Page 160 - A Course in Linear Algebra with Applications

144 Chapter Five: Basis and Dimension

Lemma 5.3.4
If U is a subspace of a vector space V, then distinct cosets of
U are disjoint. Thus V is the disjoint union of all the distinct
cosets of U.

Proof
Suppose that cosets v + U and w + U both contain a vector
x: we will show that these cosets are the same. By hypothesis
there are vectors ui, U2 in U such that

X = V + U i = W + U2-

Hence v = w + u where u = 112 — Ui G U, and consequently
v + U = (w + u) + U = 'w + U, since u + U = U, as claimed.
Finally, V is the union of all the cosets of U since v € v + U.

The set of all cosets of U in V is written

V/U.

A good way to think about V/U is that its elements arise by
identifying all the elements in a coset, so that each coset has
been "compressed" to a single vector.
The next step in the construction is to turn V/U into a
vector space by defining addition and scalar multiplication on
it. There are natural definitions for these operations, namely

(v + U) + (w + U) = (v + w) + U

c(v + U) = (cv) + U

where v, w € V and c is a scalar.
Although these definitions look natural, some care must
be exercised. For a coset can be represented by any of its
vectors, so we must make certain that the definitions just given
do not depend on the choice of v and w in the cosets v + U
and w + U.