﻭﻓﺔ ﺍﻝﺘﻁﺒﻴﻕ ﺍﻝﺨﻁﻲ ، Id : E → Eﺤﻴﺙ Idﺍﻝﺘﻁﺒﻴﻕ ﺍﻝﻤﻁﺎﺒﻕ ﻓﻲ ، Eﻤﺯﻭﺩﺍ ﺒﺎﻷﺴﺎﺱ B 2ﻓﻲ ﺍﻝﻤﻨﻁﻠﻕ ،ﻭﺍﻷﺴﺎﺱ B 1
ﻓﻲ ﺍﻝﻭﺼﻭل ،ﻭﻨﻜﺘﺏ. P = M ( Id , B 2 , B 1 ) : ♦ ﻝﺘﻜﻥ Qﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل ﻤﻥ ﺍﻷﺴﺎﺱ B 2ﻨﺤﻭ ﺍﻷﺴﺎﺱ B 1
) Q = M ( Id , B 1 , B 2 ﻝﺩﻴﻨﺎ . Q = P −1 : ﺤﺫﺍﺭ :ﻝﺘﺤﻭﻴل ﻋﺒﺎﺭﺓ ﺸﻌﺎﻉ ﻤﻜﺘﻭﺏ ﻓﻲ ﺍﻷﺴﺎﺱ B 1ﻨﺤﻭ ﺍﻷﺴﺎﺱ B 2ﻨﺴﺘﻌﻤل ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل ﻤﻥ ﺍﻷﺴﺎﺱ B 2ﻨﺤﻭ ﺍﻷﺴﺎﺱ . B 1
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17
]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
}
3
f E , B 2 → F , B 2′
{
)B 1 = {1, x , x 2 , x 3 } , B 2 = 1, ( x − 1) , ( x − 1) , ( x − 1 2
f = ( Id ) F f Id E −1
3
2
−1
♦ ﺇﺫﺍ ﻜﺎﻥ E = Fﻓﺈﻥ:
1 0 P = 0 0
E , B 2 →E ,B2 f
f = ( Id ) E f Id E −1
f E , B 1 → E , B1
x3
x2
x
1
1 3 3 1
1 2 1 0
1 1 0 0
1 0 −1 Q =P = 0 0
) f ( x , y , z ) = ( x + 3 y + 2z , x − 4z , y + 3z ﺃﻭﺠﺩ Bﻤﺼﻔﻭﻓﺔ ) fﺍﻝﺘﻤﺜﻴل ﺍﻝﻤﺼﻔﻭﻓﻲ( ﻓﻲ ﺍﻝﻘﺎﻋﺩﺓ:
}) S = {w 1 = (1,1,1) , w 2 = (1,1, 0 ) , w 3 = (1, 0, 0
ﻝﺩﻴﻨﺎ ﻤﺼﻔﻭﻓﺔ fﻓﻲ ﺍﻝﻘﺎﻋﺩﺓ ﺍﻝﻤﻌﺘﺎﺩﺓ
. ( x − 1) ، ( x − 1) ، ( x − 1) ، 1 2
3
}) E = {e1 = (1, 0, 0 ) , e 2 = ( 0,1, 0 ) , e 3 = ( 0, 0,1ﻫﻲ
ﺍﻝﻐﺭﺽ ﻤﻥ ﺍﻝﺘﻁﺒﻴﻕ ﺘﺤﻭﻴل ﻋﺒﺎﺭﺓ pﺍﻝﻤﻜﺘﻭﺏ ﻓﻲ ﺍﻝﻘﺎﻋﺩﺓ B 1ﺇﻝﻰ ﻋﺒﺎﺭﺓ ﻤﻜﺘﻭﺒﺔ ﻓﻲ ﺍﻝﻘﺎﻋﺩﺓ ، B 2ﻝﻬﺫﺍ ﻨﺴﺘﻌﻤل ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل Qﻤﻥ ﺍﻷﺴﺎﺱ B 2ﻨﺤﻭ ﺍﻷﺴﺎﺱ 1 1 1 1 7 . B 1 1 2 3 1 14 = 0 1 3 2 11 0 0 1 3 3
1 0 0 0
ﻭﻤﻨﻪ )p ( x ) = 7 + 14 ( x − 1) + 11( x − 1) + 3 ( x − 1 2
3
ﻗﺎﻋﺩﺓ: ﻝﻴﻜﻥ ، f : E → Fﺤﻴﺙ Eﻭ Fﻓﻀﺎﺌﻴﻥ ﻤﻨﺘﻬﻴﻲ ﺍﻝﺒﻌﺩ. 012
ﻭﻝﻴﻜﻥ B 1′ ، B 1ﺃﺴﺎﺴﻴﻥ ﻝـ F ، Eﻋﻠﻰ ﺍﻝﺘﺭﺘﻴﺏ.
(
. A = M f , B1 , B 1′
ﻭﻝﻴﻜﻥ B 2′ ، B 2ﺃﺴﺎﺴﻴﻥ ﺁﺨﺭﻴﻥ ﻝـ F ، Eﻋﻠﻰ ﺍﻝﺘﺭﺘﻴﺏ.
(
. B = M f , B 2 , B 2′
ﻝﺩﻴﻨﺎ ﻤﺎ ﻴﻠﻲ:
−1
B = P −1AP
ﺘﻁﺒﻴﻕ :ﺃﻜﺘﺏ p ( x ) = 1 + x + 2x 2 + 3x 3ﺒﺩﻻﻝﺔ:
)
↑ ( Id )E
↓ Id E
ﻤﺜﺎل :ﻝﻴﻜﻥ f : ℝ3 → ℝ 3ﻤﻌﺭﻑ ﻜﻤﺎ ﻴﻠﻲ:
1 ) ( x − 1 2 ) ( x − 1 3 ) ( x − 1
)
f E , B 1 → F , B 1′
1
−1 1 1 −2 3 x 0 1 −3 x 2 0 0 1 x3 ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل Qﻤﻥ ﺍﻷﺴﺎﺱ B 2ﻨﺤﻭ ﺍﻷﺴﺎﺱ : B 1 1
−1
B = Q −1AP
ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل Pﻤﻥ ﺍﻷﺴﺎﺱ B 1ﻨﺤﻭ ﺍﻷﺴﺎﺱ : B 2
)( x − 1) ( x − 1) ( x − 1
↑ ( Id )F
↓ Id E
B = Q −1AP
2 −4 3 ﻭﻝﺩﻴﻨﺎ
1 1 1 ، P = 1 1 0 ﺇﺫﻥ ﻤﺼﻔﻭﻓﺔ ﺘﻐﻴﻴﺭ ﺍﻝﻘﺎﻋﺩﺓ ﻤﻥ ﺍﻷﺴﺎﺱ Sﺇﻝﻰ 1 0 0 0 0 1 −1 ﺍﻷﺴﺎﺱ Eﻫﻲ Q = P −1ﺤﻴﺙP = 0 1 −1 : 1 −1 0 ﻭﻤﻨﻪ B = P −1AP 0 0 1 1 3 2 1 1 1 4 1 0 ﺇﺫﻥ B = 0 1 −1 1 0 −4 1 1 0 = −7 0 1 1 −1 0 0 1 3 1 0 0 9 3 0
4 1 0 ﺃﻱ. B = M ( f , S , S ) = −7 0 1 : 9 3 0
: Pﻫﻲ ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل ﻤﻥ B 1ﻨﺤﻭ . B 2 : Qﻫﻲ ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل ﻤﻥ B 1′ﻨﺤﻭ . B 2′ www.math.3arabiyate.net
1 3 A = M (f , E , E ) = 1 0 0 1 ﻤﺼﻔﻭﻓﺔ ﺘﻐﻴﻴﺭ ﺍﻝﻘﺎﻋﺩﺓ ﻤﻥ ﺍﻷﺴﺎﺱ Eﺇﻝﻰ ﺍﻷﺴﺎﺱ Sﻫﻲ
18
ﺒﻨﻴﺔ ﺍﻝﻔﻀﺎﺀ ﺍﻝﺸﻌﺎﻋﻲ: ﻝﻴﻜﻥ ) • ( K , +,ﺤﻘل ﺘﺒﺩﻴﻠﻲ ،ﻭﻝﺘﻜﻥ Eﻤﺠﻤﻭﻋﺔ ﻏﻴﺭ ﺨﺎﻝﻴﺔ .
ﻤﺜﺎل :1
ﺘﻌﺭﻴﻑ :ﻨﻘﻭل ﺃﻥ Eﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ﻋﻠﻰ ﺍﻝﺤﻘل ، Kﺇﺩﺍ ﺯﻭﺩﻨﺎ E
ﻜل ﺤﻘل Kﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ﻋﻠﻰ ﻨﻔﺴﻪ ( K , +, • ) ،ﺤﻘل.
ﺒﻌﻤﻠﻴﺘﻴﻥ (1ﻋﻤﻠﻴﺔ ﺩﺍﺨﻠﻴﺔ ) : ( +ﺘﺠﻌل ﻤﻥ Eﺯﻤﺭﺓ ﺘﺒﺩﻴﻠﻴﺔ
E =K " "+ﺍﻝﺩﺍﺨﻠﻴﺔ ﻋﻠﻰ Eﻫﻲ " "+ﺍﻝﻌﻤﻠﻴﺔ ﺍﻝﺩﺍﺨﻠﻴﺔ ﻋﻠﻰ . K
(2ﺍﻝﻌﻤﻠﻴﺔ ﺍﻝﺨﺎﺭﺠﻴﺔ ) • ( :ﻤﻌﺭﻓﺔ ﻜﻤﺎ ﻴﻠﻲ : K ×E → E
" • " ﺍﻝﺨﺎﺭﺠﻴﺔ ﻋﻠﻰ Eﻫﻲ " • " ﺍﻝﻌﻤﻠﻴﺔ ﺍﻝﺨﺎﺭﺠﻴﺔ ﻋﻠﻰ . K
(α , x ) → α • x
ﺤﺎﻝﺔ ﺨﺎﺼﺔ ℝ :ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ﻋﻠﻰ . ℝ
ﺒﺤﻴﺙ : 1) ∀α , β ∈ K : ∀u ∈ E : (α + β ) u = α u + β u
ﻤﺜﺎل :2
2) ∀α ∈ K : ∀u ,v ∈ E : α (u + v ) = α u + α v
) : ℑ ( ℝ, ℝﻤﺞ ﺍﻝﺘﻁﺒﻴﻘﺎﺕ ﻤﻥ ℝﻓﻲ . ℝ
) 3) ∀α , β ∈ K : ∀u ∈ E : (αβ ) u = α ( β u
fﺘﻁﺒﻴﻕ }ℑ ( ℝ, ℝ ) = { f : ℝ → ℝ
4) 1K • u = u
ﻋﻠﻰ Eﻨﻌﺭﻑ ﺍﻝﻌﻤﻠﻴﺘﻴﻥ E ×E → E
( f .g ) → f + g ) ∀x ∈ ℝ : ( f + g )( x ) = f ( x ) + g ( x
ﺍﺼﻁﻼﺤﺎﺕ: •
ﻋﻨﺎﺼﺭ Eﺘﺴﻤﻰ ﺃﺸﻌﺔ.
•
ﻋﻨﺎﺼﺭ ) Kﺍﻝﺤﻘل( ﺘﺴﻤﻰ ﺴﻠﻤﻴﺎﺕ .
•
ﺍﻝﻌﻨﺼﺭ ﺍﻝﺤﻴﺎﺩﻱ ﻝـ " "+ﻓﻲ Eﻨﺭﻤﺯ ﻝﻪ ﺒـ. O E :
•
ﻨﺭﻤﺯ ﻝﻠﻔﻀﺎﺀ ﺍﻝﺸﻌﺎﻋﻲ ﺒـ ) • ( K , +,ﻭﻨﻘﻭل
ℝ×E → E
( λ.f ) → λ f ) )( x ) = λ f ( x
∀x ∈ ℝ : ( λ f
ﺇﺫﻥ ℝﻑ.ﺵ
ﺃﻥ E Kﻑ.ﺵ.
ﻤﺜﺎل :3ﻤﺠﻤﻭﻋﺔ ﻜﺜﻴﺭﺍﺕ ﺍﻝﺤﺩﻭﺩ ﺒﻤﻌﺎﻤﻼﺕ ﻤﻥ ﺍﻝﺤﻘل : K
ﻗﻭﺍﻋﺩ ﺍﻝﺤﺴﺎﺏ ﻓﻲ ﻑ.ﺵ :
] = ℝ ∨ ℂ) , E = K [x
ﻝﻴﻜﻥ K , Eﻑ .ﺵ α , β ∈ K , u ,v ∈ E ، ) 1) ∀α ∈ K , ∀u ∈ E :αu = O E ⇒ (α = O K ) ∨ (u = O E
ﻨﺯﻭﺩ Eﺒـ :
) α −1 (α .α −1 = 1Kد ⇒ ) (α u = O E ) ∧ (α ≠ O K ⇒ α u = O E ⇒ α −1 (α u ) = α −1O E
)
] + : K [x ]× K [x ] → K [x
(
⇒ αα −1 u = O E ⇒ 1K u = O E ⇒ u = O E 2) α (u − v ) = α (u + ( −v ) ) = α u − α v
3) (α − β ) u = α u − β u
www.math.3arabiyate.net
] E = K [ xﻫﻭ ﻑ.ﺵ
1
(K ( p ,q ) → p + q ) ( p + q )( x ) = p ( x ) + q ( x ] K × K [x ] → K [x (λ, p ) → λ p ) ( λ p )( x ) = λ p ( x
â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
:4 ﻤﺜﺎل
؟u1 , u 2 ﻫﻭ ﻋﺒﺎﺭﺓ ﺨﻁﻴﺔ ﻓﻲv (1, −1)
.ﺵ. ﻑE K ﻤﺞ ﻭX ≠ φ
∃α , β ∈ ℝ : v = α u 1 + β u 2
H = ℑ ( X , E ) = {f : X → E , f
(1, −1) = α ( 5, 2 ) + β (1, 0 ) = ( 5α + β , 2α )
f , g ∈ H , ( f + g )( x ) = f ( x ) + g ( x )
1 α =− 5 + = 1 α β 2 ⇒ 2α = −1 β = 7 2
λ ∈ K , f ∈ H , ( λ f ( x ) ) = λ if ( x ) ﺵ. ﻑKH
}
ﺇﺫﻥ
ﺵ. ﻑℝ ﻫﻭℑ ([a , b ] , ℝ )
1 7 v = u 1 + u 2 ﻭﻤﻨﻪ 2 2
:5ﻤﺜﺎل ﺠﺩﺍﺀ ﻑ ﺸﻌﺎﻋﻴﺔ ﻋﻠﻰ ﻨﻔﺱ ﺍﻝﺤﻘل
:ﺍﻝﻔﻀﺎﺀ ﺍﻝﺸﻌﺎﻋﻲ ﺍﻝﺠﺯﺌﻲ
.ﺵ. ﻑK
E n ,............, E 3 , E 2 , E 1
E ⊇ F ≠ φ ﺵ ﻭﻝﻨﻜﻥ. ﻑE , K ﻝﻴﻜﻥ
E = E 1 × E 2 × E 3 × ..................E n
ﺝ ﻤﻥ ﺍﻝﻔﻀﺎﺀ ﺍﻝﺸﻌﺎﻋﻲ.ﺵ. ﻑF ﻨﻘﻭل ﺃﻥ:ﺘﻌﺭﻴﻑ
X ∈ E ⇔ X = ( x 1 , x 2 ,..........., x n ) / x i ∈ E i , i =1....n
: ( ﺇﺫﺍ ﻜﺎﻥK , +, • )
X ,Y ∈ E , X +Y = ( x 1 , x 2 ,..., x n ) + ( y 1 , y 2 ,..., y n )
F ≠φ ♦
= ( x 1 + y 1 , x 2 + y 2 , ... , x n + y n )
K ﺵ ﻋﻠﻰ ﺍﻝﺤﻘل. ( ﺒﺩﻭﺭﻫﺎ ﻑF , +, • ) ♦
λ ∈ K , X ∈ E : λ X = ( λ x 1 , λ x 2 ,........... λ x n )
(
O E = O E1 ,O E 2 ,................,O E n
:ﺝ.ﺵ.ﺘﻤﻴﻴﺯ ﺍﻝﻑ E ⊇F ﻭ
)
n
ﺵ. ﻑK ﻫﻭE = ΠE i
ﺵ. ﻑK ( E , + , • )
i =1
.ﺵ. ﻑℝ ﻫﻭE = ℝ ← E i = ℝ n
iF ≠ φ ( I ) i∀u ,v ∈ F :u + v ∈ F ⇔ (ﺝ.ﺵ. ﻑF ) i∀α ∈ K , ∀u ∈ F , α u ∈ F
.ﺵ. ﻑℝ ﻫﻭℝ 2 , n = 2 .ﺵ. ﻑℝ ﻫﻭℝ 3 , n = 3
iF ≠ φ ⇔ i∀u ,v ∈ F , ∀ α , β ∈ K , α u + β v ∈ F
:ﺍﻝﻌﺒﺎﺭﺓ ﺍﻝﺨﻁﻴﺔ ﻝﻤﺠﻤﻭﻋﺔ ﺃﺸﻌﺔ
( II )
ﺵ. ﻑE , K ﻝﻴﻜﻥ
α1 , α 2 ∈ K , u1 , u 2 ∈ E ﻭﻝﻴﻜﻥ
iF ≠ φ ( III ) ⇔ i∀u ,v ∈ F , ∀α ∈ K , α u + v ∈ F
ﻴﺴﻤﻰ ﻋﺒﺎﺭﺓ ﺨﻁﻴﺔ ﻓﻲ ﺍﻝﺸﻌﺎﻋﻴﻥv = α1 u 1 + α 2 u 2 ∈ E ﺍﻝﺸﻌﺎﻉ u 2 , u1
:ﻭﻨﻌﻤﻡ ﺫﻝﻙ ﻜﻤﺎ ﻴﻠﻲ
O E ∈ F ﺘﺄﻜﺩ ﺃﻥF ≠ φ ﻹﺜﺒﺎﺕ:ﻤﻼﺤﻅﺔ
E ( ﻤﺠﻤﻭﻋﺔ ﺃﺸﻌﺔ ﻤﻥu i )i ∈I ﻝﺘﻜﻥ
: ( ﻜل ﻋﺒﺎﺭﺓ ﻤﻥ ﺍﻝﺸﻜلu i ) ﻨﺴﻤﻲ ﻋﺒﺎﺭﺓ ﺨﻁﻴﺔ ﻓﻲ ﺍﻷﺸﻌﺔ
∑α u i
i
, αi ∈ K
i ∈I
2
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â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
E ﺝ ﻤﻥ.ﺵ. ﻑF2 ، F1 :ﺘﻌﺭﻴﻑ
{( x , y ) ∈ ℝ
u ∈ F1 ⇔ u = ( −2 y , y ) , y ∈ ℝ
(ﺵ. )ﻑE ﺝ ﻤﻥ.ﺵ. ﻑF1 + F2 ﺇﻥ:ﻤﻼﺤﻅﺔ
v ∈ F2 ⇔ u = ( −2 y ′, y ′ ) , y ′ ∈ ℝ
α u + v = ( −2α y , α y ) + ( −2 y ′, y ′ )
E ﺝ ﻤﻥ.ﺵ. ﻑF1 , F2 ,..........., Fp ، ﺵ. ﻑK , E :ﺘﻌﻤﻴﻡ
= ( −2α y − 2 y ′, α y + y ′ )
ﻭﺍﻝﻤﻌﺭﻑ ﺒـH = F1 + F2 + ........ + Fp ﺝ.ﺵ.ﻨﺴﻤﻲ ﺍﻝﻑ
= ( −2 (α y + y ′ ) , α y + y ′ ) ∈ F1
H = {w ∈ E / w = u 1 + u 2 + ......... + u p , u i ∈ Fi } F1 , F2 ,..........., Fp ﺒﻤﺠﻤﻭﻉ
F2 = {( x , y , z ) ∈ ℝ3 / x + 2 y + z = 3}
φ ≠ X ⊂ E ، ﺵ. ﻑK E
K = ℝ , E = ℝ [x ]
V ect ( x ) ﺒـX ﻨﺴﻤﻲ ﻤﺠﻤﻭﻋﺔ ﻜل ﺍﻝﻌﺒﺎﺭﺍﺕ ﺍﻝﺨﻁﻴﺔ ﻝﻌﻨﺎﺼﺭ OE ∈ F ⇒ F ≠ φ ?
p , q ∈ F ,α ∈ ℝ ⇒α p + q ∈ F
v ,w ∈V ect ( x ) , α ∈ K , (α .v + w ) ∈V ect ( x ) ♦
i ∈I
)
= α p ′ (0) + q ′ ( 0)
♦
α U +V = ∑ (α λi ) u i + ∑ β j v j
= (α p + q )′ ( 0 ) = A ′ ( 0 )
j ∈J
= ∑ λ ′u i + ∑ β j v j i ∈I
(A = α p + q )
p ( 0 ) = p ′ ( 0 ) q ′ ( 0 ) = q ′ ( 0 ) A ( 0 ) = (α p + q )( 0 ) = α p ( 0 ) + q ( 0 )
. X ﺍﻝﻤﻭﻝﺩ ﺒـ
j ∈J
(3
F = { p ∈ E / p ( 0 ) = p ′ ( 0 )}
V ect ( x ) = ∑ λi u i , λi ∈ K , u i ∈ X i ∈I ﻴﺴﻤﻰ ﺍﻝﻔﻀﺎﺀ ﺍﻝﺠﺯﺌﻲE ﺝ ﻤﻥ.ﺵ. ﻑV ect ( x ) ♦ ﺇﻥ
V = ∑ β j v j (v j ∈ X
(2
O ℝ3 = ( 0, 0, 0 ) ∉ F2 ⇒ ﺝ.ﺵ. ﻝﻴﺱ ﻑF2
E ﺍﻝﻔﻀﺎﺀ ﺍﻝﺸﻌﺎﻋﻲ ﺍﻝﺠﺯﺌﻲ ﺍﻝﻤﻭﻝﺩ ﺒﺠﺯﺀ ﻤﻥ
)
(1
u ,v ∈ F1 , α ∈ ℝ ⇒ α u + v ∈ F1
F1 + F2 = { u + v / u ∈ F1 ,v ∈ F2 }
i ∈I
}
/ x + 2y = 0
O E = ( 0, 0 ) ∈ F1 ⇒ F1 ≠ φ
: ﺍﻝﻤﺠﻤﻭﻋﺔ ﺍﻝﺘﺎﻝﻴﺔF2 ﻭF1 ﻨﺴﻤﻲ ﻤﺠﻤﻭﻉ
U = ∑ λi u i (u i ∈ X
2
E = ℝ 2 :ﺃﻤﺜﻠﺔ
j ∈J
V ect ( F ) = F ، ﺝ.ﺵ. ﻑF :ﺤﺎﻝﺔ ﺨﺎﺼﺔ
:ﺨﻭﺍﺹ
∑α u
ﺵ. ﻑE , K :ﻤﺜﺎل
i
X = { u1 ,u 2 } ♦
p ∈ℕ*
∈ F ﺝ ﻓﺈﻥ.ﺵ. ﻑF ﻭu i ∈ F (1
E ﺝ ﻤﻥ.ﺵ. ﻑG ﻭG ﺝ ﻤﻥ.ﺵ. ﻑF (2
V ect ( X ) = { α1 u1 + α 2 u 2 , α1 , α 2 ∈ K } X = { u 1 , u 2 ,........., u p }
i
i ∈I
E ﺝ ﻤﻥ.ﺵ. ﻑF ﻓﺈﻥ
♦
ﻫﻭ ﺃﻴﻀﺎF1 ∩ F2 ﻓﺈﻥE ﺝ ﻤﻥ.ﺵ. ﻑF2 ﻭF1 (3
p V ect ( X ) = ∑ α i u i , α i ∈ K i =1 V ect ( X ) ﻴﺭﻤﺯ ﻓﻲ ﻫﺫﻩ ﺍﻝﺤﺎﻝﺔ ﻝـ
ﺝ.ﺵ.ﻑ .ﺝ.ﺵ. ﻋﻠﻰ ﻋﺩﺩ ﻜﻴﻔﻲ ﻤﻥ ﻑ3 ﺘﻌﻤﻡ ﺍﻝﻨﺘﻴﺠﺔ E ﺝ ﻤﻥ ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ.ﺵ. ﻑFi ، i ∈ I
V ect ( X ) = u 1 , u 2 ,............, u p ﺒـ
.ﺝ.ﺵ.ﻫﻭ ﺃﻴﻀﺎ ﻑ
∩F
i
ﻓﺈﻥ
i ∈I
3
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â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
E = ℝ 2 , u = (1, 2, 0 ) , v ( 0, 0,1) , X = {u ,v } ♦ V ect ( X ) = u ,v
u = u 1 + u 2 ⇔ ( x , y ) = ( x 1 , − x 1 ) + ( x 2 , 2x 2 )
= { α u + β v , α , β ∈ ℝ}
1 x 1 = ( 2x − y ) + = x x x 1 3 2 ⇔ ⇔ − x 1 + 2x 2 = y x = 1 ( x + y ) 2 3 ﻤﻭﺠﻭﺩﺍﻥ ﺩﺍﺌﻤﺎx 1 , x 2
= { (α , 2α , 0 ) + ( 0, 0, β ) , α , β ∈ ℝ} = { (α , 2α , β ) , α , β ∈ ℝ}
. ﻤﻭﺠﻭﺩﺍﻥ ﺩﺍﺌﻤﺎu 2 ( x 2 , 2x 2 ) , u1 ( x 1 , − x 1 ) ⇐
:ﺍﻝﻤﺠﻤﻭﻉ ﺍﻝﻤﺒﺎﺸﺭ ﻝﻔﻀﺎﺀﺍﺕ ﺸﻌﺎﻋﻴﺔ ﺠﺯﺌﻴﺔ E ﺝ ﻤﻥ.ﺵ. ﻑF2 ﻭF1 ، ﺵ. ﻑK , E ﻝﻴﻜﻥ
:ﺍﻝﻔﻀﺎﺀﺍﺕ ﺍﻝﺸﻌﺎﻋﻴﺔ ﺍﻝﺠﺯﺌﻴﺔ ﺍﻹﻀﺎﻓﻴﺔ
ﺝ.ﺵ. ﻑH ﻨﻌﻠﻡ ﺃﻥ، H = F1 + F2 ﻨﻀﻊ
ﺝ.ﺵ. ﻑF ,G ، ﺵ. ﻑE ﻝﻴﻜﻥ
ﻭﻨﻜﺘﺏF2 ﻭF1 ﻤﺠﻤﻭﻉ ﻤﺒﺎﺸﺭ ﻝـH ﻨﻘﻭل ﺃﻥ:ﺘﻌﺭﻴﻑ
F ≡ )ﺃﻭF ﺝ ﺇﻀﺎﻓﻲ ﻝـ.ﺵ. ﻑG ﻨﻘﻭل ﺃﻥ:ﺘﻌﺭﻴﻑ
ﻴﻜﺘﺏ ﻋﻠﻰ ﺸﻜل ﻭﺤﻴﺩH ﺇﺫﺍ ﻜﺎﻥ ﻜل ﻋﻨﺼﺭ ﻤﻥ، H = F1 ⊕ F2
E = F ⊕ G : ﺇﺫﺍ ﺘﺤﻘﻕ، ( G ﺝ ﺇﻀﺎﻓﻲ ﻝـ.ﺵ.ﻑ
: ﺤﻴﺙw = u + v :
(w ∈ H , u ∈ F1 , v ∈ F2 )
:ﺘﻤﻴﻴﺯ ﺍﻝﻤﺠﻤﻭﻉ ﺍﻝﻤﺒﺎﺸﺭ
E = ℝ 2 :ﻤﺜﺎل F = {( x , y , z ) / x + y = z } ,G = {( x , y , z ) / x = y = z }
H = F1 + F2 ﻭ، E ﺝ ﻤﻥ.ﺵ. ﻑF2 , F1 ﻝﻴﻜﻥ
H = F1 + F2 H = F1 ⊕ F2 ⇔ F1 ∩ F2 = { O E }
ℝ = F ⊕ G ⇔ ﺇﻀﺎﻓﻴﺎﻥG ﻭF 3
F ∩ G = { O E } 3 ℝ ⊂ F + G
E = ℝ 2 :ﻤﺜﺎل
♦
u ( x , y , z ) ∈ F ∩ G ⇒ u ∈ F ∧ u ∈G ⇒x +y =z ∧ x = y =z ⇒ 2x = x ⇒ x = 0 ⇒ y = z = 0
( x , y , z ) = ( 0, 0, 0 ) = O ℝ
{( x , y ) ∈ ℝ = {( x , y ) ∈ ℝ
}
F1 =
2
/x + y =0
F2
2
/ 2x − y = 0
}
؟E = F1 ⊕ F2 ﻫل
3
F ∩ G = {O E } ﺃﻱ
i) E = F1 + F2 E = F1 ⊕ F2 ⇔ ii) F1 ∩ F2 = {O E }
w ( x , y , z ) ∈ ℝ ♦ ﻝﻴﻜﻥ 3
∃u ∈ F ,v ∈G : w = u + v
?
ii) u ( x , y ) ∈ F1 ∩ F2 ⇒ u = O ℝ2
u ( x 1 , y 1 , x 1 + y 1 ) ∈ F , v ( x 2 , x 2 , x 2 ) ∈G u +v = w ⇔ ( x 1 + x 2 , y 1 + x 2 , x 1 + y 1 + x 2 )
u ∈ F1 ∩ F2 ⇒ (u ∈ F1 ) ∧ (u ∈ F2 )
x 1 + x 2 = x .................. (1) y 1 + x 2 = y ................. ( 2 ) x 1 + y 1 + x 2 = z .......... ( 3) x 1 = z − y ( 3) − ( 2 ) ⇒ x 2 = x + y − z y = z − x 1
x + y = 0 ⇒ x = y = 0 ⇒ u = O ℝ2 2x − y = 0 ii) ℝ 2 = F1 + F2 i ℝ 2 ⊃ F1 + F2 ﻭﺍﻀﺢ ?
i F1 + F2 ⊃ ℝ 2
∀u ( x , y ) ∈ ℝ 2 , ∃u1 ∈ F1 , u 2 ∈ F2 : u = u1 + u 2 4
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]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﻭﻤﻨﻪ uﻭ vﻤﺴﺘﻘﻼﻥ ﺨﻁﻴﺎ
u ( z − y , z − x , 2z − x − y ) ∈ F v ( x + y − z , x + y − z , x + y − z ) ∈G
ℝ 2 [ x ] .3ﻑ.ﻙ.ﺡ ﺍﻝﺘﻲ ﺩﺭﺠﺘﻬﺎ ≥ 2
u +v = ( x , y , z ) = w
u = x 2 , v = x , w =1 α u + β v + δ w = Oℝ ⇒ α = β = δ = 0
ﻤﻼﺤﻅﺔ :ﻓﻲ ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ، Eﻜل ﻑ.ﺵ.ﺝ ﻴﻤﻠﻙ ﺇﻀﺎﻓﻴﺎ.
α x + β x + δ .1 = O ℝ , ∀x ∈ ℝ 2
x =0:δ =0
x =1 : α + β +δ = 0 ⇒α = β = δ = 0 x = −1 : α − β + δ = 0 ﻭﻤﻨﻪ w , v , uﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ
ﺍﻝﻌﺎﺌﻠﺔ ﺍﻝﺤﺭﺓ – ﺍﻝﻌﺎﺌﻠﺔ ﺍﻝﻤﻭﻝﺩﺓ – ﺍﻷﺴﺱ ﻓﻲ ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ: ﺍﻝﻌﺎﺌﻠﺔ ﺍﻝﺤﺭﺓ: ﻝﻴﻜﻥ K , Eﻑ.ﺵ ،
*p ∈ℕ
} S = {u 1 , u 2 ,........, u p
ﻤﻼﺤﻅﺎﺕ ﻋﺎﻤﺔ:
ﺘﻌﺭﻴﻑ :ﻨﻘﻭل ﺃﻥ ﺍﻝﻌﺎﺌﻠﺔ Sﺤﺭﺓ ،ﺃﻭ ﺃﻥ ﺍﻷﺸﻌﺔ u1 , u 2 ,......., u p
، S = { u } (1
ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ ﺇﺫﺍ ﺘﺤﻘﻕ
∀α1 , α 2 ,..., α p ∈ K ,
)ﺍﻝﺸﻌﺎﻉ uﻤﺴﺘﻘل ﺨﻁﻴﺎ ⇔ ( u ≠ φ
α1u1 + α 2u 2 + ... + α pu p = 0 ⇒ α1 = α 2 = ... = α p = 0
S = { u ,v } (2
ﺃﻱ ﻻ ﺘﻘﺒل ﺍﻝﻤﻌﺎﺩﻝﺔ ﺇﻻ ﺍﻝﺤل ﺍﻝﻤﻌﺩﻭﻡ
Sﻤﻘﻴﺩﺓ ⇔ u ,vﻤﺭﺘﺒﻁﺎﻥ ﺨﻁﻴﺎ
) (α ,α ,..., α ) = ( 0, 0,..., 0 p
2
⇔ ) ∃ λ ∈ K * : (u = λ v ) ∨ (v = λ u
1
ﻭﺇﺫﺍ ﻭﺠﺩ α iﻏﻴﺭ ﻤﻌﺩﻭﻡ ﺒﺤﻴﺙ = O E
Sﺤﺭﺓ ⇔ u ≠ O E
i =p
i
∑α u i
ﻤﺜﺎل :ﻓﻲ ، ℝ3
ﻨﻘﻭل ﺃﻥ ﺍﻝﻌﺎﺌﻠﺔ
) u (1, 2 ) , v ( −1, −2
u ,vﻤﺭﺘﺒﻁﺎﻥ ﺨﻁﻴﺎ ﻷﻥ u + v = O ℝ2 ⇔ v = −u
i =1
Sﻤﺭﺘﺒﻁﺔ ،ﺃﻭ ﺃﻥ ﺍﻷﺸﻌﺔ u1 , u 2 ,..., u pﻤﺭﺘﺒﻁﺔ ﺨﻁﻴﺎ.
(3
*
p ∈ℕ
} S = {u 1 , u 2 ,........, u pﻤﺭﺘﺒﻁﺔ ﺨﻁﻴﺎ
⇔ ﺃﺤﺩ ﻋﻨﺎﺼﺭ ﺍﻝﻌﺎﺌﻠﺔ ﻴﻜﺘﺏ ﻋﻠﻰ ﺸﻜل ﻋﺒﺎﺭﺓ ﺨﻁﻴﺔ ﻓﻲ ﺒﺎﻗﻲ
ﺃﻤﺜﻠﺔ: .1
2
ﺍﻝﻌﻨﺎﺼﺭ .
u = (1,3) , v = ( 2, −1) , E = ℝ ?
⇒α = β = 0
2
ﻤﺜﺎل:
α u + β v = Oℝ
♦
) (α ,3α ) + ( 2β , − β ) = ( 0, 0
.w = u + 2v
♦ ] w = 3x 2 − 2x , v = x , u = x 2 , E = ℝ 2 [ x ﻭﺍﻀﺢ ﺃﻥ w = 3u − 2vﻭﻤﻨﻪ } { u ,v ,wﻤﺭﺘﺒﻁﺔ ﺨﻁﻴﺎ .
v = e x , u = sin x , E = ℑ ( ℝ, ℝ ) .2 ?
α u + β v = 0 ⇒α = β = 0 α sin x + β e x = 0 , ∀x ∈ ℝ x = 0 ⇒ β = 0 π x = ⇒ α = 0
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w = ( 5, 6, −2 ) , v ( 2, 2, −1) , u (1, 2, 0 ) , E = ℝ
} { u ,v ,wﻤﻘﻴﺩﺓ ﺃﻱ u , v , wﻤﺭﺘﺒﻁﺔ ﺨﻁﻴﺎ ﻷﻥ
α + 2 β = 0 ⇒ 3α − β = 0 ⇒ α = β = 0 ﻭﻤﻨﻪ uﻭ vﻤﺴﺘﻘﻼﻥ ﺨﻁﻴﺎ ،ﺃﻭ } S = { u ,vﺤﺭﺓ.
2
3
5
(4
ﻜل ﻋﺎﺌﻠﺔ ﺠﺯﺌﻴﺔ ﻤﻥ ﻋﺎﺌﻠﺔ ﺤﺭﺓ ﻫﻲ ﺒﺩﻭﺭﻫﺎ ﻋﺎﺌﻠﺔ ﺤﺭﺓ .
(5
ﻜل ﻋﺎﺌﻠﺔ ﺘﺸﻤل ﻋﺎﺌﻠﺔ ﻤﻘﻴﺩﺓ ﻫﻲ ﺒﺩﻭﺭﻫﺎ ﻤﻘﻴﺩﺓ .
(6
ﻜل ﻋﺎﺌﻠﺔ ﺘﺸﻤل O Eﻫﻲ ﺒﺎﻝﻀﺭﻭﺭﺓ ﻤﻘﻴﺩﺓ .
]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﺍﻝﻌﺎﺌﻠﺔ ﺍﻝﻤﻭﻝﺩﺓ: ﻝﻴﻜﻥ K , Eﻑ.ﺵ ،ﻭ
*p ∈ℕ
(α , −α ) + ( 0, 3 β ) = ( 0, 0 ) ⇒
} S = {u 1 , u 2 ,........, u pﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ
ﻤﻥ ، Eﻨﻘﻭل ﺃﻥ ﺍﻝﻌﺎﺌﻠﺔ Sﺘﻭﻝﺩ ﺍﻝﻔﻀﺎﺀ Eﺇﺫﺍ ﻜﺎﻥ ﻜل ﺸﻌﺎﻉ ﻤﻥ
{ u ,vﺘﻭﻝﺩ ℝ 2
• } = ( x , y ) ∈ ℝ2
Eﻴﻜﺘﺏ ﻋﻠﻰ ﺸﻜل ﻋﺒﺎﺭﺓ ﺨﻁﻴﺔ ﻓﻲ ﺃﺸﻌﺔ ، S i =p
∀v ∈ E : ∃α1 , α 2 ,..., α p ∈ K / v = ∑ α i u i
ﺃﻱ
wﻜﻴﻔﻲ ∃α , β ∈ ℝ , w = α u + β v ،
) ( x , y ) = (α , −α ) + ( 0,3β
i =1
ﺃﻤﺜﻠﺔ: v (0, −1) , u (1, 2) , E = ℝ 2 (1
ﻭﻤﻨﻪ ﻓﻌﻼ
ﻝﻴﻜﻥ w = ( x , y ) ∈ ℝ 2ﻜﻴﻔﻲ ∃α , β ∈ ℝ : w = α u + β v
}
α = x α = x ⇒ y +x α + 3 β = y β = 3 { u ,vﺃﺴﺎﺱ ﻝـ . ℝ 2
ﺘﻌﺭﻴﻑ ﻤﻜﺎﻓﺊ ﻷﺴﺎﺱ ﻓﻲ ﻑ.ﺵ:
α = x α = x ⇔ 2 α − β = y β = 2x − y )w = x (1, 2 ) + ( 2x − y )( 0, −1
ﻝﻴﻜﻥ K , Eﻑ.ﺵ ،
*p ∈ℕ
} S = {u 1 , u 2 ,........, u pﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ
ﻤﻥ E
u ,vﻴﻭﻝﺩﺍﻥ . ℝ 2
Sﺃﺴﺎﺱ ﻝـ Eﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ﻜل ﺸﻌﺎﻉ ﻤﻥ Eﻴﻜﺘﺏ ﻋﻠﻰ ﺸﻜل ﻋﺒﺎﺭﺓ ﺨﻁﻴﺔ ﻓﻲ ﻋﻨﺎﺼﺭ Sﻭﻫﺫﻩ ﺍﻝﻜﺘﺎﺒﺔ ﻭﺤﻴﺩﺓ k =n
u = x 2 v = x ، E = ℝ 2 [ x ] (2 w = 1
∀v ∈ E , ∃! α1 , α 2 ,..., α n ∈ K : v = ∑ α k u k k =1
ﺘﺭﻤﻴﺯ :ﺘﺴﻤﻰ α1 , α 2 ,..., α nﻤﺭﻜﺒﺎﺕ vﻓﻲ ﺍﻷﺴﺎﺱ S
ﺇﻥ } { u ,v ,wﺘﻭﻝﺩ E
ﺃﻤﺜﻠﺔ:
∀p ∈ E : p = a x 2 + b x + c
E = ℝ n (1 ♦ E = ℝ2 ← n = 2 )v = ( x , y ) = ( x ,0 ) + ( 0, y ) = x (1, 0 ) + y ( 0,1
ﻤﻼﺤﻅﺔ :ﻜل ﻋﺎﺌﻠﺔ ﺘﺸﻤل ﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓ ﻫﻲ ﺒﺩﻭﺭﻫﺎ ﻤﻭﻝﺩﺓ.
ﺇﺫﻥ ) e 2 ( 0,1) , e1 (1, 0ﻴﻭﻝﺩﺍﻥ ℝ 2
ﺍﻷﺴﺎﺱ:
( x , y ) = x e1 + y e 2 ⇒ (α , β ) = ( 0, 0 ) ⇒ α = β = 0 ﻭﻤﻨﻪ } { e1 , e 2ﺃﺴﺎﺱ ﻝـ ℝ 2
ﻭ e 2 , e1ﻤﺴﺘﻘﻼﻥ ﺨﻁﻴﺎ.
ﺘﻌﺭﻴﻑ :ﻨﻘﻭل ﻋﻥ ﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ ﺃﻨﻬﺎ ﺃﺴﺎﺱ ﻓﻲ ﺍﻝﻔﻀﺎﺀ Eﺇﺫﺍ ﻜﺎﻨﺕ ﺤﺭﺓ ﻭﻤﻭﻝﺩﺓ ﻓﻲ ﺁﻥ ﻭﺍﺤﺩ.
ﻴﺴﻤﻰ ﻫﺫﺍ ﺍﻷﺴﺎﺱ ﺍﻝﻁﺒﻴﻌﻲ )ﺃﻭ ﺍﻝﻘﺎﻨﻭﻨﻲ( ﻝﻠﻔﻀﺎﺀ . ℝ 2
ﻤﻭﻀﻭﻋﺔ :ﻓﻲ ﻜل ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ﻏﻴﺭ ﻤﻌﺩﻭﻡ ﺘﻭﺠﺩ ﺃﺴﺱ.
♦ E = ℝ3 ← n = 3 ) u ( x , y , z ) = ( x , 0, 0 ) + ( 0, y , 0 ) + ( 0, 0, z
ﻤﺜﺎلV ( 0,3) , u (1, −1) , E = ℝ 2 :
}
)= x (1, 0, 0 ) + y ( 0,1, 0 ) + z ( 0, 0,1
{ u ,vﺃﺴﺎﺱ ﻝـ ℝ 2 •
}
ﻭﻤﻨﻪ ) e 3 ( 0, 0,1) , e 2 ( 0,1, 0 ) , e1 (1, 0, 0ﺘﻭﻝﺩ ℝ3
{ u ,vﺤﺭﺓ
ﻭﻫﻲ ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ ﻭﻤﻨﻪ } { e1 , e 2 , e 3ﺃﺴﺎﺱ ﻝـ ℝ3
α u + β v = Oℝ ⇒ α = β = 0
ﻴﺴﻤﻰ ﺍﻝﻘﺎﻨﻭﻥ ﺍﻝﻁﺒﻴﻌﻲ ﻝـ . ℝ3
2
www.math.3arabiyate.net
2
α e1 + β e 2 = O ℝ
6
]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﺘﻌﺭﻴﻑ ﺒﻌﺩ ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ): (Dimension n
v ∈ℝ
ﻝﻴﻜﻥ K , Eﻑ.ﺵ ﻤﻨﺘﻬﻲ ﺍﻝﺒﻌﺩ
)v = x 1 (1, 0,..., 0 ) + x 2 ( 0,1, 0,..., 0 ) + ... + x n ( 0, 0,..., 0,1
ﻤﺒﺭﻫﻨﺔ ﻭﺘﻌﺭﻴﻑ :ﻜل ﺍﻷﺴﺱ ﻝﻬﺎ ﻨﻔﺱ ﺍﻝﻌﺩﺩ ﻤﻥ ﺍﻝﻌﻨﺎﺼﺭ ) ، (Card
ﺍﻷﺸﻌﺔ:
ﻨﺴﻤﻲ ﺒﻌﺩ Eﻋﺩﺩ ﻋﻨﺎﺼﺭ ﺃﺴﺎﺱ ﻜﻴﻔﻲ ﻤﺎ ،ﻭﻨﻜﺘﺏ
) e1 (1, 0,...,0 ) + e 2 ( 0,1, 0,..., 0 ) + ... + e n ( 0, 0,..., 0,1ﺘﻭﻝﺩ
) ، dim K E = card ( Bﺤﻴﺙ : Bﺃﺴﺎﺱ ﻤﺎ .
ℝ nﻭﻫﻲ ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ ،ﻓﻬﻲ ﺇﺫﻥ ﺃﺴﺎﺱ ﻝـ ، ℝ nﺘﺴﻤﻰ ﺍﻷﺴﺎﺱ
ﻤﻼﺤﻅﺔ :ﺇﻥ ﺒﻌﺩ ﻑ.ﺵ ﻴﺘﻌﻠﻕ ﺒﺎﻝﺤﻘل ﺍﻝﻤﺴﺘﺨﺩﻡ
ﺍﻝﻘﺎﻨﻭﻨﻲ .
♦ ℂﻑ.ﺵ ﻋﻠﻰ ) ℂﻜل ﺤﻘل ﻑ.ﺵ ﻋﻠﻰ ﻨﻔﺴﻪ( ﺒﻌﺩﻩ ، 1 dim ℂ ℂ = 1
E = ℝ n [ x ] (2 p = a0 + a1x + a2 x 2 + ... + an x n , ai ∈ ℝ
)( z ∈ ℂ
p ∈ ℝ n [x ] ,
z = z ⋅1
u = 1∈ ℂ = E ,
♦ E = ℂﻫﻭ ﻑ.ﺵ ﻋﻠﻰ ﺍﻝﺤﻘل K = ℝ z = α .1 + β . i , α , β ∈ ℝ , z = α + β . i , z ∈ ℂ = E
ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ 1, x , x 2 ,..., x nﺘﻭﻝﺩ ] ℝ n [ xﻭﻫﻲ ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ ﻭﻤﻨﻪ ﺍﻝﻌﺎﺌﻠﺔ } { 1, x , x 2 ,..., x nﺃﺴﺎﺱ ﻝـ ] . ℝ n [ x
} { u1 ,u 2 } = { 1, iﺃﺴﺎﺱ ﻝﻠﻔﻀﺎﺀ ℂﻋﻠﻰ ﺍﻝﺤﻘل ، ℝ dim ℝ ℂ = 2 ♦ ∞dim ℚ ℂ = +
ﺍﻝﻔﻀﺎﺀ ﺍﻝﺸﻌﺎﻋﻲ ﺍﻝﻤﻨﺘﻪ ﺍﻝﺒﻌﺩ: ﺘﻌﺭﻴﻑ :ﻨﻘﻭل ﻋﻥ ﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ﺃﻨﻪ ﻤﻨﺘﻬﻲ ﺍﻝﺒﻌﺩ ﺇﺫﺍ ﻗﺒل ﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓ
ﺃﻤﺜﻠﺔ:
ﻤﻨﺘﻬﻴﺔ . ﻤﻭﻀﻭﻋﺔ K , E :ﻑ.ﺵ ،
*p ∈ℕ
n ≥ 1 ، E = ℝ n (1ﻫﻲ ﻑ.ﺵ ﻤﻨﺘﻪ ﺍﻝﺒﻌﺩ ﻤﻊ . dim ℝ n = n
} S = {u 1 , u 2 ,........, u pﻋﺎﺌﻠﺔ
E = ℝ n [ x ] (2ﺒﻌﺩﻩ ﻤﻨﺘﻬﻲ ﻤﻊ dim ℝ n [ x ] = n + 1
ﺃﺸﻌﺔ ﻤﻥ E
ﺃﺴﺎﺴﻪ } { 1, x , x ,..., x ﻤﻥ ﺃﺠل ] { 1, x , x } : ℝ [ xﺍﻷﺴﺎﺱ ﺍﻝﻘﺎﻨﻭﻨﻲ } {2,3x + 1, x − 2ﺃﺴﺎﺱ ﺁﺨﺭ. n
ﺃ( ﺇﺫﺍ ﻜﺎﻨﺕ Sﺘﻭﻝﺩ Eﻓﺈﻥ ﻜل ﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ ﺒﻬﺎ ﺃﻜﺜﺭ ﻤﻥ p
2
ﻋﻨﺼﺭ ﻓﻬﻲ ﺒﺎﻝﻀﺭﻭﺭﺓ ﻤﻘﻴﺩﺓ.
S ′ﻤﻘﻴﺩﺓ ⇒ Card ( S ′ ) ≥ p + 1
2
2
2
S′⊆E
ﺏ( ﺇﺫﺍ ﻜﺎﻨﺕ Sﺤﺭﺓ ﻓﺈﻥ ﻜل ﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ ﺒﻬﺎ ﺃﻗل ﻤﻥ pﺸﻌﺎﻉ
ﻨﺘﻴﺠﺔ ﻫﺎﻤﺔ:
ﻻ ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﻭﻥ ﻤﻭﻝﺩﺓ .
ﻝﻴﻜﻥ K , Eﻑ.ﺵ ﺒﻌﺩﻩ ﻤﻌﻠﻭﻡ ) ∞= n ≺ +
( dim E
ﻭﻝﺘﻜﻥ Bﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ ﻤﻥ Eﻤﻊ Card B = n
ﻨﻅﺭﻴﺔ :ﻝﻴﻜﻥ K , Eﻑ.ﺵ ﻤﻨﺘﻬﻲ ﺍﻝﺒﻌﺩ )ﻴﻤﻠﻙ ﻤﺠﻤﻭﻋﺔ ﻤﻭﻝﺩﺓ ﻤﻨﺘﻬﻴﺔ( } S = { e1 , e 2 ,...., e mﻤﻭﻝﺩﺓ B = {u 1 , u 2 ,........, u p } ،ﺤﺭﺓ
Bﺃﺴﺎﺱ ⇐ Bﺤﺭﺓ ⇐ Bﻤﻭﻝﺩﺓ .
ﻭﻝﻴﺴﺕ ﻤﻭﻝﺩﺓ ،ﻋﻨﺩﺌﺫ ﻴﻤﻜﻥ ﺇﻜﻤﺎل ﺍﻝﻌﺎﺌﻠﺔ Bﺒﻌﻨﺎﺼﺭ ﻤﻥ S
ﻤﺜﻼB = {(1, 2 ) , ( 3, −1)} , dim E = 2 , E = ℝ 2 :
ﻝﻠﺤﺼﻭل ﻋﻠﻰ ﺃﺴﺎﺱ ﻝـ . E
ﺒﻤﺎ ﺃﻥ Card B = dim Eﻴﻜﻔﻲ ﻹﺜﺒﺎﺕ ﺃﻥ Bﺃﺴﺎﺱ ﺃﻥ ﻨﺩﺭﺱ
ﺍﻻﺴﺘﻘﻼل ﺍﻝﺨﻁﻲ ﻝﻠﺸﻌﺎﻋﻴﻥ )u (1, 2 ) , v ( 3, −1
α u + β v = OE ⇔ α = β = 0
) (α , 2α ) + ( 3β , − β ) = ( 0, 0 α + 3 β = 0 ⇒ ⇒α =β =0 2α − β = 0 www.math.3arabiyate.net
7
]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﻤﻼﺤﻅﺎﺕ: (1ﻝﻴﻜﻥ K , Eﻑ.ﺵ ﺒﻌﺩﻩ nﻤﻨﺘﻬﻲ ،
ﻝﻴﻜﻥ ﺍﻝﻔﻀﺎﺀ ﺍﻹﻀﺎﻓﻲ G
ﺇﻥ ﻜل ﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ ﻤﻜﻭﻨﺔ ﺃﻜﺜﺭ ﻤﻥ nﺸﻌﺎﻉ ﻫﻲ ﺒﺎﻝﻀﺭﻭﺭﺓ
Gﺇﻀﺎﻓﻲ ﻝـ E = F ⊕ G ⇔ F dim E = dim F + dim G ⇒ dim G = dim E − dim F = 2 − 1 = 1 ﻭﻤﻨﻪ dim G = 1
ﻤﺭﺘﺒﻁﺔ ﺨﻁﻴﺎ . ﻭﻤﻨﻪ ﺍﻷﺴﺎﺱ ﻋﺒﺎﺭﺓ ﻋﻥ ﻋﺎﺌﻠﺔ ﺤﺭﺓ ﺃﻋﻅﻤﻴﺔ. (2ﻜل ﻋﺎﺌﻠﺔ ﺒﻬﺎ ﺃﻗل ﻤﻥ nﺸﻌﺎﻉ ﻻ ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﻭﻥ ﻤﻭﻝﺩﺓ ،
ﺇﺫﻥ Gﻤﻭﻝﺩ ﺒﺸﻌﺎﻉ ﻭﺍﺤﺩ
ﻓﺈﻥ ﻜل ﺃﺴﺎﺱ ﻓﻬﻭ ﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓ ﻭﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓ ﺃﺼﻐﺭﻴﺔ ،
ﺒﺤﻴﺙ wﻤﺴﺘﻘل ﺨﻁﻴﺎ ﻤﻊ . u
ﺒﻤﻌﻨﻰ ﺃﻥ ﺃﻗل ﻋﺩﺩ ﻤﻥ ﺍﻷﺸﻌﺔ ﺍﻝﻤﻭﻝﺩﺓ ﻫﻭ . n
ﻝﻨﺤﺼل ﻋﻠﻰ } { u ,wﺃﺴﺎﺱ ﻝـ E = ℝ 2
) w ( 0,1) , u (1, 2
ﺒﻌﺩ ﻑ.ﺵ.ﺝ: ﻝﻴﻜﻥ K , Eﻑ.ﺵ ﺒﻌﺩﻩ nﻤﻨﻬﻲ
) , F (1, 2
) ∞( dim E ≺ +
F ⊂ Eﻑ.ﺵ.ﺝ ﻤﻥ E
1) dim F ≤ dim E 2) dim F = dim E ⇔ E = F G , Fﻑ.ﺵ.ﺝ ﻤﻥ E
ﻨﻌﻠﻡ ﺃﻥ F + Gﻫﻭ ﺃﻴﻀﺎ ﻑ.ﺵ.ﺝ
) dim ( F + G ) = dim F + dim G − dim ( F ∩ G ﺘﻁﺒﻴﻕ K , E :ﻑ.ﺵ ﺒﻌﺩﻩ ﻤﻨﺘﻪ G , Fﻑ.ﺵ.ﺝ ﻤﻨﻪ ﺤﻴﺙ . E = F ⊕ G } 1) F ∩ G = { O E E = F ⊕G ⇔ 2) E = F + G } F ∩ G = {O E
) ⇔ dim E = dim F + dim G − dim ( F ∩ G ⇔ dim E = dim F + dim G ﻤﺜﺎلF = {( x , y ) / 2x − y = 0} :
♦ ﺘﻌﻴﻴﻥ ﺃﺴﺎﺱ ﻝـ : F u = ( x , y ) = ( x , 2x ) , x ∈ ℝ = x (1, 2 ) , x ∈ ℝ
ﺇﺫﻥ ) u1 = (1, 2ﻴﻭﻝﺩ ، Fﻭ ﻤﻨﻪ } { uﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓ ﻭﻫﻲ ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ
≠ O ℝ2 ) G = w
) (u ≠ O ℝ2
ﻭﻤﻨﻪ } { uﺃﺴﺎﺱ ﻝـ Fﺇﺫﻥ . dim F = 1
www.math.3arabiyate.net
8
)( 0,1
.G
(w
â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
( ( 0, 0, 0 ) ) = ( 0,1) ≠ ( 0, 0 ) = O
ﺍﻝﺘﻁﺒﻴﻘﺎﺕ ﺍﻝﺨﻁﻴﺔ
ℝ2
. ﻝﻴﺱ ﺨﻁﻲf ﺇﺫﻥ
ϕ : ℝ [x ] → ℝ [x ]
. ﺘﻁﺒﻴﻕf : E → F ﻭﻝﻴﻜﻥ، ﺵ. ﻑK F , E ﻝﻴﻜﻥ
(3
:ﺘﻌﺭﻴﻑ
p → ϕ ( p ) = p′
⇔ ﺨﻁﻲf
ϕ ( p + q ) = ( p + q )′ = p ′ + q ′
f ∀u ,v ∈ E , ∀α ∈ K : f
ϕ (α p ) = (α p )′ = α p ′
( u + v ) = f (u ) + f (v ) ( α u ) = α f (u )
:ﺘﻌﺭﻴﻑ ﻤﻜﺎﻓﺊ ∀u ,v ∈ E , ∀α , β ∈ K : f (α u + β v ) = α f (u ) + β f (v )
E = ℓ ( [ a , b ] , ℝ ) , F = ℝ (4
ϕ :E → ℝ f → ϕ (f
:ﺘﻌﺭﻴﻑ ﻤﻜﺎﻓﺊ
b
) = ∫a f (t ) dt
∀u ,v ∈ E , ∀α ∈ K : f (α u +v ) = α f (u ) + f (v )
ﺨﻁﻲϕ
( f (O ) = O ) ⇐ ( ﺨﻁﻲf ) :ﺨﺎﺼﻴﺔ . ( ﻝﻴﺱ ﺨﻁﻲf ) ⇐ ( f (O ) ≠ O ) :ﻋﻜﺱ ﺍﻝﻨﻘﻴﺽ E
b
ϕ (α f + g ) = ∫ α f (t ) + g (t ) dt a
b
b
a
a
F
E
= α ∫ f (t ) dt + ∫ g (t ) dt
F
:ﺘﺭﻤﻴﺯ F ﻭE ( ﻨﺭﻤﺯ ﻝﻤﺠﻤﻭﻋﺔ ﺍﻝﺘﻁﺒﻴﻘﺎﺕ ﺍﻝﺨﻁﻴﺔ ﺒﻴﻥ ﺍﻝﻔﻀﺎﺌﻴﻥ1
ℓ ( E , F ) ﺒـ
. ﺨﻁﻲf : E → F ﻝﻴﻜﻥ:ﺨﻭﺍﺹ . F ﺝ ﻤﻥ.ﺵ. ﻑf ( E 1 ) ⇐ E ﺝ ﻤﻥ.ﺵ. ﻑE 1 . E ﺝ ﻤﻥ.ﺵ. ﻑf
−1
( F1 ) ⇐ F
ﺝ ﻤﻥ.ﺵ. ﻑF1
ﻤﺞ ﺍﻝﺘﻤﺎﺘﻼﺕℓ ( E , E ) = ℓ ( E ) : E = F ( ﺇﺫﺍ ﻜﺎﻥ2
• •
.ﺍﻝﺩﺍﺨﻠﻴﺔ E ﻤﺞ ﺍﻷﺸﻜﺎل ﺍﻝﺨﻁﻴﺔ ﻋﻠﻰℓ ( E , K ) = E ′ : F = K
. ﻤﺞ ﺍﻝﺘﺸﺎﻜﻼﺕ ﺍﻝﺩﺍﺨﻠﻴﺔ: GL ( E ) (4
:ﻨﻭﺍﺓ ﻭﺼﻭﺭﺓ ﺘﻁﺒﻴﻕ ﺨﻁﻲ
. ﻤﺘﻘﺎﺒل+ ﺍﻝﺘﺸﺎﻜل ≡ ﺨﻁﻲ
ﺨﻁﻲf : E → F
:ﺃﻤﺜﻠﺔ
ﺍﻝﻤﻌﺭﻑ ﺒـE ﺍﻝﺠﺯﺀ ﻤﻥf ﻨﺴﻤﻲ ﻨﻭﺍﺓ: f ﺘﻌﺭﻴﻑ ﻨﻭﺍﺓ ker f = f
−1
({O }) = {x ∈ E F
/ f (x ) = OF }
. E ﺝ ﻤﻥ.ﺵ.ﺍﻝﻨﻭﺍﺓ ﻫﻲ ﻑ
•
. ker f ≠ φ ⇐ f (O E ) = O F
•
، f 1 : ℝ2 → ℝ2
. ﺨﻁﻲf
(1
(x , y ) → (x + y , y − x ) u ( x , y ) , y ( x ′, y ′ ) , α ∈ ℝ f (α u + v ) = f
( (α x + x ′, α y + y ′) )
= (α x + x ′ + α y + y ′, α y + y ′ − α x − x ′ )
ﺍﻝﻤﻌﺭﻑ ﺒـF ﺍﻝﺠﺯﺀ ﻤﻥf ﻨﺴﻤﻲ ﺼﻭﺭﺓ: f ﺘﻌﺭﻴﻑ ﺼﻭﺭﺓ
= (α x + α y , α y − α x ) + ( x ′ + y ′, y ′ − x ′ )
Im f = f ( E ) = {f ( x ) , x ∈ E }
= α ( x + y , y − x ) + ( x ′ + y ′, y ′ − x ′ )
= {y ∈ F / ∃ x ∈ E :f ( x ) = y } . F ﺝ ﻤﻥ.ﺵ. ﻫﻲ ﻑIm f
(3
= α f (u ) + f (v )
•
f : ℝ3 → ℝ 2
(2
( x , y , z ) → ( x + y − z ,1) 9
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â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
f : ℝ 2 → ℝ 2 :ﻤﺜﺎل
( x , y ) → ( 0, x ) f ∈L
( ﺨﻁﻲf ) f ∈ L
(E , F )
2
2
ﻝﻴﻜﻥ
2
ker f = {O E } ⇔ ﻤﺘﺒﺎﻴﻥf -1
= {( 0, y ) , y ∈ ℝ}
Im f = F ⇔ ﻏﺎﻤﺭf -2
= { y ( 0,1) , y ∈ ℝ}
F ﺘﺒﻘﻰ ﺤﺭﺓ ﻓﻲE ﻤﺘﺒﺎﻴﻥ ⇔ ﺼﻭﺭﺓ ﻜل ﻋﺎﺌﻠﺔ ﺤﺭﺓ ﻤﻥf -3
= ( 0,1)
ﻫﻲ ﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓE ﻏﺎﻤﺭ ⇔ ﺼﻭﺭﺓ ﻋﺎﺌﻠﺔ ﻤﻭﻝﺩﺓ ﻝـf -4
{ = {( u ,v ) ∈ ℝ = {( u ,v ) ∈ ℝ = {( u ,v ) ∈ ℝ
}
Im f = y = ( u ,v ) ∈ ℝ 2 / ∃ ( x , y ) ∈ ℝ 2 : ( 0, x ) = ( u ,v )
Im f ﻝـ
ﻓﺈﻥE { ⇔ ﺘﻭﻝﺩu1 ,u 2 ,........,u n } Im f = f (u 1 ) , f (u 2 ) ,...., f (u n )
:ﻨﻅﺭﻴﺔ ﺍﻝﺒﻌﺩ
/ ∃ ( x , y ) ∈ ℝ 2 : ( 0, x ) = ( u ,v )
2
/ ∃( x , y ) ∈ ℝ 2
2
/ u = 0 v ∈ℝ
}
= ( 0,1)
: ﻝﺩﻴﻨﺎ، f ∈ L ( E , F ) dim E = dim ker f + dim Im f
} :u = 0 ∧v = x }
2
= {( 0,v ) , v ∈ ℝ }
ﻭﻝﻴﻜﻥ، ﻤﻨﺘﻪdim E = n ﺵ ﻤﻊ. ﻑF ﻭE ﻝﻴﻜﻥ
ker f ∩ Im f = ( 0,1) f : ℝ3 → ℝ 2
ﻤﻨﺘﻬﻲdim E = dim F ﻤﻊf ∈ L ( E , F ) :ﻨﺘﻴﺠﺔ ( ﺘﻘﺎﺒﻠﻲ )ﺘﺸﺎﻜلf ⇔ ﻏﺎﻤﺭf ⇔ ﻤﺘﺒﺎﻴﻥf ker f
dim E = n . F ﻭE ⇐ ﻻ ﻴﻭﺠﺩ ﺃﻱ ﺘﺸﺎﻜل ﺒﻴﻥdim F = m n ≠ m
. ﺍﻝﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎS ﺃﻜﺒﺭ ﻋﺩﺩ ﻤﻥ ﺃﺸﻌﺔS ﻨﺴﻤﻲ ﺭﺘﺒﺔ
ker f = (1, −1, 0 ) dim ker f = 1 ﺃﺴﺎﺱ ﻝﻠﻨﻭﺍﺓ ﻭ
) = dim ( Im f )
10
{(1, −1, 0 )}
Im f = {(u ,v ) ∈ ℝ 2 / ∃ ( x , y , z ) ∈ ℝ 3 , f ( x , y , z ) = (u ,v )}
:( ﺭﺘﺒﺔ ﺘﻁﺒﻴﻕ ﺨﻁﻲ2 ﺒﻌﺩ ﺼﻭﺭﺘﻪf : E → F ﻨﺴﻤﻲ ﺭﺘﺒﺔ ﺘﻁﺒﻴﻕ ﺨﻁﻲ
(x , y , z ) → (x + y − z , x + y + z ) = {( x , y , z ) / x + y + z = 0 ∧ x + y − z = 0}
= x (1, −1, 0 ) , x ∈ ℝ
:( ﺭﺘﺒﺔ ﺠﻤﻠﺔ ﺃﺸﻌﺔ1 ، E ﻋﺎﺌﻠﺔ ﺃﺸﻌﺔ ﻤﻥS = { u1 ,u 2 ,........,u n } ، ﺵ. ﻑE
:ﻤﺜﺎل
y = −x ⇒ 2(x + y ) = 0 ⇒ x + y = 0 ⇒ z = 0 u ( x , −x , 0 ) ∈ ker f , x ∈ ℝ
: Rang ﺍﻝﺭﺘﺒﺔ
rg ( f
2
{( x , y ) ∈ ℝ / f ( x , y ) = ( 0, 0 )} = {( x , y ) ∈ ℝ / ( 0, x ) = ( 0, 0 )} = {( x , y ) ∈ ℝ / x = 0 , ∀y }
ker f =
:ﺨﻭﺍﺹ ﺍﻝﺼﻭﺭﺓ ﻭﺍﻝﻨﻭﺍﺓ ﻭﻋﻼﻗﺘﻬﺎ ﺒﺎﻝﺘﻁﺒﻴﻕ
(ℝ )
x z
x + y − z = u = (u ,v ) ∈ ℝ 2 / ∃ ( x , y , z ) ∈ ℝ 3 , x + y + z = v u +v u v +y = x = ,y = 2 2 2 v −u v −u = z = 2 2
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â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
:ﻤﺜﺎل
⇒ f (u ) = u − g (u )
ϕ : ℝ 3 [x ] → ℝ
( g (u ) = 0 ) ﻷﻥ
p → p ( 2)
⇒ u ∈ Im f
K = ℝ ﻭE ﺨﻁﻲ ﺒﻴﻥϕ ⇔ ϕ ∈ ( ℝ 3 [ x ])
*
Im f ⊂ ker g
•
. ﺨﻁﻲp ﺇﺫﻥ
∃u ∈ E : f (u ) = v ⇐ v ∈ Im f ﻝﻴﻜﻥ ⇒ g ( f (u ) ) = g (v )
⇒ (g f
: ker ϕ ♦ ﺘﻌﻴﻴﻥ
ker ϕ = { p ∈ E / ϕ ( p ) = 0}
)(u ) = g (v )
= { p ∈ E / ϕ ( 2 ) = 0}
g f = 0 ﺒﻤﺎ ﺃﻥ
p (x ) ∈ E ⇔ p (x ) = a x 3 +b x + c x + d
⇒ g (v ) = 0 v ∈ ker g
p ( 2) = 8 a + 4b + 2c + d
ﺃﻱ
p ∈ ker ϕ ⇔ p ( 2 ) = 0 ⇔ 8 a + 4 b + 2 c + d = 0 ⇔ d = −8 a − 4 b − 2 c
. ker g = Im f
:ﻭﻤﻨﻪ
p ( x ) = a x 3 + b x 2 + c x + ( −8 a − 4 b − 2 c )
(
) (
)
= a x 3 − 8 + b x 2 − 4 + c ( x − 2) , a,b ,c ∈ ℝ
: rg ( f ) + rg ( g ) = n ﺘﺒﻴﻴﻥ ﺃﻥ-2
ker ϕ = x 3 − 8, x 2 − 4, x − 2
:ﻝﺩﻴﻨﺎ
(ﻭﻫﺫﻩ ﺍﻷﺸﻌﺔ ﻤﺴﺘﻘﻠﺔ ﺨﻁﻴﺎ )ﺒﺩﺭﺠﺎﺕ ﻤﺨﺘﻠﻔﺔ ﻤﺜﻨﻰ ﻤﺜﻨﻰ
n = dim E = dim ker g + dim Im g
dim ker ϕ = 3
= dim Im f + dim Im g
: Im ϕ ♦ ﺘﻌﻴﻴﻥ
= rg ( f ) + rg ( g )
dim ( Im ϕ ) = dim E − dim ( ker ϕ ) = 4 − 3 = 1 dim ( Im ϕ ) = dim ℝ = 1 ﻭ، ﺝ.ﺵ. ﻑIm ϕ ⊆ ℝ ﺒﻤﺎ ﺃﻥ Im ϕ = ℝ : ﻓﺈﻥ
:ﺘﻤﺭﻴﻥ
( dim E
= n ) ﺵ ﺒﻌﺩﻩ ﻤﻨﺘﻬﻲ. ﻑE ﻝﻴﻜﻥ
. g f = 0 , f + g = id E : ﺒﺤﻴﺙf , g ∈ L
(E )
ﻭﻝﻴﻜﻥ
. ker g = Im f : ﺒﺭﻫﻥ ﺃﻥ-1 . rg ( f ) + rg ( g ) = n : ﺒﺭﻫﻥ ﺃﻥ-2
:ﺍﻝﺤل : ker g = Im f ﺘﺒﻴﻴﻥ ﺃﻥ-1 ker g ⊂ Im f •
g (u ) = 0 ⇐ u ∈ ker g ﻝﻴﻜﻥ id (u ) = u ﻝﻜﻥ ⇒ ( f + g )(u ) = u
11
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]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﺍﻝﻤﺼﻔﻭﻓﺎﺕ ﻝﻴﻜﻥ Kﺤﻘﻼ ﺘﺒﺩﻴﻠﻲ ( K = ℂ ∨ K = ℝ ) ، ﺘﻌﺭﻴﻑ :ﻨﺴﻤﻲ ﻤﺼﻔﻭﻓﺔ ﻤﻥ ﺍﻝﻨﻭﻉ ) ( m , nﺃﻭ ) ( m × n ﺘﻁﺒﻴﻕ ﻤﻥ } E = {1, 2,3,...., m } × {1, 2,3,...., nﻨﺤﻭ
ﻨﺭﻤﺯ ﻝﻤﺠﻤﻭﻋﺔ ﺍﻝﻤﺼﻔﻭﻓﺎﺕ ﻤﻥ ﺍﻝﻨﻭﻉ m × nﺒﻤﻌﺎﻤﻼﺕ ﻤﻥ K
ﻜل . K
ﺒﺎﻝﺭﻤﺯ : ﻤﺜﻼ:
1 2 A • ) ∈ Μ 2×2 ( ℝ 3 6 2 5 3 7 B • ) ∈ Μ 2×4 ( ℝ − 1 2 0 − 1 3 0 i • ) C i + 1 j j 2 ∈ Μ 3×3 ( ℂ 1 −3 0 2π i 2π 2π = e 3 = cos + i sin j 3 =1 3 3
ﻨﺭﻤﺯ ﻝﻠﻤﺼﻔﻭﻓﺔ ﺒـ . M , C , B , A A : E →K
( i , j ) → ai j ﻨﻜﺘﺏ ﻋﻨﺎﺼﺭ ﺍﻝﻤﺼﻔﻭﻓﺔ ﻋﻠﻰ ﺸﻜل ﺠﺩﻭل
a11 a12 a13 .......... a1n a21 a22 a23 .......... a2 n A ⋮ am 1 am 2 am 3 .......... amn ai jﺍﻝﻌﻨﺼﺭ ﺍﻝﻤﻭﺠﻭﺩ ﻓﻲ ﺍﻝﻤﺼﻔﻭﻓﺔ Aﻓﻲ ﺍﻝﻤﻭﻀﻊ : ﺘﻘﺎﻁﻊ ﺍﻝﺴﻁﺭ iﻤﻊ ﺍﻝﻌﻤﻭﺩ . j : iﺭﻗﻡ ﺍﻝﺴﻁﺭ ،
: jﺭﻗﻡ ﺍﻝﻌﻤﻭﺩ .
ﻋﺩﺩ ﺍﻷﺴﻁﺭ ﻫﻭ ، mﻋﺩﺩ ﺍﻷﻋﻤﺩﺓ ﻫﻭ . n
• •
• •
1 A ﻤﻥ ﺍﻝﻨﻭﻉ . 2 × 2 2 1 3 B ﻤﻥ ﺍﻝﻨﻭﻉ . 2 × 3 0 −1
3 − 1 2 3 1 C 2 ﻤﻥ ﺍﻝﻨﻭﻉ . 3 × 1 0 ) D (1, 2, −5ﻤﻥ ﺍﻝﻨﻭﻉ . 1× 3
•
H = 5ﻤﻥ ﺍﻝﻨﻭﻉ . 1× 1
•
1 i 0 M iﻤﻥ ﺍﻝﻨﻭﻉ . 3 × 3 i + 1 −1 1 1 i 2
ﻨﺴﻤﻲ ﺍﻝﻤﺼﻔﻭﻓﺔ ﺍﻝﻤﻌﺩﻭﻤﺔ ﻓﻲ ) Μ m ×n ( Kﺍﻝﻤﺼﻔﻭﻓﺔ
)
(
O α i jﺒﺤﻴﺙ ∀i , j : α i j = O k
0 ﻤﺜﺎل = : 0
O Μ 2×1
0 0 = 0 0 0 0
,
O Μ3×2
ﻤﻨﻘﻭل ﻤﺼﻔﻭﻓﺔ: ﻝﺘﻜﻥ ) ∈ Μ m ×n ( K
)
1≤i ≤ m 1≤ j ≤ n
(
A αi j
ﻨﺴﻤﻲ ﻤﻨﻘﻭل Aﺍﻝﻤﺼﻔﻭﻓﺔ t Aﻭﺍﻝﻤﻌﺭﻓﺔ ﻜﻤﺎ ﻴﻠﻲ: ﻤﺼﻔﻭﻓﺔ ﻤﻥ ﺍﻝﻨﻭﻉ n × mﻭﺍﻝﻨﺎﺘﺠﺔ ﺒﺘﺒﺩﻴل ﺃﺴﻁﺭ ﻭﺃﻋﻤﺩﺓ A
ﻤﺜﺎل: 3 ♦ 7
7 ♦ 0
−1 1 ←A 7 −1 2 1 3 5 ← B 2 5 0
1 ♦ C (1, 2 ) ← C 2
t
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j
ﺍﻝﻤﺼﻔﻭﻓﺔ ﺍﻝﻤﻌﺩﻭﻤﺔ ﻓﻲ ) Μ m ×n ( K
1≤ i ≤ m 1≤ j ≤ n
ﺃﻤﺜﻠﺔ:
) Μ m ×n ( K
12
1 A 3 1 t B 3 7 t
â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
−4 8 2 7
:ﺨﻭﺍﺹ t
(A + B ) = t A + tB t (λ A ) = λ . t A t
: ﺍﻝﻀﺭﺏ ﻓﻲ ﺴﻠﻤﻴﺔ-2
( )
A ai j
1≤ i ≤ m 1≤ j ≤ n
t
( A .B ) = t B .t A
.4
ﻤﺭﺒﻌﺔ ﺇﺫﺍ ﻜﺎﻥ ﻋﺩﺩ ﺃﺴﻁﺭﻫﺎ ﻴﺴﺎﻭﻱ ﻋﺩﺩA ﻨﻘﻭل ﻋﻥ ﻤﺼﻔﻭﻓﺔ
:ﻤﺜﺎل
. ( n = m ) ﺃﻋﻤﺩﺘﻬﺎ
9 −3 3 −1 3A ← A , λ =3 6 15 2 5
(ﺃﺜﺭ ﻤﺼﻔﻭﻓﺔ )ﺨﺎﺹ ﺒﺎﻝﻤﺼﻔﻭﻓﺔ ﺍﻝﻤﺭﺒﻌﺔ
:ﺨﻭﺍﺹ
n × n ﻤﺭﺒﻌﺔ ﻤﻥ ﺍﻝﻨﻭﻉA ∈ Μ n ( K ) ﻝﺘﻜﻥ
ﺍﻝﻤﺯﻭﺩ ﺒﺎﻝﺠﻤﻊ ﻭﺍﻝﻀﺭﺏ ﻓﻲ ﺴﻠﻤﻴﺔ ﻝﻬﺎ ﺒﻨﻴﺔΜ m ×n ( K ) ﺇﻥ
( )
A = ai j
: ﻭﺃﺴﺎﺴﻪ ﺍﻝﻘﺎﻨﻭﻨﻲ، m × n ﺒﻌﺩﻩ ﻫﻭK ﺵ ﻋﻠﻰ.ﻑ 1 0 ⋯ 0 0 1 ⋯ 0 0 0 ⋯ 0 0 0 ⋯ 0 0 0 ⋯ 0 0 0 ⋯ 0 E1 , E2 , ⋯ , E mn ⋮ ⋮ ⋮ 0 0 0 0 0 0 0 0 0 0 0 1
a b a, b , c , d ∈ ℝ = Μ 2×2 ( ℝ ) :ﻤﺜﺎل c d dim ( Μ 2×2 ( ℝ ) ) = 4 1 0 0 1 0 0 0 0 E1 , E2 , E3 , E4 0 0 0 0 1 0 0 1
n
Tr ( A ) = a11 + a22 + a33 + ......... + ann = ∑ akk
: ﺍﻝﺠﻤﻊ-1 : ﺤﻴﺙA , B ∈ Μ m ×n ( K ) ﻝﺘﻜﻥ
ﻭﺍﻝﻤﻌﺭﻓﺔC = A iB ﺍﻝﻤﺼﻔﻭﻓﺔB ﻭA ﻨﺴﻤﻲ ﺠﺩﺍﺀ:ﺘﻌﺭﻴﻑ
( )
. ﻤﺠﻤﻭﻋﺔ ﻋﻨﺎﺼﺭ ﻗﻁﺭﻫﺎA ﻨﺴﻤﻲ ﺃﺜﺭ ﺍﻝﻤﺼﻔﻭﻓﺔ
ﻋﻤﻠﻴﺎﺕ ﻋﻠﻰ ﺍﻝﻤﺼﻔﻭﻓﺎﺕ
A ∈ Μ k ×l ( K ) , B ∈ Μ l ×n ( K ) ﻝﺘﻜﻥ k × n ﻤﻥ ﺍﻝﻨﻭﻉC c i j
1≤i , j ≤ n
k =1
: ﺠﺩﺍﺀ ﻤﺼﻔﻭﻓﺘﻴﻥ-3
)
t ( A ) = A .3
:ﺍﻝﻤﺼﻔﻭﻓﺔ ﺍﻝﻤﺭﺒﻌﺔ
1≤ j ≤ n
(
.2
∈ Μ m ×n ( K ) , λ ∈ K
λ A = λ ( ai j )1≤i ≤ m ﻨﻌﺭﻑ:ﺘﻌﺭﻴﻑ
C ∈ Μk n (K )
.1
( )
A ai j
: ﺒـ
1≤ i ≤ m 1≤ j ≤ n
( )
, B bi j
1≤ i ≤ m 1≤ j ≤ n
ﺍﻝﻤﻌﺭﻓﺔ ﻜﻤﺎC ﺍﻝﻤﺼﻔﻭﻓﺔB ﻭA ﻨﺴﻤﻲ ﻤﺠﻤﻭﻉ ﺍﻝﻤﺼﻔﻭﻓﺘﻴﻥ
l
C i j = ∑ ai k i b k j
( )
∀i , j : c i j = ai j + bi j ﺤﻴﺙC c i j
k =1
j ﻓﻲ ﺍﻝﻌﻤﻭﺩA ﻤﻥ ﺍﻝﻤﺼﻔﻭﻓﺔi ﻫﻭ ﻨﺎﺘﺞ ﻀﺭﺏ ﺍﻝﺴﻁﺭc i j
1≤i ≤ m 1≤ j ≤ n
: ﻴﻠﻲ
:ﻤﺜﺎل
. ﻤﻌﺎﻤﻼ ﻤﻌﺎﻤﻼB ﻤﻥ
1 −2 5 −2 1 5 0 −2 5 2 A ,B ,D ,F 3 7 0 1 3 2 7 −1 1 0
1 7 0 7 1 3 C −2 6 , B , A :ﻤﺜﺎل −1 3 2 4 5 1
13
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]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﻜل ﻤﻥ B iAﻭ A iBﻭ C iAﻤﻌﺭﻓﺔ ،
1 −1 0 A 2 6 1 3 5 7
A iCﻏﻴﺭ ﻤﻌﺭﻓﺔ .
0 7 3 −1 1 3 1× 0 + 3 × ( −1) 1× 7 + 3 × 3 A iB = 2 4 2 × 0 + 4 × ( −0 ) 2 × 7 + 4 × 2 −3 10 = −4 26
1 = ( 42 − 3 + 0 ) − ( 0 + 5 − 14 ) = 48 7
6 5
det ( A ) = 2 3
ﺤﺴﺎﺏ ﻤﻘﻠﻭﺏ ﻤﺼﻔﻭﻓﺔ:
ﺨﺎﺼﻴﺔ: ♦
1 −1 0
A i( B iC ) = A iB + A iC
ﺘﻌﺭﻴﻑ :ﻝﺘﻜﻥ ) A ∈ Μ n ( K
( B iC )iA = B iA + C iA
ﻨﻘﻭل ﻋﻥ Aﺃﻨﻬﺎ ﻗﺎﺒﻠﺔ ﻝﻠﻘﻠﺏ ﺇﺫﺍ ﻭﺠﺩﺕ A ′ ∈ Μ nﺒﺤﻴﺙ:
♦ ﺤﺎﻝﺔ ﺍﻝﻤﺼﻔﻭﻓﺔ Aﻤﺭﺒﻌﺔ ﻤﻥ ﺍﻝﻨﻭﻉ n × n A n = A n −1 iA ,............, A 3 = A 2 iA , A 2 = A iA
A iA ′ = A ′iA = I n
ﺤﻴﺙ I nﻫﻲ ﻤﺼﻔﻭﻓﺔ ﺍﻝﻭﺤﺩﺓ ﻓﻲ ﺍﻝﻔﻀﺎﺀ Μ n ×nﻭﻫﻲ ﻤﻌﺭﻓﺔ ﺒـ: 1 0 ⋯ 0 0 1 0 In ⋮ ⋱ 0 ⋯ ⋯ 1
ﺤﺴﺎﺏ ﺍﻝﻤﺤﺩﺩﺍﺕ ﻤﻥ ﺭﺘﺒﺔ ﺩﻨﻴﺎ : ﺤﺎﻝﺔ ﺍﻝﻤﺼﻔﻭﻓﺎﺕ ﻤﻥ ﺍﻝﻨﻭﻉ 2x2 a b A ﻝﺘﻜﻥ ) ∈ Μ 2×2 ( K c d ﻨﺴﻤﻲ ﻤﺤﺩﺩ Aﺍﻝﻌﺩﺩ = ab − bc
ﻤﺜﺎل: a b c d
= ) det ( A
1 −5 3 5 A = , B ﻤﺜﺎل : 3 7 −2 13 1 −5 = ) det ( A = 22 , det ( B ) = 49 3 7
ﺤﺎﻝﺔ ﺍﻝﻤﺼﻔﻭﻓﺎﺕ ﻤﻥ ﺍﻝﻨﻭﻉ 3x3 a11 a12 a13 ﻝﺘﻜﻥ ) A a21 a22 a23 ∈ Μ 3×3 ( K a 31 a32 a33 ﻨﺴﻤﻲ ) det ( Aﺍﻝﻌﺩﺩ ﺍﻝﻤﻌﺭﻑ ﺒـ:
1 0 0 1 0 ، I 2 ﻓﻲ I 3 0 1 0 : Μ 3×3 ﻓﻲ : Μ 2×2 0 1 0 0 1 ﻨﺭﻤﺯ ﻝﻠﻤﻘﻠﻭﺏ ﺇﻥ ﻭﺠﺩ ﺒـ A −1 ﺸﺭﻁ ﻻﺯﻡ ﻭﻜﺎﻓﻲ ﻝﻭﺠﻭﺩ A −1
) A −1ﻤﻭﺠﻭﺩﺓ(
⇔ ) ( det ( A ) ≠ 0
ﻗﺎﻋﺩﺓ ﺤﺴﺎﺏ ﻤﻘﻠﻭﺏ ﻤﺼﻔﻭﻓﺔ: -1ﺍﻝﺘﺄﻜﺩ ﻤﻥ ﺃﻥ . det ( A ) ≠ 0 -2ﺘﻌﻴﻴﻥ ﻤﺼﻔﻭﻓﺔ ﺍﻝﻤﻜﻤﻤﺎﺕ ﺍﻝﺠﺒﺭﻴﺔ ﺍﻝﻤﻘﺎﺒﻠﺔ ﻨﺭﻤﺯ ﻝﻬﺎ ﺒـ . Aɶ 1 t ɶ = A −1 A -3 ) det ( A
) det ( A ) = ( a11 a22 a33 + a12 a23 a31 + a13 a21 a32 ) − ( a31 a22 a13 + a32 a23 a11 + a33 a21 a12
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)= ℂ ∨ ℝ
(K
14
]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﺘﻌﻤﻴﻡ ﺤﺴﺎﺏ ﺍﻝﻤﺤﺩﺩﺍﺕ:
ﺃ -ﺤﺎﻝﺔ ﻤﺼﻔﻭﻓﺔ A ∈ Μ 2×2 a b ɶ d −c → t Aɶ = d −b A = →A = c d −b a −c a d −b 1 d −b 1 = A −1 = det ( A ) −c a a d − b c −c a
ﺤﺴﺎﺏ ﻤﺤﺩﺩ ﻤﻥ ﺍﻝﺭﺘﺒﺔ
a21 a22 a23 .......... a2 n ⋮ ai 1 ai 2 ai 3 .......... ain an 1 an 2 an 3 .......... an n
ﺤﺴﺎﺏ ) det ( Aﻴﺘﻡ ﻭﻓﻕ ﻁﺭﻴﻘﺘﻴﻥ:
ﺃ -ﻨﺸﺭ ﺍﻝﻤﺤﺩﺩ ﻭﻓﻕ ﺴﻁﺭ: ﻨﺸﺭ ﻤﺤﺩﺩ Aﻭﻓﻕ ﺍﻝﺴﻁﺭ i 0ﻴﻌﻁﻰ ﺒﺎﻝﻘﺎﻋﺩﺓ:
ai 0 j ∆i 0 j
) (
: aɶi , j ، Aɶ = aɶi jﻴﺴﻤﻰ ﺍﻝﻤﻜﻤﻡ ﺍﻝﺠﺒﺭﻱ ﻝـ ai j
∆i j
= ) det ( A
⋮
a11 a12 a13 ﻝﺘﻜﻥ A a21 a22 a23 a 31 a32 a33 .1ﺍﻝﺘﺄﻜﺩ ﻤﻥ ﺃﻥ . det ( A ) ≠ 0 .2ﻨﺤﺴﺏ ﻤﺼﻔﻭﻓﺔ ﺍﻝﻤﻜﻤﻤﺎﺕ Aɶ 1≤i , j ≤3
n
a11 a12 a13 .......... a1n
ﺏ -ﺤﺎﻝﺔ ﻤﺼﻔﻭﻓﺔ A ∈ Μ 3×3
i+j
) ( A ∈ Μ n ×n
)aɶi j = ( −1
n
det ( A ) = ∑ ( −1) 0
i +j
j =1
ﺤﻴﺙ : ∆ i 0 jﺍﻝﻤﺤﺩﺩ ﺍﻝﻨﺎﺘﺞ ﻤﻥ ﻤﺤﺩﺩ Aﺒﺤﺫﻑ ﺍﻝﺴﻁﺭ i 0ﻭﺍﻝﻌﻤﻭﺩ j
ﺤﻴﺙ : ∆ i jﺍﻝﻤﺤﺩﺩ ﺍﻝﻨﺎﺘﺞ ﻤﻥ ﻤﺤﺩﺩ Aﺒﺤﺫﻑ ﺍﻝﺴﻁﺭ iﻭﺍﻝﻌﻤﻭﺩ j
ﺏ -ﻨﺸﺭ ﺍﻝﻤﺤﺩﺩ ﻭﻓﻕ ﻋﻤﻭﺩ:
ﻤﺜﻼ:
ﻨﺸﺭ ﻤﺤﺩﺩ Aﻭﻓﻕ ﺍﻝﻌﻤﻭﺩ j 0ﻴﻌﻁﻰ ﺒﺎﻝﻘﺎﻋﺩﺓ: a23
a22
a33
a32
a12
a11
a32
a32
= ∆11 = ∆ 23
ai j 0 ∆i j 0
)aɶ11 = ( −1
1+1
2 +3
)aɶ23 = ( −1
1 t ɶ -1ﻨﻁﺒﻕ ﺍﻝﻘﺎﻋﺩﺓ A : ) det ( A
= A −1
i + j0
n
)det ( A ) = ∑ ( −1 i =1
ﺤﻴﺙ : ∆ i j 0ﺍﻝﻤﺤﺩﺩ ﺍﻝﻨﺎﺘﺞ ﻤﻥ ﻤﺤﺩﺩ Aﺒﺤﺫﻑ ﺍﻝﺴﻁﺭ iﻭﺍﻝﻌﻤﻭﺩ j 0
ﻤﺜﺎل:
1 2 3 1 2 3 A 3 0 1 , A = 3 0 1 −1 1 3 −1 1 3 ﻨﻨﺸﺭﻩ ﻭﻓﻕ ﺍﻝﺴﻁﺭ : i 0 = 2 a2 j ∆ 2 j
2+ j
3
)det ( A ) = ∑ ( −1 j =1
3
1
−1 3
a22
2+ 2
)+ ( −1 2
2 3 1 3 1
−1 1
a23
a21 2+3
2 +1
)= ( −1
)+ ( −1
= −9 + 0 − 3 = −12
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15
â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
g : ℝ 2 → ℝ3
:ﻤﺼﻔﻭﻓﺔ ﺘﻁﺒﻴﻕ ﺨﻁﻲ . ﻋﻠﻰ ﺍﻝﺘﺭﺘﻴﺏm ﻭn ﺒﻌﺩﺍﻫﻤﺎ، ﻓﻀﺎﺌﻴﻥ ﺸﻌﺎﻋﻴﻥF ﻭE ﺘﻁﺒﻴﻕ ﺨﻁﻲf : E → F
(x , y ) → (x + y , 2 x − y , x − 3 y ) B E = {(1, 0 ) , ( 0,1)} B F = (1, 0, 0 ), ( 0,1, 0 ), ( 0, 0,1) v2 v3 v 1 f f
2
1
ﺍﻝﺘﻲ ﺃﻋﻤﺩﺘﻬﺎ ﻫﻲ ﺼﻭﺭ ﺃﺴﺎﺱ ﺍﻝﻤﻨﻁﻠﻕ ﻤﻜﺘﻭﺒﺔ ﻓﻲ ﺃﺴﺎﺱ ﺍﻝﻭﺼﻭل f (u 1 ) = α11v 1 + α 21v 2 + α 31v 3 + .......... + α m 1v m
3
2
f (u 2 ) = α12v 1 + α 22v 2 + α 32v 3 + .......... + α m 2v m
3
1 1 ⇒ M = 2 −1 1 −3
⋮
f (u j ) = α1 jv 1 + α 2 jv 2 + α 3 jv 3 + .......... + α mjv m ⋮
h : ℝ 3 [x ] → ℝ 3 [x ] p → p′
♦
B E = {1, x , x 2 , x 3 } , B F = {1, x , x 2 , x 3 }
↓ α11 α 21 M (f , B E , B F ) = ⋮ αm1
h (1) = 0 = 0i1 + 0ix + 0ix 2 + 0ix 3 h ( x ) = 1 = 1i1 + 0ix + 0ix 2 + 0ix 3
( ) = 2x = 0i1 + 2ix + 0ix h ( x ) = 3x = 0i1 + 0ix + 3ix 2
3
2
2
0 0 M (h ) = 0 0
+ 0i x
2
f (u n ) = α1nv 1 + α 2 nv 2 + α 3nv 3 + .......... + α mnv m f (u1 ) f (u 2 ) ⋯ f (u j ) ⋯ f (u n )
dim E = dim F = 4 ﻷﻥ4 × 4 ﻤﻥ ﺍﻝﻨﻭﻉh ﺇﻥ ﻤﺼﻔﻭﻓﺔ
h x
F ﺃﺴﺎﺱ ﻝـB F = { v 1 ,v 2 ,⋯ ,v m }
ﺍﻝﻤﺼﻔﻭﻓﺔB F ﻭB E ﺒﺎﻝﻨﺴﺒﺔ ﻝﻸﺴﺎﺴﻴﻥf ﻨﺴﻤﻲ ﻤﺼﻔﻭﻓﺔ:ﺘﻌﺭﻴﻑ
( (1, 0 ) ) = (1, 2,1) = v + 2v + v ( ( 0,1) ) = (1, −1, −3) = v −v + 3v 1
E ﺃﺴﺎﺱ ﻝـB E = { u1 , u 2 ,⋯ , u n }
3
↓
⋯
↓
⋯
α12 α 22
⋯
α1 j α2 j
⋯
⋮
⋯ ⋮
αm 2
⋯
⋮
⋯ ⋮
α mj
⋯
↓
α1n v 1 α 2 n v 2
⋮ ⋮ α m n v m
+ 0i x 3
m × n : M ﻨﻭﻉ ﺍﻝﻤﺼﻔﻭﻓﺔ
1 0 0 0 2 0 :ﻭﻤﻨﻪ 0 0 3 0 0 0
k : ℝ → ℝ3 t → ( t , 2t , 6t )
E ﺒﻌﺩ: M ﻋﺩﺩ ﺃﻋﻤﺩﺓ، F ﺒﻌﺩ: M ﻋﺩﺩ ﺃﺴﻁﺭ
. 3 × 2 ﻤﺼﻔﻭﻓﺔ ﻤﻥ ﺍﻝﻨﻭﻉ، ﺨﻁﻲf : ℝ 2 → ℝ 3 :ﻤﺜﻼ
♦
3 × 1 ﻤﻥ ﺍﻝﻨﻭﻉk ﻤﺼﻔﻭﻓﺔ B E = { 1} , B F = {(1, 0, 0 ) , ( 0,1, 0 ) , ( 0, 0,1)} f (1) = (1, 2, 6 ) = 1e1 + 2 e 2 + 6 e 3
1 M ( k , B E , B F ) = 2 :ﻭﻤﻨﻪ 6
16
dim E = n ﺘﻘﺎﺒﻠﻪ f ∈ L ( E , F ) ﻜل dim F = m . Μ m ×n ﻤﺼﻔﻭﻓﺔ ﻤﻥ ﺍﻝﻨﻭﻉ
ﻭﻁﻠﺏ ﺍﻝﺘﻁﺒﻴﻕ ﺍﻝﺨﻁﻲΜ m ×n ﺇﺫﺍ ﻋﻠﻤﺕ ﻤﺼﻔﻭﻓﺔ ﻤﻥ ﺍﻝﻨﻭﻉ
♦
♦
:ﺍﻝﻤﻘﺎﺒل ﻝﻬﺎ ﻓﺈﻥ ﻋﺒﺎﺭﺘﻪ ﺘﻌﻁﻰ ﺒﺎﻝﻘﺎﻋﺩﺓ
f : E ( dim E = n ) → F ( dim F = m ) f (X ) = A t X a11 a12 a13 .......... a1n x 1 a a a .......... a2 n x 2 f ( X ) = 21 22 23 ⋮ ⋮ am 1 am 2 am 3 .......... amn x n www.math.3arabiyate.net
]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل )ﺘﻐﻴﻴﺭ ﺍﻝﻘﺎﻋﺩﺓ( :
1 0 −1 A 2 0 5 3 1 6 ﻝﻨﺒﺤﺙ ﻋﻥ ﺍﻝﺘﻁﺒﻴﻕ ﺍﻝﺨﻁﻲ ﺍﻝﻤﻘﺎﺒل ﻝﻬﺎ.
Eﻓﻀﺎﺀ ﺸﻌﺎﻋﻲ ،ﺒﻌﺩﻩ . n
} B 2 = { v 1 ,v 2 ,⋯ ,v n } ، B 1 = { u1 , u 2 ,⋯ , u n ﺃﺴﺎﺴﻴﻥ ﻝـ . E ﺘﻌﺭﻴﻑ :ﻨﺴﻤﻲ ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل ﻤﻥ ﺍﻷﺴﺎﺱ B 1ﻨﺤﻭ ﺍﻷﺴﺎﺱ : B 2
ﺇﻨﻬﺎ ﻤﻥ ﺍﻝﻨﻭﻉ 3 × 3
ﺍﻝﻤﺼﻔﻭﻓﺔ Pﺍﻝﺘﻲ ﺃﻋﻤﺩﺘﻬﺎ ﻫﻲ ﻋﻨﺎﺼﺭ ﺍﻷﺴﺎﺱ B 2ﻤﻜﺘﻭﺒﺔ ﻓﻲ
ﻝﻨﻔﺭﺽ ﺃﻥ Aﻤﺭﻓﻘﺔ ﺒﺎﻝﺘﻁﺒﻴﻕ: 3
, X ( x , y , z ) ∈ ℝ3
)
f :ℝ → ℝ 3
ﺍﻷﺴﺎﺱ . B 1 v 1 = α11u1 + α 21u 2 + α 31u 3 + ⋯ + α n 1u n
(
f (x , y , z ) = t A. t X
v 2 = α12u1 + α 22u 2 + α 32u 3 + ⋯ + α n 2 u n
x −z 1 0 −1 x 2 0 5 y = 2 x + 5 z 3 1 6 z 3 x + y + 6 z ) f (x , y , z ) = (x − z , 2 x + 5 z , 3x + y + 6 z
⋮ v i = α1i u 1 + α 2i u 2 + α 3i u 3 + ⋯ + α n i u n ⋮ v n = α1nu1 + α 2 nu 2 + α 3nu 3 + ⋯ + α n n u n
1 2 ، M = ﻭﻝﻴﻜﻥ fﺍﻝﺘﻁﺒﻴﻕ ﺍﻝﺨﻁﻲ ﺤﻴﺙ: ♦ ﻝﺘﻜﻥ 3 4 ) f : M 2×2 ( ℝ ) → M 2×2 ( ℝ X → f ( X ) = MX
2 1 0 1 = 4 0 0 3 2 0 1 0 = 4 0 0 0 2 0 0 2 = 4 1 0 4
0 = 1B1 + 0B 2 + 3B3 + 0B 4 0 1 = 0B1 + 1B 2 + 0B3 + 3B 4 3 0 = 2B1 + 0B 2 + 4B3 + 0B 4 0 1 2 0 0 0 2 f ( B 4 ) = MB 4 = = = 0B1 + 2B 2 + 0B3 + 4B 4 3 4 0 1 0 4
0 2 0 1 0 2 ﺇﺫﻥ: 0 4 0 3 0 4
1 0 M (f , B , B ) = 3 0
1 f ( B1 ) = MB1 = 3 1 f ( B 2 ) = MB 2 = 3 1 f ( B3 ) = MB3 = 3
↓
⋯
⋮ ⋮ α n n u n
ﺇﻴﺠﺎﺩ ﻤﺼﻔﻭﻓﺔ fﺒﺎﻝﻨﺴﺒﺔ ﻝﻠﻘﺎﻋﺩﺓ ﺍﻝﻤﻌﺘﺎﺩﺓ 1 0 0 1 0 0 0 0 B = B1 = , B2 = , B3 = , B4 = 0 0 0 0 1 0 0 1 ﻝﺩﻴﻨﺎ:
vn
⋯
α1n u1 α 2 n u 2
⋯
v2
v1
↓
⋯
↓
↓
⋯
α1i α 2i
⋯ ⋯
α12 α 22
⋮
⋮
⋮
⋮
⋯
α ni
⋯
αn 2
⋯
vi
α11 α 21 P = ⋮ α n1
ﻨﻭﻉ ﺍﻝﻤﺼﻔﻭﻓﺔ n × n : M
ﺘﻌﺭﻴﻑ :2ﻨﺴﻤﻲ ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل ﻤﻥ ﺍﻷﺴﺎﺱ B 1ﻨﺤﻭ ﺍﻷﺴﺎﺱ : B 2 ﻤﺼﻔﻭﻓﺔ ﺍﻝﺘﻁﺒﻴﻕ ﺍﻝﺨﻁﻲ ، Id : E → Eﺤﻴﺙ Idﺍﻝﺘﻁﺒﻴﻕ ﺍﻝﻤﻁﺎﺒﻕ ﻓﻲ ، Eﻤﺯﻭﺩﺍ ﺒﺎﻷﺴﺎﺱ B 2ﻓﻲ ﺍﻝﻤﻨﻁﻠﻕ ،ﻭﺍﻷﺴﺎﺱ B 1
ﻓﻲ ﺍﻝﻭﺼﻭل ،ﻭﻨﻜﺘﺏ. P = M ( Id , B 2 , B 1 ) : ♦ ﻝﺘﻜﻥ Qﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل ﻤﻥ ﺍﻷﺴﺎﺱ B 2ﻨﺤﻭ ﺍﻷﺴﺎﺱ B 1
) Q = M ( Id , B 1 , B 2 ﻝﺩﻴﻨﺎ . Q = P −1 : ﺤﺫﺍﺭ :ﻝﺘﺤﻭﻴل ﻋﺒﺎﺭﺓ ﺸﻌﺎﻉ ﻤﻜﺘﻭﺏ ﻓﻲ ﺍﻷﺴﺎﺱ B 1ﻨﺤﻭ ﺍﻷﺴﺎﺱ B 2ﻨﺴﺘﻌﻤل ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل ﻤﻥ ﺍﻷﺴﺎﺱ B 2ﻨﺤﻭ ﺍﻷﺴﺎﺱ . B 1
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17
]â‡<ê×éfu<<<<<<<<<<<<<<<<<<<<<<<<êŞ}<¢]<–<êÞ^nÖ]<ðˆ¢]
}
3
f E , B 2 → F , B 2′
{
)B 1 = {1, x , x 2 , x 3 } , B 2 = 1, ( x − 1) , ( x − 1) , ( x − 1 2
f = ( Id ) F f Id E −1
3
2
−1
♦ ﺇﺫﺍ ﻜﺎﻥ E = Fﻓﺈﻥ:
1 0 P = 0 0
E , B 2 →E ,B2 f
f = ( Id ) E f Id E −1
f E , B 1 → E , B1
x3
x2
x
1
1 3 3 1
1 2 1 0
1 1 0 0
1 0 −1 Q =P = 0 0
) f ( x , y , z ) = ( x + 3 y + 2z , x − 4z , y + 3z ﺃﻭﺠﺩ Bﻤﺼﻔﻭﻓﺔ ) fﺍﻝﺘﻤﺜﻴل ﺍﻝﻤﺼﻔﻭﻓﻲ( ﻓﻲ ﺍﻝﻘﺎﻋﺩﺓ:
}) S = {w 1 = (1,1,1) , w 2 = (1,1, 0 ) , w 3 = (1, 0, 0
ﻝﺩﻴﻨﺎ ﻤﺼﻔﻭﻓﺔ fﻓﻲ ﺍﻝﻘﺎﻋﺩﺓ ﺍﻝﻤﻌﺘﺎﺩﺓ
. ( x − 1) ، ( x − 1) ، ( x − 1) ، 1 2
3
}) E = {e1 = (1, 0, 0 ) , e 2 = ( 0,1, 0 ) , e 3 = ( 0, 0,1ﻫﻲ
ﺍﻝﻐﺭﺽ ﻤﻥ ﺍﻝﺘﻁﺒﻴﻕ ﺘﺤﻭﻴل ﻋﺒﺎﺭﺓ pﺍﻝﻤﻜﺘﻭﺏ ﻓﻲ ﺍﻝﻘﺎﻋﺩﺓ B 1ﺇﻝﻰ ﻋﺒﺎﺭﺓ ﻤﻜﺘﻭﺒﺔ ﻓﻲ ﺍﻝﻘﺎﻋﺩﺓ ، B 2ﻝﻬﺫﺍ ﻨﺴﺘﻌﻤل ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل Qﻤﻥ ﺍﻷﺴﺎﺱ B 2ﻨﺤﻭ ﺍﻷﺴﺎﺱ 1 1 1 1 7 . B 1 1 2 3 1 14 = 0 1 3 2 11 0 0 1 3 3
1 0 0 0
ﻭﻤﻨﻪ )p ( x ) = 7 + 14 ( x − 1) + 11( x − 1) + 3 ( x − 1 2
3
ﻗﺎﻋﺩﺓ: ﻝﻴﻜﻥ ، f : E → Fﺤﻴﺙ Eﻭ Fﻓﻀﺎﺌﻴﻥ ﻤﻨﺘﻬﻴﻲ ﺍﻝﺒﻌﺩ. 012
ﻭﻝﻴﻜﻥ B 1′ ، B 1ﺃﺴﺎﺴﻴﻥ ﻝـ F ، Eﻋﻠﻰ ﺍﻝﺘﺭﺘﻴﺏ.
(
. A = M f , B1 , B 1′
ﻭﻝﻴﻜﻥ B 2′ ، B 2ﺃﺴﺎﺴﻴﻥ ﺁﺨﺭﻴﻥ ﻝـ F ، Eﻋﻠﻰ ﺍﻝﺘﺭﺘﻴﺏ.
(
. B = M f , B 2 , B 2′
ﻝﺩﻴﻨﺎ ﻤﺎ ﻴﻠﻲ:
−1
B = P −1AP
ﺘﻁﺒﻴﻕ :ﺃﻜﺘﺏ p ( x ) = 1 + x + 2x 2 + 3x 3ﺒﺩﻻﻝﺔ:
)
↑ ( Id )E
↓ Id E
ﻤﺜﺎل :ﻝﻴﻜﻥ f : ℝ3 → ℝ 3ﻤﻌﺭﻑ ﻜﻤﺎ ﻴﻠﻲ:
1 ) ( x − 1 2 ) ( x − 1 3 ) ( x − 1
)
f E , B 1 → F , B 1′
1
−1 1 1 −2 3 x 0 1 −3 x 2 0 0 1 x3 ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل Qﻤﻥ ﺍﻷﺴﺎﺱ B 2ﻨﺤﻭ ﺍﻷﺴﺎﺱ : B 1 1
−1
B = Q −1AP
ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل Pﻤﻥ ﺍﻷﺴﺎﺱ B 1ﻨﺤﻭ ﺍﻷﺴﺎﺱ : B 2
)( x − 1) ( x − 1) ( x − 1
↑ ( Id )F
↓ Id E
B = Q −1AP
2 −4 3 ﻭﻝﺩﻴﻨﺎ
1 1 1 ، P = 1 1 0 ﺇﺫﻥ ﻤﺼﻔﻭﻓﺔ ﺘﻐﻴﻴﺭ ﺍﻝﻘﺎﻋﺩﺓ ﻤﻥ ﺍﻷﺴﺎﺱ Sﺇﻝﻰ 1 0 0 0 0 1 −1 ﺍﻷﺴﺎﺱ Eﻫﻲ Q = P −1ﺤﻴﺙP = 0 1 −1 : 1 −1 0 ﻭﻤﻨﻪ B = P −1AP 0 0 1 1 3 2 1 1 1 4 1 0 ﺇﺫﻥ B = 0 1 −1 1 0 −4 1 1 0 = −7 0 1 1 −1 0 0 1 3 1 0 0 9 3 0
4 1 0 ﺃﻱ. B = M ( f , S , S ) = −7 0 1 : 9 3 0
: Pﻫﻲ ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل ﻤﻥ B 1ﻨﺤﻭ . B 2 : Qﻫﻲ ﻤﺼﻔﻭﻓﺔ ﺍﻻﻨﺘﻘﺎل ﻤﻥ B 1′ﻨﺤﻭ . B 2′ www.math.3arabiyate.net
1 3 A = M (f , E , E ) = 1 0 0 1 ﻤﺼﻔﻭﻓﺔ ﺘﻐﻴﻴﺭ ﺍﻝﻘﺎﻋﺩﺓ ﻤﻥ ﺍﻷﺴﺎﺱ Eﺇﻝﻰ ﺍﻷﺴﺎﺱ Sﻫﻲ
18
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