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