VECTOR SPACES
-JASMINE GALAPON & CHRISTOPHER COLANGO
VECTOR SPACE
- Is a fundamental concept in linear algebra. It is a
mathematical structure that consists of a set of
objects called vectors, along with operations such as
addition and scalar multiplication. These operations
follow specific rules that define the properties of a
vector space.
A vector space consists of a nonempty set V, together
with two operations + (vector addition) and • (scalar
multiplication).
Suppose that v and w are two vectors in V and let a in R.
• Their sum v + w is in V.
• The scalar product of v by a, denoted by a• v is in V.
The set V is closed under vector addition and scalar
multiplication.
Consider the vector space of
2D vectors, denoted as R^2,
where R represents the set of
real numbers. In this vector
space, each vector has two
components (x, y) where x and y
are real numbers.
VECTOR SPACE
PROPERTIES
1. CLOSURE UNDER ADDITION:
Take two vectors u = (2, 3) and v = (4, 1) from
R^2.
Their sum u + v = (2+4, 3+1) = (6, 4).
Since both components of the resulting vector are
real numbers; the sum is in R^2.
VECTOR SPACE
PROPERTIES
2. CLOSURE UNDER SCALAR
MULTIPLICATION:
Consider the vector u = (2, 3) and scalar c = 2.
The scalar multiplication cu = 2(2, 3) = (4, 6), which
is also in R^2.
VECTOR SPACE
PROPERTIES
3. ASSOCIATIVITY OF ADDITION:
For three vectors u = (2, 3) , v = (4, 1) , and
w = (1, 5)
(u + v) + w = ((2+4)+1, (3+1)+5) = (7, 9)
and u + (v + w) = (2+(4+1), 3+(1+5)) = (7, 9).
Both results are the same, demonstrating
associativity.
VECTOR SPACE
PROPERTIES
4. COMMUTATIVITY OF ADDITION:
Taking the same vectors u = (2, 3) and
v = (4, 1)
we have u + v = (2+4, 3+1) = (6, 4) and v + u =
(4+2,1+3) = (6, 4).
The order of addition does not affect the result,
showing commutativity.
VECTOR SPACE
PROPERTIES
5. IDENTITY ELEMENT OF ADDITION:
The zero vector, denoted as 0, is (0, 0) . For
any vector u = (2, 3)
we have u + 0 = (2+0, 3+0) = (2, 3)
which leaves u unchanged.
VECTOR SPACE
PROPERTIES
6. EXISTENCE OF ADDITIVE INVERSES:
For the vector u = (2, 3) , its additive inverse is
-u = (-2, -3)
since u + (-u) = (2+(-2), 3+(-3)) = (0, 0).
VECTOR SPACE
7. COMPATIBILITY
OF SCALAR
PROPERTIES
MULTIPLICATION WITH FIELD MULTIPLICATION:
Taking the vector v=(4,1) , u = (2, 3) , scalar c = 2,
and scalar d = 3, we have
c (u + v) = 2((2, 3) + (4, 1))
Similarly,
= 2(6, 4)
(cd)u = (2*3) (2, 3)
= (12, 8)
= 6(2, 3)
cu + cv = 2(2, 3) + 2(4, 1)
= (12, 18)
= (4, 6) + (8, 2)
c(du) = 2(3(2, 3))
= (12, 8).
= 2(6, 9)
= (12, 18).
Topic 2. Subspaces
Definition. A nonempty subset W of a
vector space V is a subspace of V
when W is a vector space under the
operations of addition
and scalar multiplication defined in V . It
is clear from the definition that a
subspace is a vector space.
Thus, any subspace of a vector space
contains the zero vector. In
fact, the set containing the zero vector
is a subspace of any vector
space and is called the zero
subspace. Another subspace of a
vector
space is itself.
Now, one might think that to show
that a nonempty
subset of a vector space is a
subspace needs to check all the
axioms holds for that set. The next
theorem, however, tells us that it is
enough to check the closure property
for addition (Axiom 1) and the closure
property for scalar multiplication
(Axiom 6).
Theorem 4.2.1.
If W is a nonempty subset of a
vector space V , then W is a
subspace of V if and only if the
following conditions hold.
(i) If u and v are in W, then u + v is in
W.
(ii) If u is in W and c is any scalar,
then cu is in W
Proof.
If W is a subspace, then W is a vector
space and by definition, conditions (i) and
(ii) hold.
Conversely, suppose conditions (i) and (ii)
hold. We want to show
that W is a subspace. We first not that, by
condition (ii), (−1)u is
in W for any u in W. From condition (i), u
+ (−1)u is also in W. But u + (−1)u = 0,
and so 0 is in W. Then u + 0 = u for any u
in W. Finally, the other axioms hold in W
since they hold in V . Therefore, W is a
subspace.
Show that the set of 2D vectors V = {(x, y) |
x + y = 0} is a vector space under vector
addition and scalar multiplication. To prove
that V is a vector space, we need to verify
that it satisfies the vector space axioms.