1.7 Spanningsetsilinearndependencey 1.2 in text Last time we proved that if 72 148 Ju's is a set of vectors in R then is a subspace of R t.vtts.tt turn t tz the R This subspace is se important that we give it a special name Definition The subspace 5 t.vtts.tt turn t tz the R is called the subspace spanned by 73 15 z V We write span hi Jz V Span 1 We say that B is a spanning set for 5 or that 72 spans 5
in R2 Exe f t is a non Zero Vector N then spank tt ter 7 spank is a line through the origin The set J is a spanning set for or f spans Note The sets J 2J and T are all sets for as well What if instead we have two vectors VT VT What does span Tv Exi f to and T E then F t.lt tt2le t.it R tf ti taek R look like spanning 1 shape 3 So span to E Be the whole 2 dimensional plane µ z
Ex f t and f then o x spantjz t.tottzlo ti tze R a t t R ft to ter t 3 fol So spantu.ve span to a line in R oae i He Y a wt i s ird Why did that happen Note The vector Tz in the second example was a multiple of VT so it didn't contribute anything to span We found that span VT VT span span114.1 so instead of spanning a plane they only spanned a line Let's examine this in pet 2
E f f f and Vj fe then span ri V V t.l ttzl3ojtt3fe ti ta t c R Gt 3 4 55 t o ti tr t E R tf So Span T Tz Js span Exi f E Jz VT t Vj So te R g and ff spanritz T ft t ta g t.l ti ta t e R til t t t E then t t t.int c R titta Y t tatts ti ta t e R span it is is span
E Theorem Let it V be vectors in R f Ya Can be written as a linearcombination of vi f we y t then Span T V span19,5 V f there is a vector Vie VT Jz V that can be written as a linear combination of the other vectors then the set VT VT V will be called linearly dependent f this is not possible we'll say that vi V is linearly independent z f L f is linearly dependent as is linearly dependent as Note f ii V is linearly f f 3 g L tf dependent then there is a vector Vi and scalars ti tz 4 c R such that Vi t J t t ti V t fit fit t o t tktk
By moving it to the other side we get tht t t ti Ti Fi t t it Fit t t TkV 8 Thiscoefficient to So if VT Jz V is linearly dependent then the equation hit taut t.it o has a solution with at least one ti t O This can only happen when 5,5 V is linearly dependent wtf Because if t J t tzvz t t EkV 0 and some ti is non Zero then we can move tifi over to get Liv t J t t ti V t t it fit t t tktk By dividing both sides by ti we have i i Vi fe Yt ftp.jy.i ft i vi it VT can be written as a linear combination of the other vectors in VT VT V it V is linearly dependent tfei k
This provides us with a more useful definition of linear dependence Definition A set of vectors T Tz V is ii i linearly dependent if there are scalars ti tr Ek not all 0 such that Jit tart t tkfie 8 linearly independent if the only solution to the equation Jit tart t tkfie Ois t Lz the _0 called the trivial solution E s the set linearly independent Solutioni Suppose ti ta are such that This S 41 t t.e ttsf3zj o es us a systemofequations L t t Zte t 3T tz t 2T t 1 O O O
From we get L From we get tat 2T O tz o To From we get tit2tz t 34 0 t o To To thus the only solution to the equation z 41 t t.e ttsf3zj E is t tz 0 the trivial solution o is Note This example shows that linearly independent deciding whether or not a set of vectors is linearly independent boils down to solving system of equations This is the topic of chapter 2 Exercise Show that if vijz V contains 8 then this set is linearly dependent Bases The real idea behind today's lesson the simplest spanning set for a subspace 5 is one that is linearly independent
Definition f it V is a linearlyindependent set that Ex Let T f plans a subspace 5 of R then VT VT V is called a basis for S q Vj ff and 5 span vi vi B Then T V VJ is a spanning set for 5 but it is not a basis for 5 Why so T 5,5 is not iii Because it linearly independent we don't change the span span it vi is span Since neither vector in Y f other this set is linearly independent a is a basis for S By removing is a multiple of the
Definition The standard basis for R is E Ez En Where Ei has a 1 in the ith entry Ee The standard basis for R sea is E ET where E f and E E and 0 elsewhere net s E et really a basis for RZ 1 t span s R2 as any vector L f a t's independent tilt t.e fg th g can be written as ndeed if tie tze 0 then So YES t is a basis 1 t tetro Exercise Write down and draw the standard basis Ei Ez Es in R Show that ET Ez e5 is linearlyindependent and that a span LeT Ez 5 1123 i e show it really is a basisfor 1123
Proposition f D VT VT V is a basis for a subspace 5 of R then every e S can be written as t yttnt tt inexactlyoneway.pro we know it can be done in at least one way since Span 93 5 But if t J t turn and Sir t Skin then t g t t tu sie Tru T 0 since 93 is linearly independent Thus t Si ta sa tu Sk so there was really only to write 5 Bo Lines.O anes andhyperpla Linear independence gives us a vector equation for lines planes and hyperplanes in R Definition i f p Te R with 8 18 then is a line in Rh passing through p ter
ii f f e R and T Tz ER is linearly independent then t t T t et t ta e R is a plane in R passing through F ii f FelR and 5,5 Jn ER is linearly independent then Pt 4T t tat t ten Ju i ti ta ta ER is a hyperplane in R passing through p Exe Does S Spanff 9 plane or hyperplane in R l O f describe a line same Since these vectors form a linearly dependent set So fo E spanffgf.fi fff spanffgff1f
This last set is linearly independent as the vectors are not multiples of each other Thus t.fi ttzf9o r a iti.tzer linearly independent S is a plane in 1134