chapter 2 vector spaces over -...

Chapter 2

Vector Spaces Over Z2

Here we describe vector spaces over the field Z2. The field Z2 h'~s 0

and 1 as elements and is a field of hara t'ristic two. In particulcl,l'

1 + 1 = 0 in Z2.

We characterize a vector space over Z2 a' an abelian group 11 su b

that 11 + v = 0 for all v E V. Charactcriz'1tion of linearly inc! p nclcnt

sets, basis etc are also given.

2.1 Basic Properties

Here we observe that vector spaces over Z2 are characterized 1y its

additive group structure

Theorem 2.1. A non-empty set V 'is a v ctor space ov r Z2 ~f and

only ~l 11 is an abelian group unci T addition such that 'V + v = 0 faT

all v E 11.

31

2.1. 'Basic Properties

Proof. Let V be a vector space over Z2. Th n for au. v E V

v + u = (1 + l)v = Otl = O.

Since V is a vector space (' 1 +) is dearl) an abelian gr up.

Conversely suppose (Vi +) is an abelian group and v + v = 0 for

all v E V. Define scalar multiplication by

a'll = afor all v E "

and

1v = v for all v E V

Thcn the conditions of a vector space arc satisfi cl as follows:

(i) n.(u + w) = a'll + aw holds sinc for ~ = a

ex (v + w) = a= ~v + n. w

and for a = 1,

a(v + w) = v + 'LV = elU + (\"U.

Thus

0'(1l + ()) = em + 0'1'

32

Cfiapter 2. Vector Spaces Over 712

holds for all a E Z2 and u, v E V.

,-

(il) Consider (a + fJ)v for a, fJ E Z2 and LEV. SUI po C (V = tJ = 1.

Then (a + ,6)v = (1 + l)v = O· u = 0 = (v + u) = (\/J +;3/ .

Now let a = 1 and fJ = O. Then

(a + (3)v = Iv = '/I = (Y/I + th .

Jaw if a = fJ = 0, (a + f3)v = 0 = o:v + (3u.

It follow that (a + fJ)v = av + {3v for all rx, (j E Z2·

(iii) ow for, a. (3 E Z2, consider a(f3v) for u E V.

If /3 = 0, then a(f3v) = a· 0 = 0 = (cxI3)v.

If;3 = 1; then a(fJv) = av = (a/3)u.

Thus,

Cr (fJv) = (a(j)v for all a 1 Ij E Z2 and '( E V.

(iv) Also 1 . v = '/' for all v E V by lefinition.

Thus V is a vector space over 2 2. D

Remark 2.2. We observe that if V is a vector spacc' over Z2. th 11

any subgroup of (V, +) is a subspa of V.

33

2.1. 'Basic Properties

Now we proceed to des ribe linear independ nce in v tor Sp'l C'

over Z2.

Theorem 2.3. Let 11 be a vector space over Z2 and

VI, V2, ... ,Vn E V. Then any non zero element which 1,S a lin a1'

combination of 'Ul, V2, ... ,Vn is of the form V-i l + 'Ui2 + ... + V'i,. wh Tf'

Proof. Any linear combination of VI, V2, ... " 'Un is glV 11 by

alVl + cx?v? + ... + a v where a· E Z2. When Cl I'_ ~ n n t 0, th

corresponding term IS zero and can be omitted. If (1', 1= O. tIl 11

The above linear combination reduces to Ui l + 'Ui2 + ' .. + V-i,. wher'

V'i] , V12 ; ... ,'Ui,,. are elements among VI, V2, ... ,Un for whi h the

coefficient is non-zero. o

Remark 2.4. Any linear combination of S = {VI, U2, ... , Vn } m y be

identified with a subset of S.

For A = {VI, ... ,v j .}, we denote [A] = VI + V2 + ... + V/,,

The following theorem characterizes linearly indep nd nt sets.

Cliapter 2. Vector Spaces Over'l2

Theorem 2.5. Let V be a vector space over Z2 and

'Ul, 'U2;···, 'Un E V. Then {ali U2,···, Un} is a linearly inrl p nd nt

set if and only ~f [A] =f 0 for any non-empty set A ~ {'U]. '1/2· ... , un}·

Proof. Suppose {'LLl; 'IL2, ... i 'U 71 } is linearly jndependent. L t

Sjnce {Ul i 1L2i ... , un} i linearly independent,

alUl + a2U2 + ... + anUn = 0 gives a.; = 0 for i = 1. 2.... , '/I.

So, for any subset A of {'Ul,'LL2 .... ,'Un},

[A] = Ui + 'Uj + ... =f 0 wh re'LL·i 'Uj, ... E A.

Conversely suppose that,

[A] =f 0 for any non-empty subset A ~ {7J,1 i ... , nil}'

Let

Then, L U'i = 0 = [A]. So by hypothesis, A = 0. That is, CYi = 0lL;EA

for every ,t:. So, {Ul, ... , 'Un} is linecI,rly independent.

35

o

2.1. 'Basic Properties

Note 2.6

If {UI;1L2, ... 'un} is linearly indqenclcnt.

distinct.

Theorem 2.7. Let V be a vector spac 0'/ T Z2. A non- mpty set

S = {VI; V2, ... i Vn } ~ V is a basis of V if and only if for any

non-zero v E V,

faT a unique subset {'(til' 'Ll 12 ··· . ,Hi,.} of S.

Proof. If S is a basis, then clearly the cOnc.htioll holds. ow a8S11111('

that S satisfies the given condition. Then span S = V. So it remains

to prove that S is linearly independent.

Let

Now 0 E V can be written as 0 = OUI +OV2 +... +Ovn- By uniqu 'ness

of the linear combination given in hypoth si. it follows that ai = 0

for every i. So S is a linearly ind pendent s t. Thus S is a basis of

v.

36

o

Chapter 2. 'lJector Spaces Over'l2

2.2 Combinatorial Properties

In this section we describ e, eral ombinatorial prop rties of

vector spac saver 2 2 . vVe provide cal ulatiol1s for th numb r of

elements in a vector space V over Z2 the number of choi es of basis)

complements etc.

Theorem 2.8. If V is a vector spac of dimension n over 2 2, then

the number of elements in V is 2/1.

Proof. Let dim V = n and let {Vj, 'U2, .... 'Un} be a basi. of V. Th n

any ·u E V can be uniquely written a'

For each a'i, there are two choices 0 or 1. Therefore, total numb l' of

choices of 'U is 2T1• Thus number of elements of V is 211

• o

Now we relate the structure of vect r spac""s ov l' 2 2 with

symmetric difference of sets.

Definition 2.9. Let 5 be any 'et. For subsets A. n of 5, 1 t

AlJ.B = (A U B) - (A n B).

37

2.2. Comvinatoria( Properties

That i A6.B is the symmetric difference of 1 and B.

Theorem 2.10 (d. [2]). For any nonernpty t til, t of all8lLb, t

of S is an ab lian group with respect to the op ratio'll ~ deft>ned by

A6.B = (A U B) - (A n B)

Theorem 2.11. Let V be a vector space 0'U T Z2 with a ba is. Th n

the abelian group CV, +) is isomorphic to (P(S),~) 'Uh ',- P(. ) 1, th

set of all subsets of s.

Proof. D fine ¢ : V ---7 P(S) by

where l/,l, 71.2, - .. ,1/'1' E S and v = '1/'1 + '/1,2 + ... + '1/"1"

By theorem 2.7, ¢ is well defined. Now we prav that for 'I . '1/ E V

¢(v + w) = ¢(v)~¢('/II).

Case 1. (jJ(v) n ¢( 'W) = 0.

Let 1 = '/11 + 1/,2 + ... + 1/.,,. and W = If/I + '/112 + ... + 1/'/. hell

v + 'W = U1 + U2 + ... + 'Ur + It I + 'lU2 + ... + w,

38

Cfiapter 2. 'Vector Spaces Over Z2

Sin q{L) n ¢( 'U.) = 0. u.; and Wj ar di tin t. a

¢(u + u.) = {LLj.'U2 ... . Ur.WI U2····· ud·

That is.

¢(u + w) = q{u) U ¢(w)

= ¢(v)6¢(w), since ¢(u) n (/J(w) = 0.

Case 2. ¢(v) n ¢('111) -1= 0.

Let ¢(v) {-u'J. U2····; Ur·'UI 'U2 , U } and

¢ (111 ) {/I l. /I 2· . . . . 11 t; 7 I; U2 'loS}

so that ¢(v) n ¢(w) = {Vl,'U2 '" 'us}, \V

v = 'U I + U2 + ... + 'Ur + VI + U2 + ... + l's anel

W = WI + 'W2 + ... + Wt + VI + U2 + ... + U8 so th':\,t

V + 'W = 'UI + 'U2 + ... + U r + WI + 'W2 + ... + UII

= (¢(V) U ¢(W)) - (¢(v) n ¢(UI))

= ¢(u)6¢(w).

39

2.2. C01l16inatoria{ Properties

Thus ¢ is a homomorphism.

To prove that ¢ is one to nc.

Let v = 'U1 + ... + 'U,?" W = WI + '[112 + ... + 'WI.'

Let ¢(t) = ¢(w)

where ¢(v) = {Ul;. 0 0' Ho,} and

¢(W) = {WI; . 0 0 , wdThus '/1,1 + ... +'1/., = WI + ... +'111' Hence v = 'W. Thus ¢ is onc-ro- 11('.

To prove ¢ is onto.

Let {U1,' 0.; uo,} E P(S). Then,

'u = V,l + ... + v,jO E V.

Hence corresponding to evcry 'ct {U1' U2, 0 •• , 'ill'} ther' cxi~t.',

v = 'u,] + ... + u, E V, so that ¢(v) = {'Ul,'U2 .... ,U,.}. H n' qJ

is onto. Thus ¢ is an i 'amorphism. o

Remark 2.12. Let V be a v etor space over Z2. Th n any I-dim nsiol1(\,l

subspac of V is {O, v} with u i= 0 and any 2-clim-n 'ional ~llhspa c

of V is {O,v,'w,v + w} with '/;lU E V - {O} an 1v i= 'W.

The following theorem will be us ful in omputations 0]] vector

-lO

Chapter 2. Vector Spaces Over Z2

spaces over £:2.

Theorem 2.13. Let V1/1. H 2 be proper subspaces of a vector pa e V

Proof. We need only compute the ca 'e where W1 and \IV2 are of

dimension (71, - 1). Now

number of elements in \lV1 = 2n-1 = number of el ments in \IV2 .

So, W1 U \IV2 contain atmost 2 x 2'"-1 - 1 = 2/1. - 1 clement', sin '

o

In the next theorem, we count the number of subspaccs of sp cificcl

types in V.

Theorem 2.14. Let II be of dimen ion n over Z2. Then)

1. The number of choices of basis of V is

2T1-

1 X 2"-2(22 - 1) x '" X 2'"-I'(2J.- - 1) x ... X (2"- 1)

n!

2. The number of 1-dimensional sub pa of V is 2n - 1.

(211- 1) (2n- 1 - J)

3. The number of 2-dimensional ubspaces of V is 3 .

2.2. Comvinatoria( Properties

4. The number of k-dimensional subspaces of V i

(2'1t - 1)(211-

1 - 1) (2"-'" 1 - 1)(2 k - 1)(2k - 1 - 1) (22 - 1)(2 - 1)

5. Th number of n - k dimensional subspa es of V i, same a th

number of k-dimensional subspaces of v.

Proof. Since V is of dimension n, the numb r of lements of V is

1. A basis of 1/ is any set of n linearly ind pend nt ve tors. Now

'n linearly independent vectors of V an 1c ·llOs('n as follow .

First choose any nonzero vector in V, call it VI. Th('n choose any

vector V2 =F VI· Then choose a vector V3 ~ span{VLl U2} and so

on. In general Vk is chosen such that VI,; ~ span{vL l 'U21'" l'Uk-d·

So the number of choices of a basis in that order is

Writing in reverse order th number is

12

Cliapter 2. Vector Spaces Over Z2

ow the arne basi can occur in n! differ llt permutation m

the choice d scribed above. So the number of b<-1 Cf) for \ is

217-

J X 217-

2(22 - 1) x ... x 217 -"(2" - 1) x ... X (211- 1)

n!

(2.1)

2. Anyone dimensional subspacc is d tcrmillcl by H, 11011Z r

element of V. Sinc there are 211- 1 nonz 1'0 clcments, th r

are 211 - lone-dimensional 'ubspac .

3. Any 2-dimensional subspace is of the form {O, Ii. I , Ii +l'} \Vh r

'U #- lJ and u and v are nonzero. Thus any 2-di11l(ltlSiollal sub-

space is determined by a pair of nonzero vectors. II. I' E V. These

may be chosen in (2'11 - 1)C2 ways. So the numbcr of hoic of

(211- 1)(211

- 2)the basis is (211, - 1)C2 = . = (2'11 - 1)(211 -[ - 1).

2

For the subspace {O"u, v, U + v} i there are thr cliff rent 1 es

given by {u v}, {u,u+v} {v U+I}. Thusforca h.'ubspa of

dimension two appear with three cliff r nt ba '0 '.

(211, - 1)(2n - 1 - 1)of suI 'paces is ------­

3

. th numl I'

4, Any /';-clim nsional subspa c of VI' g 'nerat cI I y a s t f k

·13

2.2. Com6inatoria{ Properties

linearly independent vector. Now th numb r of 'hoi 'CH for

a set of h: linearly independent ,vectors in ,. as in th' else ( f

proving (2.1) is

(211 - 1) x (211. - 2) x (211. - 22) X ... X (211 - 2"'-1)

k!

2k- 1(271 -k'+1 - 1) x 2k- 2(2n-k+2 - 1) x ... X 2/';-"'(211 - l)

k\

(2.2)

Now by (2.1) above the number of differcnt ba Cf) whi h ('(),n

generate the same k-dimensional ubspa e is

2/,;-1 X 2k- 2 (22 -1) x ... x (2'" - l)

k!

Therefore the number of k-dimensional u1)sp''l.·c· of 'T is

(2.:~)

2k-1(2n-k+1 - 1) x 2k-2(2n-k+2 - 1) x '" X 2"'-"'(211 - l)2/;;-1 X 2k:-2(22 - 1) x ... X (21.. - 1)

(2'1L - 1)(2'11-1 - 1) (2n - H1 - 1)(2k - 1)(2k - 1 - 1) (22 - 1)(2 - 1)

(2.4)

5. Substituting n - k in place of kin formulFl. (2. ) and 'arrying ont

the simplifications we get that the numb r of '11 - k dimcnsional

subspaces of V is the same as thc nnmbcr of k clilllCllSiolla.t

subspaces of V. o

Chapter 2. Vector Spaces Over'l2

2.3 Linear Transformations and Common Com-

plements

We show that linear tran formations of v ctar spa s i to' I 0\ '1'

Z2 are homomorphism of the additive groups (V +) to (I {11 +).

Theorem 2.15. Let i!. H/ b v ctor spaces ov r Z2· Th n, a 'II /,(/,7)

T : V -7 vV i a linear transf01'mation of V to VI 'if

VI, V2 E V. It is sufficient to prove that

Since T is a homomorphi m of additive groups (i!. +) t.o (H . +), T(O)=O.

Let (Y = O. Then (Y'/) = 0 and so.

T(O'L') = T(O) = 0 = c~T(u).

I ow let ct = 1. Then (tv = L anel 0

T(o.u) = T(l') = 1 . T(L) = Cl'T(I).

15

2.3. Linear 'Transformations mu{ Common Comp[emwts

Then T is a linear transformation. o

Definition 2.16. A linear map.f V ~ Z2

functional on V.

all d a lin ar

Proposition 2.17 . L t f be a linear functional on a vector p :tce V

of dimension '/1 over 2.2 . Then the null spac of f is of lim lloioll

n-1.

Proof. Since f is non-zero, range of f IS full of Z2. Now 1y

theorem 1.24, dim N(f) = n - 1. 0

Now we show that linear functional on V are determined by th ir

null spaces.

Theorem 2.18. Two linear functionals on V over Z2 will b qual

if and only if they have the same null space.

Proof. If two linear functionals are equal, they have the sam llull

space.

Conversely suppose that f and .9 arc lin ar functionals 811 h that

N(f) = N(g).

Cfiapter 2. Vector Spaces Over'l,2

Suppose that, N(f) = (g) = V. Then f = 9 = O. ow upp .

that N'(f) i- V. Therefore range of f i' Z2· Then, by propo iti n

above, dimension of [N(f)]' = 1 for any campI 'lllcnt [N(f)]' of (j').

Let v E V. If v E N(f) = N(g). then f(l) = 0 = 09(-1). If v ~ '(f),

then V = (v) EB N(f), where (v) is the linear sllbspac generat cl by

v. Also f(v) i- a and since N(f) = N(g). we have g(u) i- O. Th 11

f('u) = 1 and g(v) = 1. Thus f = g. 0

Theorem 2.19. Let V be an n-dimensional vector space ov r Zz.

Then every (n - 1) dimensional subspac lT' of \' will r!r termine' a.

unique non zero linear functional for which TV is the null. pac.

Proof. Let W be an ('17, - 1) dimensional fmbsI a e of V. L .t,

'1.1, E V - W. Th n V = (lL) EB Tl' '1nd so any v E V an b

written as v = w + au for ome w E TV an 1 a E Zz· D fine, f

by, f(v) = ~. In particular f(w) = afor all w E VV. Thu f be ames

a lin ar functional on V. Clearly (f) = 11'. ow I t f 9 b lin ar

fl1nctionals such that N (f) = N (g). Then by th l' m 2.1 f = g.

TIl refore f is unique. 0

47

2.3. Linear 'Transformations ana (0111111on (0111p{elllell ts

Theorem 2.20. Let V be an n-d'imensional v ct07' po, o'ver Z2.

Then the number oj linear junctional on F is 211- 1.

Proof· Let {Ul J 'U2, . .. un} be a basi of\ . Any linear functional is

detcrmined by a non zero map of {H1. 'lL2 .... 'U'I} to {O. l}. H n

the number of non zero linear functiolla,ls on V = 2/1 - 1 0

If nT] and H!2 are subspaces of V thcn we can COIl id r th

po~sjbility of having a common complement for 11'1 anel IF2.

VV give sufficient condition for the exist ncC' of com111011 c mpl ­

111l2nt::; and also determine the number of 'uch comlllon cO\llph lllC'nts.

Definition 2.21. If V = H EB lr'. then dim II" it) calleel th

co-dimension of vV, Note that, Codim lt1! = dim V - dim IV.

Theorenl 2.22. Let 111/1 and lJl'2 be s1Lbspaces ofV. Then theTe :rist.

a subspace vVo oj W such that V = H'1 EB llVo= 11 '2 EB 1Vo if and only

if dim 11'1 = dim Hi2 ·

Proof. We prove the result bj induction on 1'. wh re

r = Codim II 1 = dim \ - dim \\ L·

·1

Cnapter 2. 'Vector Spaces Over Z2

L t dimV = n. If,. = 0 then V = 1V] = lr2 and o.lro ={O}.

TIm the result hold'. Similarly if T = n. then dim 11] = dim Ir 2 = 0

o that V is a common complement of Ir1 and 1V2 so result holels.

Jow let T = 1 < n. Th n V =j=. W] U 1';V2 , by th orcm 2.13. 1100se

:c E V - (1V] U IV2). Th n if Hia = (x), the I-dimensional suhspar

generat d by x, then 1iVan 1iV] = 0 and so 1Va + IVI = vVo EB 1VI'

Also Hia + 11-11 is of dimension n and so IVa EB IF1 = V. Silllilctrly.

vVa EB 1V2 = V. Thus, vVa is a common compl 111 nt of IV1 '·mel II·2·

Now suppose that for T < m the result is tnH'.

Let T = m + 1 < n. Again by them III 2.13. V =I- Ir j U 1l"2·

A· befor , choose .c E V - (Hi] U H2)' Let. 11 { = II'] E9 (.r). ow.

Codim lV{ = Codim 11V] - 1 = 171. Al 0 I t 1V~ = lV2 E9 (.1'). Then

Codim {;Yi~ = m. Then by induction there exists lI'o such that

V = W{ E9 W~ = Hl~ E9 11r~

ow; V = n'1 EB (.1:) EB nr6= n'2 E9 (x) E9 lro. Set. 11r

o = II (~E9 (.1').

Then VI = 1V1 EB lVo = 1V2 EB 1Va. This prov s the theorem. 0

ow we show that if dim IV2 > dim Ill. thell w -an choo..e ct

49

2.3. Linear Transformations ana Common Comp(ements

complement of VV2 which can be extended to a complement of rvI .

Corollary 2.23. Let H/1 and lV2 be subspac s of an 'I/. (limen'ional

vector space V with dim lV, = r < '" = dim vV2 < n. Th n there

exist subspaces U and lVo such that V = vVl E9 U E9 lift'o = lV2 E9 no·

Proof. Choos a subspac U of dimension s - r such that

Then, dim(H/l E9 U) = s = dim W2 . Now, by theorem 2.22, thc)'

exists a subspac lVo such that, lV = [vVl E9 U] E9 rI'o = rf"2 E9 n'o· D

Now we consid r the number of common complcments of givcn

subspaces lVl and vV2 .

Theorem 2.24. Let vVl and VV2 be subspaces of a vector spac V

and let V be of finite dimension n over Z2 with vVI f=. lV2· If

dim 11 1 = dim W2 = n - 1

then the number of common complements for vVl and n'2 is 2n - 2 .

Proof. First note that since ~Vl f=. VV2 th re exists v E ~VJ with

v ~ W2· So dim(H 1 + rV2) > climl/1l2 = n - 1. Therefor

50

Chapter 2. Vector Spaces Over Z2

Let W be a common omplement for 11'1 an I 1/12. That iH.

v = W EB WI = ltV EB 1/112 . Since dim HI] = n - 1 = dim H 2 and

dim V = n, dim IIV = 1. Then,

= (n - 1) + (17, - 1) - n

= 'n - 2.

Hence, I1/V1 n 11'21 = 2n- 2 . Since dim 11 1 = dim lr2 = n - 1, allY

common campI ment of 1/V1 ancllV2 is a one-dim nsional subHpa C' (.1')

such that x tf- 1/V1 U 1'1/2 , Therefore number of common campI 111 ntH

of 1IV1 and 1112 is IV - (1/V1 U 1IV2)I· Tow

= 2/1 _ 2/1-2 .

.51

2.3. Linear 'Transformations a1la Common Comp(emwts

So,

= 211-

2 . D

Theorem 2.25. Let vVl and IV2 be d'l tinct one dimen ional

subspaces of a vector space V of dimension n ov r Z2. Then th

number of common complements for IIVl and V\/2 is 271.-2

Proof. Given dim 1111 = dim 1;1/2 = 1. Suppose n r is a sub pace f

V such that V = W E9 IIVl = IIV E9 n 2. SUPPOS(\ IVl = (11) 'mel

VV2 = ('V2). Then since V{/1 i= H12 , {vJ. ud is a linearly independent

set. Let VI, V2, ... ,V'II be a basis of V, Then, the natural map

f : V ~ V /111 rv f{ determines a linear fun tional f such that

Conversely if f is a linear fun tional on V with f (V2) i= 0 and

ker f = 1;1/, then 111 is a ammon complem nt of ~Vl = (Vl) and

1112 = (V2)' Thus, IIV E9 1;1 1 = IV EB H 2 = V if an 1only if there xists a

linear functional f with 1(Vl) i= o. 1(-L2) i= 0 and ker f = IV. In fact

52

Cliapter 2. Vector Spaces Over 7L2

if f is any linear functional on V with f(vl) #- 0 and /('02) #- 0

then ker f is a common complement of Hil = (U1) an 1 1\'2 = ('02)'

Since {VI, '02, ... , 'On} is a basis of V, any linear functional n abov

is completely determined by its values on 'O:3.V.J, ... ;'0/1' The vaIn "

of f(V3)' f(V4), ... , f(vn ) can be determined arbitrarily. Th refor

the number of such functionals is 2n- 2. Henc the theorem. D

Let VI, '02 E V, VI #- '02, ;Uj #- 0, V2 #- 0 wher Vi' an

n-dimensional vector space. Then a common campiem nt of (VI)

and ('02) is called a common campI ment of VI and 1'2.

Theorem 2.26. Let VI, '02 E V, VI #- V2, VI #- O. V2 #- O. Then

VI + '02 lies in any common complement of ('01) and ('1'2).

Proof. Let Z be a common complement of (VI) and (/2)' Sin e,

VI + V2 E V and V = Z EB (VI) = Z EB (V2), we hav V2 E Z EB (VI)'

Since 'U2 tf. Z we have '02 = Z + UI for 0111e z E Z. Th 1'efo1' ,

VI + V2 = '111 + Z + VI = z. Hence, lJ[ + /2 E Z. o

TheorelTI 2.27. Let VI #- '02 be nonzero in V. Let Z I· Z2· . .. be

53

2.3. Linear Transfonnations ana Common COlllp{e11lellts

common complements of VI and V2 in V. Th n nZi = {O. VI + (2}.

Proof. Let'Ll E 11 be nch that 1U # 0 and 'IL ~ {( 1· U2. (11 +(2}. Then

{'/II. '1J2, (j1} i . linearly ind pendent. Let .f be a linear fun tional n i/

su h that f (VI) = f (V2) = f (w) = 1. Th-n clearly 'LV ~ ker f <tnd

ker f is a common complement of VI and 'lJ2· Th n 7J.I ~ nZi. Also by

theorem 2.26, V] +V2 E n Zi' It follows that n Zi = {O, Uj +U2}. 0

Theorem 2.28. Let VI,V2,V3 be nonzero vectors in V. 1r {lj.12. /I:~}

is linearly independent, then there exists a common complpllwnt fO'/'

Proof. Since {VI. 'U2. V3} is linearly ind pend nt, there is a hcsis f

V in the form {UI, 'U2· 'U3, 1U4, WS, ... , wn}. Let./ h a linear functional

i = 4. 5 "'; n. Then ker f is a common compl ment of '( 1. U2 ''lncl

~. 0

Whenev r we talk about the common ompl 111 nt of two r mOl'

vectors, they will b assumed to be lin arly inclep 'ndent.

54

Chapter 2. Vector Spaces Over712

Theorem 2.29. Every comnwn complement of VI, L2. U3 'I, al 0 a

complement of VI + V2 + t:~.

Proof. Let TiV be a common 'omplement of (VI) (U2) and (U:3)' in"

every n - 1 dimensional subspace 1-V will determin a llni III lin nr

functional f : V ---+ !( with W = ker f there is a linear fllll tionnl I

such that filii = 0 and f(z) = 1 for every z ~ W. Jaw v]. (_.lJ:~ ~ Il'.

So f(VI) = f(V2) = f(V3) = 1. Therefore

f(VI + U2 + V3) = ill']) + f(V2) + f(v3) = 1+ 1 + 1

=1

o

Next we show that a common complement containing a given v etor

is possible.

Theorem 2.30. Let H be a common complement 01"1']. U2· '( :3. Tlwn

for any 'V ~ (VI) V2) V3)) th re exists a common compl m nt ICo of

VI, V2 V3 such that v E IVa.

55

2.3. Linear Transformations and COlllmon Comp(emwts

Proof. Since {VI, V2, V3} is linearly independent and v ~ span{ VI· 'L 2, t'3}

the set {V, VI, V2, V3} is linearly independent. Ext nd thi to ala. is

of V.

Let {'U, VI, V2, V3, 'LU5, ... , wn } be this basis. ow {v VI +U2: VI +U:l}

is linearly indep ndent. It follmvs that {v. VI +V2, VI +V3, 'W5" ... LUll}

is also linearly independent.

Let 1!Vo = span{lJ,Vl + 'U2.'Ul + V3.'LU5 .... ,'tUn }. This is all '11-1

dimensional subspace of V. Now

=\1.

Similarly,

Also V E VVo. Hence the theorem.

.56

o