Page 151 - A Course in Linear Algebra with Applications

5.3: Operations with Subspaces 135

Assume therefore that [ 7 ^ 0 and W ^ 0, and put
m = dim(U) and n = dim(W). Consider first the case where
U D W = 0. Let {ui,u 2 ,... u m } and {wi,w 2 , • •., w n } be
,
bases of U and W respectively. Then the vectors u i , . . . , u m
and w i , . . . , w n surely generate U + W. In fact these vectors
are also linearly independent: for if there is a linear relation
between them, say

CiUi H h c m u m + diwi H h d n w n = 0,

then

ciUi H h c mu m = -di)wi H h (-dn)w n ,
(
a vector which belongs to both U and W, and so to U HW,
which is the zero subspace. Consequently this vector must be
the zero vector. Therefore all the Cj and dj must be zero since
the Ui are linearly independent, as are the Wj. Consequently
the vectors u i , . . . , u m , w i , . . . , w n form a basis of U + W, so
that dim([7 + W) = m + n = dim(J7) + dim(W), the correct
formula since U n W = 0 in the case under consideration.
Now we tackle the more difficult case where U D W ^ 0.
First choose a basis for UT\W, say {zi,..., z r }. By 5.1.4 this
may be extended to bases of U and of W, say

{zi,..., z r , u,-_|_i,..., u m |

and
{zi,. . , z r , w r + i , . . . w n }
.
,
respectively. Now the vectors

.
,
Zi . . . Z r , U r + i , . . . U m , W r + 1 , . . , W n
,
generate U + W: for we can express any vector of U or W in
terms of them. What still needs to be proved is that they are