Definition {\displaystyle {\mbox{Definition}}{}_{}^{}}
v ∈ V is a linear combination of elements in S ⊂ V {\displaystyle v\in V{\mbox{ is a linear combination of elements in }}S\subset V} if ∃ u 1 , … , u n ∈ S and a 1 , … , a n ∈ F such that V = ∑ a i u i {\displaystyle {\mbox{ if }}\exists u_{1},\ldots ,u_{n}\in S{\mbox{ and }}a_{1},\dots ,a_{n}\in F{\mbox{ such that }}V=\sum a_{i}u_{i}}
Example {\displaystyle {\mbox{Example}}{}_{}^{}}
In P 3 ( R ) , {\displaystyle {\mbox{In }}P_{3}(\mathbb {R} ){\mbox{,}}} v 1 = 2 x 3 − 2 x 2 + 12 − 6 is a linear combination of: {\displaystyle v_{1}^{}=2x^{3}-2x^{2}+12-6{\mbox{ is a linear combination of:}}} u 1 = x 3 − 2 x 2 − 5 x − 3 and u 2 = 3 x 3 − 5 x 2 − 4 x − 9 {\displaystyle u_{1}^{}=x^{3}-2x^{2}-5x-3{\mbox{ and }}u_{2}=3x^{3}-5x^{2}-4x-9} but v 2 = 3 x 3 − 2 x 2 + 7 x + 8 is not. {\displaystyle {\mbox{but }}v_{2}^{}=3x^{3}-2x^{2}+7x+8{\mbox{ is not.}}}
Why? {\displaystyle {\mbox{Why?}}{}_{}^{}}
v 1 = 2 x 3 − 2 x 2 + 12 − 6 = a 1 u 1 + a 2 u 2 = a 1 ( x 3 − 2 x 2 − 5 x − 3 ) + a 2 ( 3 x 3 − 5 x 2 − 4 x − 9 ) {\displaystyle v_{1}^{}=2x^{3}-2x^{2}+12-6=a_{1}^{}u_{1}+a_{2}u_{2}=a_{1}(x^{3}-2x^{2}-5x-3)+a_{2}(3x^{3}-5x^{2}-4x-9)} v 1 = − 4 u 1 + 2 u 2 {\displaystyle v_{1}^{}=-4u_{1}+2u_{2}}
We say that a subset S ⊂ V generates or spans V if span S = { all linear combinations of elements in S } = V {\displaystyle {\mbox{We say that a subset }}S\subset V{\mbox{ generates or spans }}V{\mbox{ if span }}S=\lbrace {\mbox{ all linear combinations of elements in }}S\rbrace =V}
Examples {\displaystyle {\mbox{Examples}}{}_{}^{}}
V = M 2 × 2 ( R ) {\displaystyle V=M_{2\times 2}(\mathbb {R} )}
M 1 = ( 1 0 0 0 ) , M 2 = ( 0 1 0 0 ) , M 3 = ( 0 0 1 0 ) , M 4 ( 0 0 0 1 ) {\displaystyle M_{1}={\begin{pmatrix}1&0\\0&0\end{pmatrix}},M_{2}={\begin{pmatrix}0&1\\0&0\end{pmatrix}},M_{3}={\begin{pmatrix}0&0\\1&0\end{pmatrix}},M_{4}{\begin{pmatrix}0&0\\0&1\end{pmatrix}}}
N 1 = ( 0 1 1 1 ) , N 2 = ( 1 0 1 1 ) , N 3 = ( 1 1 0 1 ) , N 4 ( 1 1 1 0 ) {\displaystyle N_{1}={\begin{pmatrix}0&1\\1&1\end{pmatrix}},N_{2}={\begin{pmatrix}1&0\\1&1\end{pmatrix}},N_{3}={\begin{pmatrix}1&1\\0&1\end{pmatrix}},N_{4}{\begin{pmatrix}1&1\\1&0\end{pmatrix}}}
Claims {\displaystyle {\mbox{Claims}}{}_{}^{}}
Proof of 1 {\displaystyle {\mbox{Proof of 1}}{}_{}^{}}
Given any B = ( b 11 b 12 b 21 b 22 ) need to find a 1 , a 2 , a 3 , a 4 such that, {\displaystyle {\mbox{Given any }}B={\begin{pmatrix}b_{11}^{}&b_{12}\\b_{21}&b_{22}\end{pmatrix}}{\mbox{ need to find }}a_{1},a_{2},a_{3},a_{4}{\mbox{ such that,}}}
( b 11 b 12 b 21 b 22 ) = B = a 1 M 1 + a 2 M 2 + a 3 M 3 + a 4 M 4 = ( a 1 0 0 0 ) + ( 0 a 2 0 0 ) + ( 0 0 a 3 0 ) + ( 0 0 0 a 4 ) {\displaystyle {\begin{pmatrix}b_{11}^{}&b_{12}\\b_{21}&b_{22}\end{pmatrix}}=B=a_{1}M_{1}+a_{2}M_{2}+a_{3}M_{3}+a_{4}M_{4}={\begin{pmatrix}a_{1}&0\\0&0\end{pmatrix}}+{\begin{pmatrix}0&a_{2}\\0&0\end{pmatrix}}+{\begin{pmatrix}0&0\\a_{3}&0\end{pmatrix}}+{\begin{pmatrix}0&0\\0&a_{4}\end{pmatrix}}}
= ( a 1 a 2 a 3 a 4 ) ⇔ { b 11 = a 1 b 12 = a 2 b 21 = a 3 b 22 = a 4 {\displaystyle ={\begin{pmatrix}a_{1}^{}&a_{2}\\a_{3}&a_{4}\end{pmatrix}}\Leftrightarrow {\begin{cases}b_{11}=a_{1}\\b_{12}=a_{2}\\b_{21}=a_{3}\\b_{22}=a_{4}\end{cases}}} A system of 4 equations with 4 unknowns {\displaystyle {\mbox{A system of 4 equations with 4 unknowns}}{}_{}^{}} Proof of 2 {\displaystyle {\mbox{Proof of 2}}{}_{}^{}}
( b 11 b 12 b 21 b 22 ) = B = a 1 N 1 + a 2 N 2 + a 3 N 3 + a 4 N 4 = ( 0 a 1 a 1 a 1 ) + ( a 2 0 a 2 a 2 ) + ( a 3 a 3 0 a 3 ) + ( a 4 a 4 a 4 0 ) {\displaystyle {\begin{pmatrix}b_{11}^{}&b_{12}\\b_{21}&b_{22}\end{pmatrix}}=B=a_{1}N_{1}+a_{2}N_{2}+a_{3}N_{3}+a_{4}N_{4}={\begin{pmatrix}0&a_{1}\\a_{1}&a_{1}\end{pmatrix}}+{\begin{pmatrix}a_{2}&0\\a_{2}&a_{2}\end{pmatrix}}+{\begin{pmatrix}a_{3}&a_{3}\\0&a_{3}\end{pmatrix}}+{\begin{pmatrix}a_{4}&a_{4}\\a_{4}&0\end{pmatrix}}}
= ( a 2 + a 3 + a 4 a 1 + a 3 + a 4 a 1 + a 2 + a 4 a 1 + a 2 + a 3 ) ⇔ { b 11 = a 2 + a 3 + a 4 b 12 = a 1 + a 3 + a 4 b 21 = a 1 + a 2 + a 4 b 22 = a 1 + a 2 + a 3 {\displaystyle ={\begin{pmatrix}a_{2}^{}+a_{3}+a_{4}&a_{1}+a_{3}+a_{4}\\a_{1}+a_{2}+a_{4}&a_{1}+a_{2}+a_{3}\end{pmatrix}}\Leftrightarrow {\begin{cases}b_{11}=a_{2}+a_{3}+a_{4}\\b_{12}=a_{1}+a_{3}+a_{4}\\b_{21}=a_{1}+a_{2}+a_{4}\\b_{22}=a_{1}+a_{2}+a_{3}\end{cases}}}
Trick {\displaystyle {\mbox{Trick}}{}_{}^{}}
M 1 = 1 3 ( N 1 + N 2 + N 3 + N 4 ) − 3 N 1 {\displaystyle M_{1}={\frac {1}{3}}\left(N_{1}+N_{2}+N_{3}+N_{4}\right)-3N_{1}} M 2 = 1 3 ( N 1 + N 2 + N 3 + N 4 ) − 3 N 2 {\displaystyle M_{2}={\frac {1}{3}}\left(N_{1}+N_{2}+N_{3}+N_{4}\right)-3N_{2}} M 3 = 1 3 ( N 1 + N 2 + N 3 + N 4 ) − 3 N 3 {\displaystyle M_{3}={\frac {1}{3}}\left(N_{1}+N_{2}+N_{3}+N_{4}\right)-3N_{3}} M 4 = 1 3 ( N 1 + N 2 + N 3 + N 4 ) − 3 N 4 {\displaystyle M_{4}={\frac {1}{3}}\left(N_{1}+N_{2}+N_{3}+N_{4}\right)-3N_{4}}
B = b 11 M 1 + b 12 M 3 + b 21 M 3 + b 22 M 4 {\displaystyle B=b_{11}^{}M_{1}+b_{12}M_{3}+b_{21}M_{3}+b_{22}M_{4}} = b 11 ( 1 3 ( N 1 + N 2 + N 3 + N 4 ) − 3 N 1 ) + … {\displaystyle =b_{11}\left({\frac {1}{3}}\left(N_{1}+N_{2}+N_{3}+N_{4}\right)-3N_{1}\right)+\ldots }
= a linear combination of N 1 , N 2 , N 3 , N 4 {\displaystyle ={\mbox{ a linear combination of }}N_{1}^{},N_{2},N_{3},N_{4}}
Proof of 3 {\displaystyle {\mbox{Proof of 3}}{}_{}^{}}
Indeed in a 1 M 1 + a 2 M 2 + a 3 M 3 = ( a 1 a 2 a 3 0 ) lower right corner is always 0 {\displaystyle {\mbox{Indeed in }}a_{1}^{}M_{1}+a_{2}M_{2}+a_{3}M_{3}={\begin{pmatrix}a_{1}&a_{2}\\a_{3}&0\end{pmatrix}}{\mbox{ lower right corner is always }}0}
for example ( 240 157 e π ) not in span. {\displaystyle {\mbox{for example }}{\begin{pmatrix}240&157\\e&\pi \end{pmatrix}}{\mbox{ not in span.}}}
Proof of 4 {\displaystyle {\mbox{Proof of 4}}{}_{}^{}}
a 1 N 1 + a 2 N 2 + a 3 N 3 = ( a 2 + a 3 a 1 + a 3 a 1 + a 2 a 1 + a 2 + a 3 ) {\displaystyle a_{1}^{}N_{1}+a_{2}N_{2}+a_{3}N_{3}={\begin{pmatrix}a_{2}+a_{3}&a_{1}+a_{3}\\a_{1}+a_{2}&a_{1}+a_{2}+a_{3}\end{pmatrix}}}
( 240 157 e π ) is equal? { 240 = a 2 + a 3 157 = a 1 + a 3 e = a 1 + a 2 π = a 1 + a 2 + a 3 ⇒ No solution {\displaystyle {\begin{pmatrix}240&157\\e&\pi \end{pmatrix}}{\mbox{ is equal? }}{\begin{cases}240=a_{2}+a_{3}\\157=a_{1}+a_{3}\\e=a_{1}+a_{2}\\\pi =a_{1}+a_{2}+a_{3}\end{cases}}\Rightarrow {\mbox{No solution}}}
Motivation {\displaystyle {\mbox{Motivation}}{}_{}^{}}
S ⊂ V is linearly dependent if it is wasteful, {\displaystyle S\subset V{\mbox{ is linearly dependent if it is wasteful,}}} i.e. if ∃ v ∈ V such that ∃ a 1 … a n ∈ F and u 1 … u 2 ∈ S {\displaystyle {\mbox{i.e. if }}\exists v\in V{\mbox{ such that }}\exists a_{1}^{}\ldots a_{n}\in F{\mbox{ and }}u_{1}^{}\ldots u_{2}\in S} and ∃ b 1 … b m ∈ F and w 1 … w m ∈ S {\displaystyle {\mbox{ and }}\exists b_{1}^{}\ldots b_{m}\in F{\mbox{ and }}w_{1}\ldots w_{m}\in S}
so that ∑ i = 1 n a i u i = v = ∑ i = 1 m b i w i {\displaystyle {\mbox{so that }}\sum _{i=1}^{n}a_{i}u_{i}=v=\sum _{i=1}^{m}b_{i}w_{i}}
∑ a i u i − ∑ b i w i = 0 {\displaystyle \sum a_{i}u_{i}-\sum b_{i}w_{i}=0}
can be represented as ∑ c i z i = 0 {\displaystyle {\mbox{can be represented as }}\sum c_{i}z_{i}=0}
S ⊂ V is called linearly dependent if you can find {\displaystyle S\subset V{\mbox{ is called linearly dependent if you can find }}} z 1 … z n ∈ S different from each other and c 1 … c n ∈ F so that not all of which are 0 , {\displaystyle z_{1}^{}\ldots z_{n}\in S{\mbox{ different from each other and }}c_{1}^{}\ldots c_{n}\in F{\mbox{ so that not all of which are }}0,} so that ∑ c i z i = 0 otherwise, S is called linearly independent {\displaystyle {\mbox{so that }}\sum c_{i}z_{i}=0{\mbox{ otherwise, }}S{\mbox{ is called linearly independent}}}
Example 1 {\displaystyle {\mbox{Example 1}}{}_{}^{}}
In R , S = { ( 1 2 3 ) , ( 4 5 6 ) , ( 7 8 9 ) } is linearly dependent {\displaystyle {\mbox{In }}\mathbb {R} ,S=\lbrace {\begin{pmatrix}1&2&3\end{pmatrix}},{\begin{pmatrix}4&5&6\end{pmatrix}},{\begin{pmatrix}7&8&9\end{pmatrix}}\rbrace {\mbox{ is linearly dependent}}}
1 ⋅ ( 1 2 3 ) − 2 ⋅ ( 4 5 6 ) + 1 ⋅ ( 7 8 9 ) = 0 {\displaystyle 1\cdot {\begin{pmatrix}1&2&3\end{pmatrix}}-2\cdot {\begin{pmatrix}4&5&6\end{pmatrix}}+1\cdot {\begin{pmatrix}7&8&9\end{pmatrix}}=0}
Example 2 {\displaystyle {\mbox{Example 2}}{}_{}^{}}
R n , e i = ( 0 ⋮ 1 ⋮ 0 ) i t h row {\displaystyle \mathbb {R} ^{n},e_{i}={\begin{pmatrix}0\\\vdots \\1\\\vdots \\0\end{pmatrix}}i^{th}{\mbox{ row}}}
S = { e 1 , … , e n } {\displaystyle S=\lbrace e_{1}^{},\ldots ,e_{n}\rbrace }
Claim S is linearly independent {\displaystyle {\mbox{Claim }}S{\mbox{ is linearly independent}}{}_{}^{}}
( 0 ⋮ 0 ) = 0 = ∑ i = 1 n a i e i = ( a 1 a 2 ⋮ a n ) ⇒ a 1 = 0 a 2 = 0 ⋮ a n = 0 {\displaystyle {\begin{pmatrix}0\\\vdots \\0\end{pmatrix}}=0=\sum _{i=1}^{n}a_{i}e_{i}={\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix}}\Rightarrow {\begin{matrix}a_{1}=0\\a_{2}=0\\\vdots \\a_{n}=0\end{matrix}}}
not not all a i are 0 ⇒ not linearly dependent. {\displaystyle {\mbox{not not all }}a_{i}^{}{\mbox{ are }}0\Rightarrow {\mbox{ not linearly dependent.}}}
Claim S ⊂ V is linearly independent iff whenever ∑ a i u i = 0 and distinct u i ∈ S then ∀ i a i = 0 {\displaystyle {\mbox{Claim }}S\subset V{\mbox{ is linearly independent iff whenever }}\sum a_{i}u_{i}=0{\mbox{ and distinct }}u_{i}\in S{\mbox{ then }}\forall i\quad a_{i}=0}
Comments {\displaystyle {\mbox{Comments}}{}_{}^{}}
{ 0 } is linearly dependent. example 7 ⋅ 0 = 0 {\displaystyle \lbrace 0\rbrace {\mbox{ is linearly dependent. example }}7\cdot 0=0}
if u ≠ 0 assume a ⋅ u = 0 , and a ≠ 0 ⇒ a − 1 a u = 0 ⇒ u = 0 contradiction results, so no such a exists. {\displaystyle {\mbox{if }}u\neq 0{\mbox{ assume }}a\cdot u=0{\mbox{, and }}a\neq 0\Rightarrow a_{}^{-1}au=0\Rightarrow u=0{\mbox{ contradiction results, so no such }}a{\mbox{ exists.}}} So { u } is not linearly dependent, hence it is linearly independent. {\displaystyle {\mbox{ So}}{}_{}^{}\lbrace u\rbrace {\mbox{is not linearly dependent, hence it is linearly independent.}}}