Bases for Vector Spaces

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1 Bses for Vector Spces A set is independent if, roughly speking, there is no redundncy in the set: You cn t uild ny vector in the set s liner comintion of the others A set spns if you cn uild everything in the vector spce s liner comintions of vectors in the set Putting these two ides together, sis is n independent spnning set: A set with no redundncy out of which you cn uild everything Definition Let V e n F-vector spce A suset B of V is sis if it is linerly independent nd spns V The numer of elements in sis for V is clled the dimension of V, nd is denoted dimv Someone could oject t this point tht I don t know tht two ses for V might not hve different numers of elements in which cse the dimension of V wouldn t mke sense I ll show lter tht this cn t hppen Exmple Consider the stndrd sis for R 3 :,, As the nme implies, this set is sis I oserved erlier tht this set is independent Moreover, c + +c Therefore, every vector in R 3 cn e written s liner comintion of the vectors, so the set lso spns Hence Exmple The following set is sis for R 3 over R:,, Mke mtrix with the vectors s columns nd row reduce: A = Since A row reduces to the identity, it is invertile,nd there re numer of conditions which re equivlent to A eing invertile First, since A is invertile the following system hs unique solution for every (,,c): x y z c

2 But this mtrix eqution cn e written s x +y +z In other words, ny vector (,,c) R 3 cn e written s liner comintion of the given vectors: The given vectors spn R 3 Second, since A is invertile, the following system hs only x =, y =, z = s solution: x y z c This mtrix eqution cn e written s x +y +z Since the only solution is x =, y =, z =, the vectors re independent Hence, the given set of vectors is sis for R 3 The following result is cler from the lst exmple Proposition Let F e field {v,v 2,,v n } is sis for F n if nd only if A = v v 2 v n is invertile Note tht y erlier results on invertiility, this is equivlent to the following conditions (mong others): A row reduces to the identity 2 deta Thus, if you hve n vectors in F n, this gives you severl wys of checking whether or not the set is sis A sis for V is spnning set for V, so every vector in V cn e written s liner comintion of sis elements The next result sys tht such liner comintion is unique Lemm Let {v,v 2,,v n } e sis for vector spce V Every v V cn e written in exctly one wy s v = v + 2 v n v n, i F Proof Let v V Since {v,v 2,,v n } spns V, there re sclrs, 2,, n such tht Suppose tht there is nother wy to do this: v = v + 2 v n v n v = v + 2 v n v n Then v + 2 v n v n = v + 2 v n v n 2

3 Hence, Since {v,v 2,,v n } is independent, Therefore, ( )v +( 2 2 )v 2 + +( n n )v n = =, 2 2 =,, n n = =, 2 = 2,, n = n Tht is, the two liner comintions were ctully the sme This proves tht there s only one wy to write v s liner comintion of the v i s I wnt to show tht two ses for vector spce must hve the sme numer of elements I need some preliminry results, which re importnt in their own right Lemm If A is n m n mtrix with m < n, the system Ax = hs nontrivil solutions Proof Write A = 2 n n m m2 mn The condition m < n mens tht the following system hs more vriles thn equtions: 2 n n m m2 mn x x 2 x n If A row reduces to row reduced echelon mtrix R, then R cn hve t most m leding coefficients Therefore, some of the vriles x, x 2,, x n will e free vriles (prmeters); if I ssign nonzero vlues to the free vriles (eg y setting ll of them equl to ), the resulting solution will e nontrivil Theorem Let {v,v 2,,v n } e sis for vector spce V () Any suset of V with more thn n elements is dependent () Any suset of V with fewer thn n elements cnnot spn Proof () Suppose {w,w 2,,w m } is suset of V, nd tht m > n I wnt to show tht {w,w 2,,w m } is dependent Write ech w s liner comintion of the v s: w = v + 2 v n v n w 2 = 2 v + 22 v n v n w m = m v + m2 v mn v n This cn e represented s the following mtrix eqution: w w 2 w m 2 m v v 2 v n 2 22 m2 n 2n mn 3

4 Since m > n, the mtrix of s hs more columns thn rows Therefore, the following system hs nontrivil solution x =, x 2 = 2, x m = m : 2 m x 2 22 m2 x 2 n 2n mn x m Tht is, not ll the s re, ut 2 m 2 22 m2 2 n 2n mn m But then 2 m w w 2 w m 2 v v 2 v n 2 22 m2 2 n 2n mn m Therefore, In eqution form, m w w 2 w m 2 m w + 2 w m w m = This is nontrivil liner comintion of the w s which dds up to, so the w s re dependent () Suppose tht {w,w 2,,w m } is set of vectors in V nd m < n I wnt to show tht {w,w 2,,w m } does not spn V Suppose on the contrry tht the w s spn V Then ech v cn e written s liner comintion of the w s: v = w + 2 w m w m v 2 = 2 w + 22 w m w m v n = n w + n2 w nm w m In mtrix form, this is 2 n v v 2 v n w w 2 w m 2 22 n2 m 2m nm Since n > m, the coefficient mtrix hs more columns thn rows Hence, the following system hs nontrivil solution x =, x 2 = 2, x n = n : 2 n x 2 22 n2 x 2 m 2m nm x n 4

5 Thus, 2 n 2 22 n2 2 m 2m nm n Multiplying the v nd w eqution on the right y the -vector gives v v 2 v n 2 2 n w w 2 w m 2 22 n2 2 m 2m nm n n Hence, In eqution form, this is v v 2 v n 2 n v + 2 v n v n = Since not ll the s re, this is nontrivil liner comintion of the v s which dds up to contrdicting the independence of the v s This contrdiction mens tht the w s cn t spn fter ll Exmple The stndrd sis for R 3 contins 3 vectors Hence, the following set of four vectors cn t e independent:, 2 3,, 7 Likewise, the following set of two vectors cn t spn R 3 : 2, Corollry If {v,,v n } is sis for vector spce V, then every sis for V hs n elements Proof If {w,,w m } is nother sis for V, then m cn t e less thn n or {w,,w m } couldn t spn Likewise, m cn t e greter thn n or {w,,w m } couldn t e independent Therefore, m = n The Corollry shows tht the dimension of finite-dimensionl vector spce is well-defined tht is, in finite-dimensionl vector spce, ny two ses hve the sme numer of elements This is true in generl; I ll stte the relevnt results without proof () Every vector spce hs sis The proof requires set-theoretic result clled Zorn s Lemm () Two ses for ny vector spce hve the sme numer of elements Specificlly, if B nd C re ses for vector spce V, there is ijective function f : B C 5

6 I ve lredy given one exmple of n infinite sis: {,x,x 2,x 3,} This set is sis for the vector spce R[x] of polynomils with rel coefficients over the field of rel numers The next result shows tht, in principle, you cn construct sis y: () Strting with n independent set nd dding vectors () Strting with spnning set nd removing vectors Theorem Let V e vector spce () Any set of independent vectors is suset of sis for V () Any spnning set for V contins suset which is sis Prt () mens tht if S is n independent set, then there is sis T such tht S T (If S ws sis to egin with, then S = T) Prt () mens tht if S is spnning set, then there is sis T such tht T S Proof I ll only give the proof in the cse where V hs finite dimension n, though it is true for ny vector spce () Let {v,,v m } e independent If this set spns V, it s sis, nd I m done Otherwise, there is vector v V which is not in the spn of {v,,v m } I clim tht {v,v,,v m } is independent Suppose Suppose Then I cn write v + v + + m v m = v = ( v + + m v m ) Since v hs een expressed s liner comintion of the v k s, it s in the spn of the v k s, contrry to ssumption Therefore, this cse is ruled out The only other possiility is = Then v + + m v m =, so independence of the v k s implies = = m = Therefore, {v,v,,v m } is independent I cn continue dding vectors in this wy until I get set which is independent nd spns sis The process must terminte, since no independent set in V cn hve more thn n elements () Suppose {v,,v m } spns V I wnt to show tht some suset of {v,,v m } is sis If {v,,v m } is independent, it s sis, nd I m done Otherwise, there is nontrivil liner comintion v + + m v m = Assume without loss of generlity tht Then v = ( 2 v m v m ) Since v is liner comintion of the other v s, I cn remove it nd still hve set which spns V: V = (v 2,,v m ) I continue throwing out vectors in this wy until I rech set which spns nd is independent sis The process must terminte, ecuse no set contining fewer thn n vectors cn spn V 6

7 It s possile to crry out the dding vectors nd removing vectors procedures in some specific cses The lgorithms re relted to those for finding ses for the row spce nd column spce of mtrix, which I ll discuss lter Suppose you know sis should hve n elements, nd you hve set S with n elements ( the right numer ) To show S is sis, you only need to check either tht it is independent or tht it spns not oth I ll justify this sttement, then show y exmple how you cn use it I need preliminry result Proposition Let V e finite dimensionl vector spce over field F, nd let W e suspce of V If dimw = dimv, then V = W Proof Suppose dimw = dimv = n, ut V W I ll show tht this leds to contrdiction Let {x,x 2,,x n } e sis for W Suppose this is not sis for V Since it s n independent set, the previous result shows tht I cn dd vectors y, y 2, y m to mke sis for V: {x,x 2,,x n,y,y 2,y m } But this is sis for V with more thn n elements, which is impossile Therefore, {x,x 2,,x n } is lso sis for V Let x V Since {x,x 2,,x n } spns V, I cn write x s liner comintion of the elements of {x,x 2,,x n }: x = x + 2 x n x n, i F But since x, x 2, x n re in W nd W is suspce, ny liner comintion of x, x 2, x n must e in W Thus, x W Since x ws n ritrry element of V, I get V W, so W = V You might e thinking tht this result is ovious: W nd V hve the sme dimension, nd you cn t hve one thing inside nother, with oth things hving the sme size, unless the things re equl This is intuitively wht is going on here, ut this kind of intuitive resoning doesn t lwys work For exmple, the even integers re suset of the integers nd, s infinite sets, oth re of the sme order of infinity (crdinlity) But the even integers ren t ll of the integers: There re odd integers s well Corollry Let S e set of n vectors in n n-dimensionl vector spce V () If S is independent, then S is sis for V () If S spns V, then S is sis for V Proof () Suppose S is independent Consider W, the spn of S Then S is independent nd spns W, so S is sis for W Since S hs n elements, dimw = n But W V nd dimv = n By the preceding result, V = W Hence, S spns V, nd S is sis for V () Suppose S spns V Suppose S is not independent By n erlier result, I cn remove some elements of S to get set T which is sis for V But now I hve sis T for V with fewer thn n elements (since I removed elements from S, which hd n elements) This is contrdiction, nd hence S must e independent Exmple Show tht the following set is sis for R 3 :,, 2 7

8 Since I hve 3 vectors in R 3 (which hs dimension 3), I only need to show tht the set is independent So suppose x +y +z 2 This gives the mtrix eqution 2 x y z Solve the system y row-reduction: 2 The solution is x =, y =, nd z = Hence, the set is independent, nd the preceding result shows tht it s sis Note: You could lterntively show tht the set spns To do this, you d show tht for n ritrry (,,c) R 3, you cn solve the following system for x, y, nd z in terms of,, nd c: x +y +z 2 c You cn see it s lmost the sme thing, ut this will e little messier thn checking independence c 25 y Bruce Ikeng 8

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