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4.1 Vector Spaces & Subspaces
Many concepts concerning vectors in R n^ can be extended to other mathematical systems. We can think of a vector space in general, as a collection of objects that behave as vectors do in R n. The objects of such a set are called vectors.
A vector space is a nonempty set V of objects, called vectors , on which are defined two operations, called addition and multiplication by scalars (real numbers), subject to the ten axioms below. The axioms must hold for all u , v and w in V and for all scalars c and d.
- u v is in V.
- u v v u.
- u v w u v w
- There is a vector (called the zero vector) 0 in V such that u 0 u.
- For each u in V, there is vector − u in V satisfying u − u 0.
- c u is in V.
- c u v c u c v.
- c d u c u d u.
- cd u cd u .
- 1 u u.
Vector Space Examples
EXAMPLE: Let M 2 2 a^ b c d
: a, b, c, d are real
In this context, note that the 0 vector is.
EXAMPLE: Let n ≥ 0 be an integer and let
P n the set of all polynomials of degree at most n ≥ 0.
Members of P n have the form
p t a 0 a 1 t a 2 t^2 a (^) n t n
where a 0 , a 1 , … , a (^) n are real numbers and t is a real variable. The set P n is a vector space.
We will just verify 3 out of the 10 axioms here.
Let p t a 0 a 1 t a (^) n t n^ and q t b 0 b 1 t b (^) n t n. Let c be a scalar.
Axiom 1 :
The polynomial p q is defined as follows: p q t p t q t. Therefore,
p q t p t q t
__________ __________t __________t n
which is also a _____________________ of degree at most ________. So p q is in P n.
Axiom 4 : 0 0 0 t 0 t n (zero vector in P n)
p 0 t p t 0 a 0 0 a 1 0 t a (^) n 0 t n a 0 a 1 t a (^) n t n^ p t and so p 0 p
Axiom 6 : c p t c p t ________ ________t ________t n which is in P n.
The other 7 axioms also hold, so P n is a vector space.
EXAMPLE: Is H x x 1
: x is real a subspace of _______?
I.e., does H satisfy properties a, b and c?
− 0.5 0.5 1 1.5 2 x^1
0.
1
1.
2
2.
3 x^2
Graphical Depiction of H Solution:
All three properties must hold in order for H to be a subspace of R^2.
Property (a) is not true because
___________________________________________.
Therefore H is not a subspace of R^2.
Another way to show that H is not a subspace of R^2 : Let
u 0 1
and v 1 2
, then u v
and so u v 1 3
, which is ____ in H. So property (b) fails and so H is not a subspace of
R^2.
− 0.5 0.5 1 1.5 2 x^1
0.
1
1.
2
2.
3
x 2
− 0.5 0.5 1 1.5 2 x^1
0.
1
1.
2
2.
3
x 2
Property (a) fails Property (b) fails
A Shortcut for Determining Subspaces
THEOREM 1 If v 1 , … , v p are in a vector space V, then Span v 1 , … , v p is a subspace of V.
Proof: In order to verify this, check properties a, b and c of definition of a subspace.
a. 0 is in Span v 1 , … , v p since
0 _____ v 1 _____ v 2 _____ v p
b. To show that Span v 1 , … , v p closed under vector addition, we choose two arbitrary vectors in Span v 1 , … , v p :
u a 1 v 1 a 2 v 2 a (^) p v p and v b 1 v 1 b 2 v 2 b (^) p v p.
Then u v a 1 v 1 a 2 v 2 a (^) p v p (^) b 1 v 1 b 2 v 2 b (^) p v p
___ v 1 ____ v 1 ____ v 2 ____ v 2 ____ v p ____ v p (^) a 1 b 1 v 1 (^) a 2 b 2 v 2 (^) a (^) p b (^) p v p.
So u v is in Span v 1 , … , v p .
c. To show that Span v 1 , … , v p closed under scalar multiplication, choose an arbitrary number c and an arbitrary vector in Span v 1 , … , v p : v b 1 v 1 b 2 v 2 b (^) p v p.
Then c v cb 1 v 1 b 2 v 2 b (^) p v p
______ v 1 ______ v 2 ______ v p
So c v is in Span v 1 , … , v p .
Since properties a, b and c hold, Span v 1 , … , v p is a subspace of V.
