Answer:
Defintion. A subset W of a vector space V is a subspace if
(1) W is non-empty
(2) For every v, ¯ w¯ ∈ W and a, b ∈ F, av¯ + bw¯ ∈ W.
Expressions like av¯ + bw¯, or more generally
X
k
i=1
aiv¯ + i
are called linear combinations. So a non-empty subset of V is a subspace if it is
closed under linear combinations. Much of today’s class will focus on properties of
subsets and subspaces detected by various conditions on linear combinations.
Theorem. If W is a subspace of V , then W is a vector space over F with operations
coming from those of V .
In particular, since all of those axioms are satisfied for V , then they are for W.
We only have to check closure!
Examples:
Defintion. Let F
n = ai ∈ F with coordinate-wise addition and scalar
multiplication.
This gives us a few examples. Let W ⊂ F
n be those points which are zero except
in the first coordinate:
W = {(a, 0, . . . , 0)} ⊂ F
n
.
Then W is a subspace, since
a · (α, 0, . . . , 0) + b · (β, 0, . . . , 0) = (aα + bβ, 0, . . . , 0) ∈ W.
If F = R, then W0 = (a1, . . . , an) is not a subspace. It’s closed under
addition, but not scalar multiplication.
We have a number of ways to build new subspaces from old.
Proposition. If Wi for i ∈ I is a collection of subspaces of V , then
W =
\
i∈I
Wi = w¯ ∈ Wi∀i ∈ I
is a subspace.
Proof. Let ¯v, w¯ ∈ W. Then for all i ∈ I, ¯v, w¯ ∈ Wi
, by definition. Since each Wi
is
a subspace, we then learn that for all a, b ∈ F,
av¯ + bw¯ ∈ Wi
,
and hence av¯ + bw¯ ∈ W. ¤
Thought question: Why is this never empty?
The union is a little trickier.
Proposition. W1 ∪ W2 is a subspace iff W1 ⊂ W2 or W2 ⊂ W1.
i hope this helped have a nice day/night :)