Page 161 - A Course in Linear Algebra with Applications

5.3: Operations with Subspaces 145

To verify this, suppose we had chosen different represen-
tatives, say v' for v + 17 and w' for w + U. Then v' = v + Ui
and w' = w + u 2 where ui, u 2 £ U. Therefore

v' + w' = (v + w) + (ui + u 2 ) e (v + w) + U,

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so that (v + w') + U = (v + w) + U. Also cv' = cv + cu ± <E
(cu) + U and hence cv' + U = cv + U. These arguments show
that our definitions are free from dependency on the choice of
coset representatives.

Theorem 5.3.5
If U is a subspace of a vector space V over a field F, then V/U
is a vector space over F where sum and scalar multiplication
are defined above: also the zero vector is 0 + U = U and the
negative of v + U is (—v) + U.
Proof
We have to check that the vector space axioms hold for V/U,
which is an entirely routine task. As an example, let us verify
one of the distributive laws. Let v, w G V and let c E F.
Then by definition

c((v + U) + (w + U)) = c((v + w) + U) = c(v + w) + U
= (cv + cw) + U,

which by definition equals (cv+U) + (cw+U). This establishes
the distributive law. Verification of the other axioms is left to
the reader as an exercise. It also is easy to check that 0 is the
zero vector and (—v) + U the negative ofv + U.
Example 5.3.8

Suppose we take U to be the zero subspace of the vector space
V: then V/0 consists of all v + 0 = {v}, i.e., the one-element
subsets of V. While V/0 is not the same vector space as V,
the two spaces are clearly very much alike: this can be made
precise by saying that they are isomorphic (see 6.3).