16 1. Linea r Algebr a Revie w
Proof. Exercis e 4 .
Definition 1.25. I f E an d F ar e subset s o f a vecto r spac e V , w e
define th e su m
E + F = {x + y : x e E,y e F}.
Proposition 1.26. If E and F are subspaces of a vector space V,
then E - f F = span{F , F}, and therefore E + F is a subspace of V.
Moreover, dim( F + F) d i mF-f - dimF, with equality if and only if
EnF = {0}.
Proof. Th e proo f tha t E + F = span{F , F} i s lef t a s Exercis e 5 .
To prove the second statement, le t {ei}f
=1
b e a basis for E an d le t
{/i}£Li be a basis for F. The n the union {ei , e2 , . . . , e £, / i, /2 , . . . , /
m
}
spans E + F, whic h proves that dim( F + F) £+m = di m F + d i m F .
From Lemm a 1.19 , w e kno w tha t if E H F = {0} , then th e vec -
tors {ei , e2,..., e^ , / i, /2 , . . . , /
m
} for m a basi s fo r span( F + F) , an d
therefore w e have dim(E + F) = l + m = dim(E) + dim(F) . IfEnF
contains a nonzer o elemen t x , the n x ca n b e writte n a s a linea r com -
bination o f th e element s i n eac h basis :
l k
x y ^
Ci ei
= y ^
di
ji.
i=l
i = l
But thi s equalit y implie s tha t th e unio n o f th e base s i s a linearl y
dependent set , s o £ + m i s strictl y greate r tha n dim(E ' + F). Thi s
proves th e las t par t o f th e proposition .
If E an d F ar e subspace s o f V with E n F = {0 } and E + F = V ,
then F i s sai d t o b e a subspace complement o f E i n V . Propositio n
1.26 abov e show s that , i n thi s case , di m V = d i m F + dimF , an d th e
union o f an y basi s o f E wit h an y basi s o f F i s a basi s fo r V .
Definition 1.27. I f E an d F ar e subspace s o f V suc h tha t ever y
v EV ca n b e writte n uniquel y a s a su m v = x -f- y wit h x G E,y F ,
then w e say V is a direct sum o f E an d F . Thi s wil l b e denote d
V = E + F.
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