# Linear Algebra: #3 Subspaces

**V**be a vector space over a ﬁeld F and let

**W**⊂

**V**be some subset. If

**W**is itself a vector space over F, considered using the addition and scalar multiplication in

**V**, then we say that

**W**is a

*subspace*of

**V**. Analogously, a subset H of a group G, which is itself a group using the multiplication operation from G, is called a

*subgroup*of G. Subﬁelds are similarly deﬁned.

**Theorem 2**Let

**W**⊂

**V**be a subset of a vector space over the ﬁeld F. Then

**W**is a subspace of

**V**⇔ a ·

**v**+ b ·

**w**∈

**W**,

for all

**v**,

**w**∈

**W**and a, b ∈ F.

*Proof.*

The direction ‘⇒’ is trivial.

For ‘⇐’, begin by observing that 1 ·

**v**+ 1 ·

**w**=

**v**+

**w**∈

**W**, and a ·

**v**+ 0 ·

**w**= a ·

**v**∈

**W**, for all

**v**,

**w**∈

**W**and a ∈ F. Thus

**W**is closed under vector addition and scalar multiplication.

Is

**W**a group with respect to vector addition? We have 0 ·

**v**=

**0**∈

**W**, for

**v**∈

**W**; therefore the neutral element

**0**is contained in

**W**. For an arbitrary

**v**∈

**W**we have

Therefore (−1) ·

**v**is the inverse element to

**v**under addition, and so we can simply write (−1) ·

**v**= −

**v**.

The other axioms for a vector space can be easily checked.

The method of this proof also shows that we have similar conditions for subsets of groups or ﬁelds to be subgroups, or subﬁelds, respectively.

**Theorem 3**Let H ⊂ G be a (non-empty) subset of the group G. Then H is a subgroup of G ⇔ ab

^{−1}∈ H, for all a, b ∈ H.

*Proof*

The direction ‘⇒’ is trivial. As for ‘⇐’, let a ∈ H. Then aa

^{-1}= e ∈ H. Thus the neutral element of the group multiplication is contained in H. Also ea

^{-1}= a

^{-1}∈ H. Furthermore, for all a, b ∈ H, we have a(b

^{-1})

^{-1}= ab ∈ H. Thus H is closed under multiplication. The fact that the multiplication is associative (a(bc) = (ab)c for all a, b and c ∈ H follows since G itself is a group; thus the multiplication throughout G is associative.

**Theorem 4**Let

**U**,

**W**⊂

**V**be

*subspaces*of the vector space

**V**over the ﬁeld F. Then

**U**∩

**W**is also a subspace.

*Proof*

Let

**v**,

**w**∈

**U**∩

**W**be arbitrary vectors in the intersection, and let a, b ∈ F be arbitrary elements of the ﬁeld F. Then, since

**U**is a subspace of

**V**, we have a ·

**v**+ b ·

**w**∈

**U**. This follows from theorem 2. Similarly a ·

**v**+ b ·

**w**∈

**W**. Thus it is in the intersection, and so theorem 2 shows that

**U**∩

**W**is a subspace.

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