Download Example: Let be the set of all polynomials of degree n or less. (That

Example:
Let 𝑃𝑛 be the set of all polynomials of
degree n or less. (That is
𝑃𝑛 = {𝑎0 + 𝑎1 𝑥 + 𝑎2
2
𝑥
+ ⋯+
Then 𝑃𝑛 is a vector space.
𝑛
𝑎𝑛 𝑥 |𝑎𝑖
∈ ℝ} )
4.2 Subspaces
Definition 1
A subset 𝑊 of a vector space is called a subspace of 𝑉 if 𝑊
itself a vector space under the addition & scalar multiplication
defined on 𝑉. (pg. 340)
Theorem: Test for a subspaces
W≠ ∅
𝑊 is a subspace iff
i. 𝑢, 𝑣 ∈ 𝑤 ⇒ 𝑢 + 𝑣 ∈ 𝑤
ii. 𝑘 scalar, 𝑢 ∈ 𝑤 ⇒ 𝑘𝑢 ∈ 𝑤 (pg. 341)
Examples
1. 𝑉 is a vector space.
Then W={0} is a subspace of 𝑉.
2. 𝑊 is the set of all (𝑥, 𝑦, 𝑧) such that
𝑥 = 0.
3. 𝑊 is the set of all (𝑥, 𝑦, 𝑧) such that
𝑥 + 𝑦 + 𝑥 = 0.
Building Subspaces
Theorem
(pg. 345)
If 𝑤1 , 𝑤2 , … , 𝑤𝑟 are subspaces of a vector spaces 𝑉, then the
intersection of these subspaces is also a subspace of 𝑉.
Proof:
𝑟
𝑖=1 𝑤𝑖
Let 𝑢 , 𝑣 ∈
≠ ∅ (since 0 ∈ 𝑤𝑖 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑖)
𝑟
𝑖=1 𝑤𝑖
⇒ 𝑢, 𝑣 ∈ 𝑤𝑖 for each i=1,..,r
⇒ 𝑢 + 𝑣 ∈ 𝑤𝑖 for each i=1,..,r
⇒ 𝑢+𝑣 ∈
Similarly k u ∈
𝑤𝑖
wi (completing the proof)
Recall :
The vector 𝑤 is called a linear
combination of the vectors
𝑣1 , 𝑣2 , … , 𝑣𝑘 provided there exist
scalars 𝑐1 , 𝑐2 , . . . , 𝑐𝑘 such that
𝑤 = 𝑐1 𝑣1 +𝑐2 𝑣2 + ⋯ + 𝑐𝑘 𝑣𝑘 .
Theorem :
If 𝑆 = {𝑤1 , 𝑤2 , . . . , 𝑤𝑟 } is a non-empty set of vectors
in a vector space 𝑉,
then the set 𝑊 of all linear combinations of the
vectors in 𝑆 is a subspace of 𝑉.
(Note: This 𝑊 is the space spanned by the vectors
𝑤1 , 𝑤2 , . . . , 𝑤𝑟 and write 𝑊 = 𝑠𝑝𝑎𝑛(𝑠)
= 𝑠𝑝𝑎𝑛 { 𝑤1 , 𝑤2 , . . . , 𝑤𝑟 }
Proof : Refer pg. 347
Definition: Linear Independence
The vectors 𝑣1 , 𝑣2 , … , 𝑣𝑘 in a vector space 𝑉 are
said to be linearly independent provided that the
equation :
𝑐1 𝑣1 +𝑐2 𝑣2 + ⋯ + 𝑐𝑘 𝑣𝑘 = 0
has only the trivial solution.
(ie. 𝑐1 𝑣1 +𝑐2 𝑣2 + ⋯ + 𝑐𝑘 𝑣𝑘 = 0
𝑐𝑖 =0.)
Example 1
𝑛
The standard unit vectors in ℝ
𝑒1 =(1,0,0,…,0)
𝑒2 =(0,1,0,...,0)
⋮
𝑒𝑛 =(0,0,0,…,1)
are linearly independent.
Read: Example 2 (pg. 361)
Bases for vector spaces
Definition: Basis
If 𝑉 is any vector space and 𝑆 = {𝑣1 , 𝑣2 , … , 𝑣𝑛 }
is a finite set of vectors in 𝑉, then 𝑆 is called a
basis for 𝑉 provided that
a) The vectors in 𝑆 are linearly independent, and
b)The vectors in 𝑆 𝑠𝑝𝑎𝑛 𝑉(ie.every vector is a linear
(pg. 378)
combination of the vectors in S).
Example 1:
𝑒1 =(1,0,0), 𝑒2 =(0,1,0) & 𝑒3 =(0,0,1)
3
unit vectors in ℝ forms a basis for
3
ℝ . (called the standard basis for ℝ3)
Read examples 3,4 on page 379
Note :
1. Any linearly independent 𝑛 vectors in
𝑛
ℝ
is a basis for
𝑛
ℝ .
2. Any set of more than 𝑛 vectors in
linearly dependent.
𝑛
ℝ
is
Note:
𝑎 = (𝑎1 , 𝑎2 , 𝑎3 ), 𝑏 = (𝑏1 , 𝑏2 , 𝑏3 )
𝑎1 𝑏1 𝑐1
iff 𝑎2 𝑏2 𝑐2 ≠ 0
𝑎3 𝑏3 𝑐3
and 𝑐
= (𝑐1 , 𝑐2 , 𝑐3 ) are linearly independent
Proof : 𝑎, 𝑏, 𝑐 are linearly independent
⟺ 𝛼𝑎 +𝛽𝑏 +𝛾𝑐 = 0 gives 𝛼 = 𝛽 = 𝛾 = 0
𝑎1
𝑐1
𝑏1
0
⟺ 𝛼 𝑎2 + 𝛽 𝑏2 + 𝛾 𝑐2 = 0 gives 𝛼 = 𝛽 = 𝛾 = 0
𝑎3
𝑐3
𝑏3
0
𝛼
𝑎1 𝑏1 𝑐1
0
⟺ 𝑎2 𝑏2 𝑐2 𝛽 = 0 has only the trivial solutions
𝛾
𝑎3 𝑏3 𝑐3
0
𝑎1
𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑐1
𝑐2 is invertible.
𝑐3
𝑎1
⟺ 𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑐1
𝑐2 ≠ 0
𝑐3
⟺
Remark:
𝑛
Independent of 𝑛-vectors in ℝ
The 𝑛-vectors
𝑛
𝑣1 , 𝑣2 , … , 𝑣𝑛 in ℝ are linearly
independent iff the 𝑛 × 𝑛 matrix
𝐴 = [𝑣1 𝑣2 … 𝑣𝑛 ] with them as column
vectors has non-zero determinant.
Example 2:
2
𝑆 = 1, 𝑥, 𝑥 , … , 𝑥
𝑛
is a basis for the
vector space 𝑃𝑛 of polynomials of
degree 𝑛 or less.
(called as the standard basis)
Theorem: Uniqueness of basis representation
(pg. 382)
If {𝑣1 , 𝑣2 , … , 𝑣𝑛 } is a basis for a vector space 𝑉,
then every vector 𝑣 in 𝑉 can be expressed in the
form
𝑣 = 𝑐1 𝑣1 +𝑐2 𝑣2 + ⋯ + 𝑐𝑛 𝑣𝑛
in exactly one way.
Dimension
(pg.391)
Note that for any vector space
with a finite basis, any two bases
for a vector space consist of the
same number of vectors.
A non-zero vector space 𝑉 is called finitedimensional provided that a basis of 𝑉 has
only finitely many vectors.
In this case dimension of 𝑉 is the number of
vectors in a basis. (written as 𝑑𝑖𝑚𝑉)
Read: Theorem 4.5.2 (page 391)
&
Theorem 4.5.4 (page 395)
Example 1 :
Dimensions of Some Familiar Vector Spaces
𝑛
dim ℛ = 𝑛
The standard basis has 𝑛 vectors.
dim 𝑃𝑛 = 𝑛 + 1 The standard basis has 𝑛 + 1 vectors.
dim 𝑀𝑚𝑛 = 𝑚𝑛 The standard basis has 𝑚𝑛 vectors.
Read Example 6 (Bases by Inspection)
(page 395)