Matematikan eta zehazkiago aljebra linealean bektore espazioa edo espazio bektoriala hutsa ez den multzo batetik sorturiko egitura aljebraiko bat da, egitura hau aipatutako multzo ez hutsa horren eta bektore batuketa batetik (barne operazioa) edota eskalar biderketa sortzen da.
Bektore espazioaren irudikapena.
Izan bitez ( K , + , ⋅ ) {\displaystyle (K,+,\cdot )} V {\displaystyle V} V K {\displaystyle VK}
( V , + ) {\displaystyle (V,+)} Talde abeldarra da, hau da, multzoak gehiketa aplikazioa definituta du, taldearen elementu neutroa 0 V {\displaystyle 0_{V}} Kanpo aplikazioa (eragiketa) existitzen da:⋅ : K × V → V ( k , v ) → k ⋅ v {\displaystyle {\begin{aligned}\cdot :K\times V\to V\\(k,v)\to k\cdot v\\\end{aligned}}} eskalar eta bektore arteko biderketa) Gorputzeko elementuak eskalarrak deritze eta V {\displaystyle V}
Banatze propietatea eskalarren gehiketarekiko: ∀ k 1 , k 2 ∈ K , ∀ v ∈ V : ( k 1 + k 2 ) ⋅ v = k 1 ⋅ v + k 2 ⋅ v {\displaystyle \forall k_{1},k_{2}\in K,\forall v\in V:(k_{1}+k_{2})\cdot v=k_{1}\cdot v+k_{2}\cdot v}
Banatze propietatea bektoreen gehiketarekiko:∀ v 1 , v 2 ∈ V , ∀ k ∈ K : ( v 1 + v 2 ) ⋅ k = v 1 ⋅ k + v 2 ⋅ k {\displaystyle \forall v_{1},v_{2}\in V,\forall k\in K:(v_{1}+v_{2})\cdot k=v_{1}\cdot k+v_{2}\cdot k}
Elkartze propietatea: ∀ k 1 , k 2 ∈ K , ∀ v ∈ V : k 1 ( k 2 ⋅ v ) = ( k 1 ⋅ k 2 ) ⋅ v {\displaystyle \forall k_{1},k_{2}\in K,\forall v\in V:k_{1}(k_{2}\cdot v)=(k_{1}\cdot k_{2})\cdot v}
Bektorea eta gorputzeko biderketarekiko elementu neutroaren arteko produktua bektore bera da: ∀ v ∈ V : v ⋅ 1 K = v {\displaystyle \forall v\in V:v\cdot 1_{K}=v}
Z {\displaystyle \mathbb {Z} } Q {\displaystyle \mathbb {Q} } ( Z , + ) {\displaystyle (\mathbb {Z} ,+)}
Ostean, Q {\displaystyle \mathbb {Q} } Q {\displaystyle \mathbb {Q} } Q 2 {\displaystyle \mathbb {Q} ^{2}} Q {\displaystyle \mathbb {Q} }
Beraz ( K , + , ⋅ ) {\displaystyle (K,+,\cdot )} K n {\displaystyle K^{n}} K {\displaystyle K}
Espazio bektorialak: Q n {\displaystyle \mathbb {Q} ^{n}} Q {\displaystyle \mathbb {Q} } R n {\displaystyle \mathbb {R} ^{n}} R {\displaystyle \mathbb {R} } C n {\displaystyle \mathbb {C} ^{n}} C {\displaystyle \mathbb {C} }
Propietate gehiago:
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V {\displaystyle V} K {\displaystyle K}
( i ) {\displaystyle (i)} ∀ k ∈ K : k ⋅ 0 V = 0 V {\displaystyle \forall k\in K:k\cdot 0_{V}=0_{V}}
Badakigu 0 V = 0 V + 0 V {\displaystyle 0_{V}=0_{V}+0_{V}} k ⋅ 0 V = k ⋅ ( 0 V + 0 V ) = k ⋅ 0 V + k ⋅ 0 V {\displaystyle k\cdot 0_{V}=k\cdot (0_{V}+0_{V})=k\cdot 0_{V}+k\cdot 0_{V}} k ⋅ 0 V = k ⋅ 0 V + k ⋅ 0 V {\displaystyle k\cdot 0_{V}=k\cdot 0_{V}+k\cdot 0_{V}} 0 V = k ⋅ 0 V {\displaystyle 0_{V}=k\cdot 0_{V}}
( i i ) {\displaystyle (ii)} ∀ v ∈ V : v ⋅ 0 K = 0 V {\displaystyle \forall v\in V:v\cdot 0_{K}=0_{V}}
Badakigu 0 K = 0 K + 0 K {\displaystyle 0_{K}=0_{K}+0_{K}} v ⋅ 0 K = v ⋅ ( 0 K + 0 K ) = v ⋅ 0 K + v ⋅ 0 K ⟹ v ⋅ 0 K = v ⋅ 0 K + v ⋅ 0 K {\displaystyle v\cdot 0_{K}=v\cdot (0_{K}+0_{K})=v\cdot 0_{K}+v\cdot 0_{K}\Longrightarrow v\cdot 0_{K}=v\cdot 0_{K}+v\cdot 0_{K}} 0 V = v ⋅ 0 K {\displaystyle 0_{V}=v\cdot 0_{K}}
( i i i ) {\displaystyle (iii)} k ⋅ v = 0 V ⟺ k = 0 K e d o v = 0 V {\displaystyle k\cdot v=0_{V}\Longleftrightarrow k=0_{K}edov=0_{V}}
⟹ {\displaystyle \Longrightarrow }
Demagun k ∈ K − { 0 K } {\displaystyle k\in K-\{0_{K}\}} k − 1 {\displaystyle k^{-1}} k ⋅ v = 0 V {\displaystyle k\cdot v=0_{V}}
( k − 1 ⋅ k ) ⋅ v = k − 1 ⋅ 0 V ⟹ 1 K ⋅ v = 0 V ⟹ v = 0 V {\displaystyle (k^{-1}\cdot k)\cdot v=k^{-1}\cdot 0_{V}\Longrightarrow 1_{K}\cdot v=0_{V}\Longrightarrow v=0_{V}}
⟸ {\displaystyle \Longleftarrow } ( i ) {\displaystyle (i)} ( i i ) {\displaystyle (ii)}
( i v ) {\displaystyle (iv)} − v = ( − 1 K ) ⋅ v , ∀ v ∈ V {\displaystyle -v=(-1_{K})\cdot v,\forall v\in V}
Mugitu adierazpenean bektorea eskumara: 0 V = v + ( − 1 K ) ⋅ v {\displaystyle 0_{V}=v+(-1_{K})\cdot v} v = ( 1 K ) ⋅ v , ∀ v ∈ V {\displaystyle v=(1_{K})\cdot v,\forall v\in V} ( 1 K ) ⋅ v + ( − 1 K ) ⋅ v = 0 V {\displaystyle (1_{K})\cdot v+(-1_{K})\cdot v=0_{V}}
v ⋅ ( 1 K − 1 K ) = 0 V {\displaystyle v\cdot (1_{K}-1_{K})=0_{V}} 1 K − 1 K = 0 K {\displaystyle 1_{K}-1_{K}=0_{K}} v ⋅ ( 1 K − 1 K ) = v ⋅ ( 0 k ) = 0 V {\displaystyle v\cdot (1_{K}-1_{K})=v\cdot (0_{k})=0_{V}}
( v ) {\displaystyle (v)} u + w = v + w ⟹ u = v {\displaystyle u+w=v+w\Longrightarrow u=v}
( v i ) {\displaystyle (vi)} 0 v = 0 , ∀ v ∈ V {\displaystyle 0v=0,\forall v\in V}
( v i i ) {\displaystyle (vii)} λ 0 = 0 , ∀ λ ∈ K {\displaystyle \lambda 0=0,\forall \lambda \in K}
( v i i i ) {\displaystyle (viii)} ( − λ ) v = − λ v , ∀ λ ∈ K {\displaystyle (-\lambda )v=-\lambda v,\forall \lambda \in K} ∀ v ∈ K {\displaystyle \forall v\in K}
( i x ) {\displaystyle (ix)} λ v = 0 ⟺ λ = 0 {\displaystyle \lambda v=0\Longleftrightarrow \lambda =0} v = 0 {\displaystyle v=0}
( x ) λ v = μ v {\displaystyle (x)\lambda v=\mu v} v ≠ 0 ⟹ λ = μ {\displaystyle v\neq 0\Longrightarrow \lambda =\mu }
( x i ) {\displaystyle (xi)} λ v = λ w {\displaystyle \lambda v=\lambda w} λ ≠ 0 ⟹ v = w {\displaystyle \lambda \neq 0\Longrightarrow v=w}
Kanpo estekak
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Datuak: Q125977
Multimedia: Vector spaces / Q125977
