Page 164 - A Course in Linear Algebra with Applications

148 Chapter Five: Basis and Dimension

Proof
If U = 0, then dim([7) = 0 and V/0 = {{v} | v G V}, which
clearly has the same dimension as V. Thus the formula is
valid in this case.
Now let U ^ 0 and choose a basis {ui,..., u m } of U. By
5.1.4 we may extend this to a basis

{ui,..., u m , u m + i , . . . u n }
,

of V. Here of course m — dim([/) and n = dim(V). A typical
n
element v of V has the form v = ^ Cjiij, where the c^ are

scalars. Next
n n
v + C/=( Yl CiUi)+U= ^ Ci(ui + U),
i=m+l i=m+l

m
c u
.
since Yl i i ^ U. Hence u m + i + U, .., u n + U generate
i=l
the quotient space V/U.
n
On the other hand, if Yl Ciiyn + U) — 0 v/u = U, then
i=m+l
n
Y CjUj e U, so that this vector is a linear combination of
i=m+l
Ui,..., u m . Since the Uj are linearly independent, it follows
that c m + i — • • • = c n = 0. Therefore u m + i + U, ..., u n + U
form a basis of V/U and hence
dim(V/J7) = n-m = dim(V) - dim(C7).

Exercises 5.3

1. Find three distinct subspaces U, V, W of R 2 such that
2
n = u®v = v®w = weu.