Definition: Let ( u i ) = ( u 1 , u 2 , … , u n ) be a sequence of vectors in V {\displaystyle {\mbox{Definition: Let }}(u_{i})=(u_{1},u_{2},\ldots ,u_{n}){\mbox{ be a sequence of vectors in }}V} .
A sum of the form: {\displaystyle {\mbox{A sum of the form:}}{}_{}^{}}
a i ∈ F , ∑ i = 1 n a i u i = a 1 u 1 + a 2 u 2 + … + a n u n {\displaystyle a_{i}\in F,\sum _{i=1}^{n}a_{i}u_{i}=a_{1}u_{1}+a_{2}u_{2}+\ldots +a_{n}u_{n}}
is called a Linear Combination of the u i {\displaystyle {\mbox{is called a Linear Combination of the }}u_{i}^{}} .
span ( u i ) := { The set of all possible linear combinations of the u i } {\displaystyle {\mbox{span}}(u_{i}^{}):=\lbrace {\mbox{ The set of all possible linear combinations of the }}u_{i}^{}\rbrace }
If S ⊂ V is any subset, {\displaystyle {\mbox{If }}{\mathcal {S}}\subset V\ {\mbox{ is any subset, }}}
span ( S ) := { The set of all linear combination of vectors in S } = { ∑ i = 0 n a i u i , a i ∈ F , u i ∈ S } {\displaystyle {\mbox{span}}({\mathcal {S}}):=\lbrace {\mbox{The set of all linear combination of vectors in }}{\mathcal {S}}\rbrace =\left\lbrace \sum _{i=0}^{n}a_{i}u_{i},\quad a_{i}\in F,u_{i}\in {\mathcal {S}}\right\rbrace }
span ( S ) always contains 0 even if S = ∅ {\displaystyle {\mbox{span}}({\mathcal {S}}){\mbox{ always contains }}0{\mbox{ even if }}{\mathcal {S}}=\emptyset }
Theorem
∀ S ⊂ V , span ( S ) is a subspace of V {\displaystyle \forall {\mathcal {S}}\subset V{\mbox{, span}}({\mathcal {S}}){\mbox{ is a subspace of }}V}
Proof: {\displaystyle {\mbox{Proof:}}{}_{}^{}}
1. 0 ∈ span ( S ) {\displaystyle 0\in {\mbox{ span}}({\mathcal {S}})} . 2. Let x ∈ span ( S ) ⇒ x = ∑ i = 1 n a i u i , u i ∈ S , {\displaystyle {\mbox{Let }}x\in {\mbox{ span}}({\mathcal {S}})\Rightarrow x=\sum _{i=1}^{n}a_{i}u_{i}{\mbox{, }}u_{i}\in {\mathcal {S}}{\mbox{, }}}
and let y ∈ span ( S ) ⇒ y = ∑ i = 1 m b i v i , v i ∈ S {\displaystyle {\mbox{and let }}y\in {\mbox{ span}}({\mathcal {S}})\Rightarrow y=\sum _{i=1}^{m}b_{i}v_{i}{\mbox{, }}v_{i}\in {\mathcal {S}}}
x + y = ∑ i = 1 n a i u i + ∑ i = 1 m b i v i = ∑ i = 1 max ( m , n ) c i w i {\displaystyle x+y=\sum _{i=1}^{n}a_{i}u_{i}+\sum _{i=1}^{m}b_{i}v_{i}=\sum _{i=1}^{{\mbox{max}}(m,n)}c_{i}w_{i}}
where c i = ( a 1 + b 1 , a 2 + b 2 , … , a max ( m , n ) + b max ( m , n ) ) and w i ∈ S {\displaystyle \qquad {\mbox{ where }}c_{i}=(a_{1}+b_{1},a_{2}+b_{2},\ldots ,a_{{\mbox{max}}(m,n)}+b_{{\mbox{max}}(m,n)}){\mbox{ and }}w_{i}\in {\mathcal {S}}}
3. c x = c ∑ i = 1 n a i u i = ∑ i = 1 n ( c a i ) u i ∈ span ( S ) {\displaystyle cx=c\sum _{i=1}^{n}a_{i}u_{i}=\sum _{i=1}^{n}(ca_{i})u_{i}\in {\mbox{ span}}({\mathcal {S}})}
Example 1.
Let P 3 ( R ) = { a x 3 + b x 2 + c x + d } ⊂ P ( R ) , where a , b , c , d ∈ R {\displaystyle {\mbox{Let }}P_{3}(\mathbb {R} )=\lbrace ax^{3}+bx^{2}+cx+d\rbrace \subset P(\mathbb {R} ){\mbox{, where }}a,b,c,d\in \mathbb {R} } .
u 1 = x 3 − 2 x 2 − 5 x − 3 u 2 = 3 x 3 − 5 x 2 − 4 x − 9 v = 2 x 3 − 2 x 2 + 12 x − 6 {\displaystyle {\begin{matrix}u_{1}^{}&=&x^{3}-2x^{2}-5x-3\\u_{2}^{}&=&3x^{3}-5x^{2}-4x-9\\v_{}^{}&=&2x^{3}-2x^{2}+12x-6\end{matrix}}}
Let W = span ( u 1 , u 2 ) , {\displaystyle {\mbox{Let }}W={\mbox{span}}(u_{1}^{},u_{2}^{}){\mbox{,}}}
Does/Is v ∈ W ? {\displaystyle {\mbox{Does/Is }}v\in W{\mbox{ ?}}}
v ∈ W if it is a linear combination of span ( u 1 , u 2 ) {\displaystyle v\in W{\mbox{ if it is a linear combination of span}}(u_{1}^{},u_{2}^{})}
v = a 1 u 1 + a 2 u 2 for some a 1 , a 2 ∈ R {\displaystyle v=a_{1}u_{1}+a_{2}u_{2}{\mbox{ for some }}a_{1},a_{2}\in \mathbb {R} }
If ∃ a 1 , a 2 ∈ R {\displaystyle {\mbox{If }}\exists a_{1},a_{2}\in \mathbb {R} }
2 x 3 − 2 x 2 + 12 x − 6 = a 1 ( x 3 − 2 x 2 − 5 x − 3 ) + a 2 ( 3 x 3 − 5 x 2 − 4 x − 9 ) = ( a 1 + 3 a 2 ) x 3 + ( − 2 a 1 − 5 a 2 ) x 2 + ( − 5 a 1 − 4 a 2 ) x + ( − 3 a 1 − 9 a 2 ) {\displaystyle {\begin{matrix}2x^{3}-2x^{2}+12x-6&=&a_{1}^{}(x^{3}-2x^{2}-5x-3)+a_{2}^{}(3x^{3}-5x^{2}-4x-9)\\\ &=&(a_{1}^{}+3a_{2}^{})x^{3}+(-2a_{1}^{}-5a_{2}^{})x^{2}+(-5a_{1}^{}-4a_{2}^{})x+(-3a_{1}^{}-9a_{2}^{})\end{matrix}}}
Need to solve { 2 = a 1 + 3 a 2 − 2 = − 2 a 1 − 5 a 2 12 = − 5 a 1 − 4 a 2 − 6 = − 3 a 1 − 9 a 2 {\displaystyle {\mbox{Need to solve}}{\begin{cases}2=a_{1}^{}+3a_{2}^{}\\-2=-2a_{1}^{}-5a_{2}^{}\\12=-5a_{1}^{}-4a_{2}^{}\\-6=-3a_{1}^{}-9a_{2}^{}\end{cases}}}
Solve the four equations above and we will get a 1 = − 4 and a 2 = 2 {\displaystyle {\mbox{Solve the four equations above and we will get }}a_{1}^{}=-4{\mbox{ and }}a_{2}^{}=2}
Check if a 1 = − 4 and a 2 = 2 holds for all 4 equations. {\displaystyle {\mbox{Check if }}a_{1}^{}=-4{\mbox{ and }}a_{2}^{}=2{\mbox{ holds for all 4 equations.}}}
Since it holds, v ∈ W {\displaystyle {\mbox{Since it holds, }}v\in W}
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