Page 152 - A Course in Linear Algebra with Applications

136 Chapter Five: Basis and Dimension

linearly independent. Suppose that in fact there is a linear
relation

r m n
3 j+
]PeiZ;+ ^ C U 5Z dk ™ k = 0
i=l j=r+l fc=r+l

where the e^, Cj, d k are scalars. Then
n r m
Y^ d w k = ^2(-ei)zi+ X (-CJ)UJ,
k
fc=r+l i=l j = r + l

which belongs to both U and VF and so to U f) W. The
vector ^ dfcWfc is therefore expressible as a linear combination
of the Zi since these vectors are known to form a basis of
the subspace U D W. However Zi,..., z r , w r + i , . . . , w n are
definitely linearly independent. Therefore all the dj are zero
and our linear relation becomes
r m
J2e iZi+ ^ CjUj = 0.

But z i , . . . , z r , u r + i , , u m are linearly independent, so it fol-
lows that the Cj and the e^ are also zero, which establishes
linear independence.
We conclude that the vectors z i , . . . , z r , u r + i , . . . , u m ,
w r + i , . . . , w n form a basis of U + W. A count of the basis
vectors reveals that dim(C7 + W) equals

r + (m — r) + (n — r) = m + n — r
= dim(U) + dim(W) - dim(U D W).

Example 5.3.2
Suppose that U and W are subspaces of R 10 with dimensions
6 and 8 respectively. Find the smallest possible dimension for
unw.