Vector Spaces
Real Vector Space:
The vector spaces with real scalars are called real vector spaces.
Complex Vector Space:
The vector spaces with complex scalars are called complex vector spaces.
Note:
For now, we will be concerned exclusively with real vector spaces.
Real Vector Spaces:
The following defines the notion of a vector space 𝑽, where 𝑲 is the field
of scalars.
Definition:
Let 𝑽 be a non-empty set of objects, on which two operations are defined:
i. Vector Addition:
ii. Scalar Multiplication:
then 𝑽 is called a vector space over the field 𝑲, if the following axioms hold, for
⃗ ,𝒗
any 𝒖 ⃗⃗⃗ , 𝒘
⃗⃗⃗⃗ ∈ 𝑽:
⃗ ,𝒗
1) Closure Law of Addition: If 𝒖 ⃗ ∈ 𝑽, then 𝒖
⃗ +𝒗
⃗ ∈ 𝑽. (𝑽 is closed under
addition)
2) Associative Law of Addition: ⃗ + (𝒗
𝒖 ⃗⃗⃗ ) = (𝒖
⃗ +𝒘 ⃗ +𝒗
⃗ )+𝒘
⃗⃗⃗
3) Additive Identity: There is a vector in 𝑽, called the zero vector, denoted by ⃗𝟎
⃗ ∈𝑽
such that, for any 𝒖
⃗ + ⃗𝟎 = ⃗𝟎 + 𝒖
𝒖 ⃗ =𝒖
⃗
⃗ ∈ 𝑽, there is a vector in 𝑽, denoted by −𝒖
4) Additive Inverse: For each 𝒖 ⃗ , called
⃗ , such that
a negative of 𝒖
⃗ + (−𝑢
𝑢 ⃗ ) = (−𝑢
⃗)+𝑢
⃗ =0
⃗ +𝒗
5) Commutative Law of Addition: 𝒖 ⃗ =𝒗
⃗ +𝒖
⃗
⃗ is any
6) Scalar Multiplication: If 𝒌 is any scalar element of the field 𝑲 and 𝒖
⃗ is in 𝑽. (Closed under Scalar Multiplication)
vector in V, then 𝒌𝒖
7) For any scalar 𝒌 ∈ 𝑲, we have 𝒌(𝒖 ⃗ ) = 𝒌𝒖
⃗ +𝒗 ⃗ + 𝒌𝒗
⃗
8) For any scalars 𝒂, 𝒃 ∈ 𝑲, we have (𝒌 + 𝒎)𝒖
⃗ = 𝒌𝒖
⃗ + 𝒎𝒖
⃗
9) For any scalars 𝒂, 𝒃 ∈ 𝑲, we have ⃗ ) = (𝒌𝒎)𝒖
𝒌(𝒎𝒖 ⃗
⃗ =𝒖
10) Multiplicative Identity: For the unit scalar 𝟏 ∈ 𝑲, we have 𝟏 𝒖 ⃗
Example # 1: Let
𝑽 = 𝑹𝟐 = {(𝒙, 𝒚); 𝒙, 𝒚 ∈ 𝑹}
Prove that 𝑽 is a vector space under the usual (standard) operations of addition
and scalar multiplication defined by:
⃗ = (𝒖𝟏 , 𝒖𝟐 ) + (𝒗𝟏 , 𝒗𝟐 ) = (𝒖𝟏 + 𝒗𝟏 , 𝒖𝟐 + 𝒗𝟐 ) − − − (𝟏)
⃗ +𝒗
𝒖
⃗ = 𝒌(𝒖𝟏 , 𝒖𝟐 ) = (𝒌𝒖𝟏 , 𝒌𝒖𝟐 ) − − − (𝟐)
𝒌𝒖
Solution:
1. Closure Law of Addition: 𝑽 is closed under addition [as defined in equation (1)]
2. Associative Law of Addition: Let
⃗ = (𝑢1 , 𝑢2 ),
𝑢 ⃗⃗⃗ = (𝑣1 , 𝑣2 ),
𝑣 ⃗⃗ = (𝑤1 , 𝑤2 )
𝑤
then
(𝑢 ⃗⃗ = ( (𝑢1 , 𝑢2 ) + (𝑣1 , 𝑣2 ) ) + (𝑤1 , 𝑤2 )
⃗ + 𝑣) + 𝑤
(𝑢 ⃗⃗ = (𝑢1 + 𝑣1 , 𝑢2 + 𝑣2 ) + (𝑤1 , 𝑤2 )
⃗ + 𝑣) + 𝑤
(𝑢 ⃗⃗ = (𝑢1 + 𝑣1 + 𝑤1 , 𝑢1 + 𝑣1 + 𝑤2 )
⃗ + 𝑣) + 𝑤
(𝑢 ⃗⃗ = (𝑢1 + (𝑣1 + 𝑤1 ), 𝑢2 + (𝑣2 + 𝑤2 ))
⃗ + 𝑣) + 𝑤
(𝑢
⃗ + 𝑣) + 𝑤
⃗⃗ = (𝑢1 , 𝑢2 ) + (𝑣1 + 𝑤1 , 𝑣2 + 𝑤2 )
⃗ +𝒗
(𝒖 ⃗ )+𝒘 ⃗ + (𝒗
⃗⃗⃗ = 𝒖 ⃗⃗⃗ )
⃗ +𝒘
⃗ = (𝑢1 , 𝑢2 ),
3. Additive Identity: Let 𝑢 ⃗0 = (0,0)
⃗ + ⃗0 = (𝑢1 , 𝑢2 ) + (0, 0) = (𝑢1 , 𝑢2 ) = 𝑢
𝑢 ⃗
⃗ = (𝑢1 , 𝑢2 ), then there exist −𝑢
4. Additive Inverse: Let 𝑢 ⃗ = (−𝑢1 , −𝑢2 ),
⃗ ) = (𝑢1 + (−𝑢1 ), 𝑢2 + (−𝑢2 )) = (𝑢1 − 𝑢1 , 𝑢2 − 𝑢2 ) = (0, 0) = ⃗0
⃗ + (−𝑢
𝑢
5. Commutative Law of Addition: Let 𝑢 ⃗ = (𝑢1 , 𝑢2 ), 𝑣 = (𝑣1 , 𝑣2 ), then
⃗⃗⃗
⃗ + 𝑣 = (𝑢1 , 𝑢2 ) + (𝑣1 , 𝑣2 )
𝑢
⃗ + 𝑣 = (𝑢1 + 𝑣1 , 𝑢2 + 𝑣2 )
𝑢
⃗ + 𝑣 = (𝑣1 + 𝑢1 , 𝑣2 + 𝑢2 )
𝑢
⃗ + 𝑣 = (𝑣1 , 𝑣2 ) + (𝑢1 , 𝑢2 )
𝑢
⃗ +𝒗
𝒖 ⃗ =𝒗
⃗ +𝒖
⃗
6. Scalar Multiplication: 𝑽 is closed under scalar multiplication. [as defined
in equation (2)]
7. ⃗ + 𝑣 ) = 𝑘( (𝑢1 , 𝑢2 ) + (𝑣1 , 𝑣2 ) )
𝑘(𝑢
⃗ + 𝑣 ) = 𝑘(𝑢1 + 𝑣1 , 𝑢2 + 𝑣2 )
𝑘(𝑢
⃗ + 𝑣 ) = (𝑘𝑢1 + 𝑘𝑣1 , 𝑘𝑢2 + 𝑘𝑣2 )
𝑘(𝑢
⃗ + 𝑣 ) = (𝑘𝑢1 , 𝑘𝑢2 ) + (𝑘𝑣1 , 𝑘𝑣2 )
𝑘(𝑢
⃗ + 𝑣 ) = 𝑘(𝑢1 , 𝑢2 ) + 𝑘(𝑣1 , 𝑣2 ))
𝑘(𝑢
𝒌(𝒖 ⃗ ) = 𝒌𝒖
⃗ +𝒗 ⃗ + 𝒌𝒗
⃗
8. (𝑘 + 𝑚)𝑢
⃗ = (𝑘 + 𝑚)(𝑢1 , 𝑢2 )
(𝑘 + 𝑚)𝑢
⃗ = (𝑘𝑢1 + 𝑚𝑢1 , 𝑘𝑢2 + 𝑚𝑢2 )
(𝑘 + 𝑚)𝑢
⃗ = (𝑘𝑢1 , 𝑘𝑢2 ) + (𝑚𝑢1 , 𝑚𝑢2 )
(𝑘 + 𝑚)𝑢
⃗ = 𝑘(𝑢1 , 𝑢2 ) + 𝑚(𝑢1 , 𝑢2 )
(𝑘 + 𝑚)𝑢
⃗ =𝑘𝑢
⃗ +𝑚𝑢
⃗
9. ⃗ ) = 𝑘(𝑚(𝑢1 , 𝑢2 )) = (𝑘 𝑚 𝑢1 , 𝑘 𝑚 𝑢2 )
𝑘( 𝑚 𝑢
⃗ ) = 𝑘𝑚(𝑢1 , 𝑢2 ) = 𝑘 𝑚 (𝑢
𝑘( 𝑚 𝑢 ⃗)
10. ⃗ = 1(𝑢1 , 𝑢2 ) = (𝑢1 , 𝑢2 ) = 𝑢
1𝑢 ⃗
Since the set 𝑽 satisfies all the properties, so 𝑽 is a vector space.
Example # 2: Let 𝑽 = 𝑹𝟑 = {(𝒙, 𝒚, 𝒛) | 𝒙, 𝒚, 𝒛 ∈ 𝑹 }, prove that 𝑽 is a vector space
under the usual (standard) operations of addition and scalar multiplication
defined by:
⃗ = (𝒖𝟏 , 𝒖𝟐 , 𝒖𝟑 ) + (𝒗𝟏 , 𝒗𝟐 , 𝒗𝟑 ) = (𝒖𝟏 + 𝒗𝟏 , 𝒖𝟐 + 𝒗𝟐 , 𝒖𝟑 + 𝒗𝟑 ) − − − (𝟏)
⃗ +𝒗
𝒖
⃗ = 𝒌(𝒖𝟏 , 𝒖𝟐 , 𝒖𝟑 ) = (𝒌𝒖𝟏 , 𝒌𝒖𝟐 , 𝒌𝒖𝟑 ) − − − (𝟐)
𝒌𝒖
Solution: Students will do themselves
Example # 3: Let 𝑉 = 𝑅2 , under the usual (standard) operations of addition
defined by:
⃗ + ⃗𝒗 = (𝒖𝟏 , 𝒖𝟐 ) + (𝒗𝟏 , 𝒗𝟐 ) = (𝒖𝟏 + 𝒗𝟏 , 𝒖𝟐 + 𝒗𝟐 ) − − − (𝟏)
𝒖
And if 𝒌 is any scalar element of field 𝑲, then define
⃗ = 𝒌(𝒖𝟏 , 𝒖𝟐 ) = (𝒌𝒖𝟏 , 𝟎) − − − (𝟐)
𝒌𝒖
Check whether 𝑽 is vector space or not?
Solution: All properties of addition are satisfied. (Check it by yourself)
Let us check the properties of scalar multiplication.
⃗ = (𝑢1 , 𝑢2 ) in 𝑽, then 𝑘⃗⃗⃗𝑢 = 𝑘(𝑢1 , 𝑢2 ) = (𝑘𝑢1 , 0) ∈ 𝑉.
6. Let 𝑢
7. Let ⃗ = (𝑢1 , 𝑢2 ),
𝑢 𝑣 = (𝑣1 , 𝑣2 )
⃗ + 𝑣 ) = 𝑘( (𝑢1 , 𝑢2 ) + (𝑣1 , 𝑣2 ) )
𝑘(𝑢
⃗ + 𝑣 ) = 𝑘(𝑢1 + 𝑣1 , 𝑢2 + 𝑣2 )
𝑘(𝑢
⃗ + 𝑣 ) = (𝑘 𝑢1 + 𝑘 𝑣1 , 0)
𝑘(𝑢
⃗ + 𝑣 ) = (𝑘 𝑢1 , 0) + (𝑘 𝑣1 , 0)
𝑘(𝑢
⃗ + 𝑣 ) = 𝑘(𝑢1 , 0) + 𝑘(𝑣1 , 0)
𝑘(𝑢
⃗ = (𝑢1 , 𝑢2 ),
Since 𝑢 ⃗ ≠ (𝑢1 , 0),
𝑣 = (𝑣1 , 𝑣2 ) and 𝑢 𝑣 ≠ (𝑣1 , 0), therefore,
⃗ + 𝑣) ≠ 𝑘 𝑢
𝑘(𝑢 ⃗ + 𝑘 ⃗⃗⃗𝑣
As the 7th property does not satisfy. So, it is not a vector space.
Example 4:
Check whether 𝑽 is vector space or not?
𝑽 = Set of all pairs of real numbers of the form (𝒙, 𝟎). i.e., {(𝑥, 0); 𝑥 ∈ 𝑅}
with the standard operations on 𝑹𝟐 .
⃗ = (𝒖𝟏 , 𝒖𝟐 ) + (𝒗𝟏 , 𝒗𝟐 ) = (𝒖𝟏 + 𝒗𝟏 , 𝒖𝟐 + 𝒗𝟐 ) − − − (𝟏)
⃗ +𝒗
𝒖
⃗ = 𝒌(𝒖𝟏 , 𝒖𝟐 ) = (𝒌𝒖𝟏 , 𝒌𝒖𝟐 ) − − − (𝟐)
𝒌𝒖
Solution:
⃗ = (𝑢1 , 0),
1. Closure Law of Addition: Let 𝑢 𝑣 = (𝑣1 , 0) ∈ 𝑉, then
(𝑢
⃗ + 𝑣) = (𝑢1 + 𝑣1 , 0) ∈ 𝑉
So, 𝑽 is closed under addition.
2. Associative Law of Addition:
⃗ = (𝑢1 , 0), 𝑣 = (𝑣1 , 0), 𝑤
Let 𝑢 ⃗⃗ = (𝑤1 , 0) ∈ 𝑉, then
𝑢⃗ + (𝑣 + 𝑤⃗⃗ ) = (𝑢1 , 0) + ( (𝑣1 , 0) + (𝑤1 , 0) )
⃗ + (𝑣 + 𝑤
𝑢 ⃗⃗ ) = (𝑢1 , 0) + (𝑣1 + 𝑤1 , 0)
⃗ + (𝑣 + 𝑤
𝑢 ⃗⃗ ) = (𝑢1 + 𝑣1 + 𝑤1 , 0)
⃗ + (𝑣 + 𝑤
𝑢 ⃗⃗ ) = (𝑢1 + 𝑣1 , 0) + (𝑤1 , 0)
⃗ + (𝑣 + 𝑤
𝑢 ⃗⃗ ) = (𝑢
⃗ + 𝑣) + 𝑤
⃗⃗
3. Additive Identity: Let ⃗ = (𝑢1 , 0), then
𝑢
⃗ + ⃗0 = (𝑢1 , 0) + (0,0) = (𝑢1 , 0) = 𝑢
𝑢 ⃗
4. Additive Inverse: Let 𝑢 ⃗ = (𝑢1 , 0), then
⃗ + (−𝑢
𝑢 ⃗ ) = (𝑢1 , 0) + (−𝑢1 , 0)
⃗
= (𝑢1 − 𝑢1 , 0) = (0,0) = 0
5. Commutative law of Addition: Let 𝑢 ⃗ = (𝑢1 , 0), 𝑣 = (𝑣1 , 0) ∈ 𝑉, then
⃗ + 𝑣 = (𝑢1 , 0) + (𝑣1 , 0)
𝑢
⃗ + 𝑣 = (𝑢1 + 𝑣1 , 0)
𝑢
𝑢
⃗ + 𝑣 = (𝑣1 + 𝑢1 , 0)
⃗ + 𝑣 = (𝑣1 , 0) + (𝑢1 , 0)
𝑢
𝑢
⃗ +𝑣 =𝑣+𝑢 ⃗
6. Scalar Multiplication: Let 𝑢⃗ = (𝑢1 , 0) ∈ 𝑉, then
⃗ = (𝑘𝑢1 , 𝑘0) = (𝑘𝑢1 , 0) ∈ 𝑉
𝑘𝑢
7. ⃗ + 𝑣) = 𝑘((𝑢1 , 0) + (𝑣1 , 0)) = 𝑘(𝑢1 + 𝑣1 , 0) = (𝑘𝑢1 + 𝑘𝑣1 , 0)
𝑘(𝑢
= (𝑘𝑢1 , 0) + (𝑘𝑣1 , 0) = 𝑘(𝑢1 , 0) + 𝑘(𝑣1 , 0)
= (𝑘𝑢
⃗ + 𝑘𝑣)
8. (𝑘 + 𝑚)𝑢
⃗ = (𝑘 + 𝑚)(𝑢1 , 0)
= ((𝑘 + 𝑚)𝑢1 , 0) = (𝑘𝑢1 + 𝑚𝑢1 , 0)
= (𝑘𝑢1 , 0) + (𝑚𝑢1 , 0)
= 𝑘(𝑢1 , 0) + 𝑚(𝑢1 , 0) = 𝑘𝑢
⃗ + 𝑚𝑢
⃗
9. ⃗ ) = 𝑘(𝑚𝑢1 , 0) = (𝑘𝑚𝑢1 , 0)
𝑘(𝑚𝑢
= 𝑘𝑚(𝑢1 , 0) = (𝑘𝑚)𝑢
⃗
10. ⃗ = 1(𝑢1 , 0) = (𝑢1 , 0) = 𝑢
1𝑢 ⃗
Since all properties of vectors space are satisfied by the set 𝑽. Therefore, 𝑽 is a
vector space.
Example # 5: Check whether 𝑽 is a vector space or not.
𝑽 = Set of all pairs of real numbers of the form (𝒙, 𝒚), where 𝒙 ≥ 𝟎, i.e.
𝑽 = {(𝒙, 𝒚); 𝒙 ≥ 𝟎, 𝒚 ∈ 𝑹}
with standard operations on 𝑹𝟐 .
Solution:
As
𝑉 = {(𝑥, 𝑦); 𝑥 ≥ 0, 𝑦 ∈ 𝑅}
1. Let ⃗ = (𝑢1 , 𝑢2 ),
𝑢 𝑣 = (𝑣1 , 𝑣2 ) ∈ 𝑉
(𝑢
⃗ + 𝑣) = (𝑢1 + 𝑣1 , 𝑢2 + 𝑣2 ) ∈ 𝑉
Because 𝑢
⃗ 1 + 𝑣1 ≥ 0. So, V is closed under addition.
2. 𝑢⃗ +𝑣 =𝑣+𝑢 ⃗ (Easy to verify)
3. 𝑢⃗ + (𝑣 + 𝑤 ⃗⃗ ) = (𝑢⃗ + 𝑣) + 𝑤
⃗⃗ ) (Easy to verify)
4. Let 𝑢 ⃗ = (𝑢1 , 𝑢2 ), 𝑢⃗ +0⃗ = (𝑢1 , 𝑢2 ) + (0,0) = (𝑢1 , 𝑢2 ) = 𝑢
⃗
5. Let 𝑢 ⃗ = (𝑢1 , 𝑢2 ), Then there doesn’t exist −𝑢 ⃗ = (−𝑢1 , −𝑢2 ) because
𝑢
⃗ 1 should be positive.
Hence, 5th property fails, So, 𝑽 is not vector space.
Example 6: Show that the set of all pairs of real numbers of the form (𝑥, 1) with the
operations
(𝑥, 1) + (𝑥 ′ , 1) = (𝑥 + 𝑥 ′ , 1)
and
𝑘(𝑥, 1) = (𝑘 2 𝑥, 1)
is not a vector space.
Example # 7: Determine whether the set of all triples of real numbers with
standard vector addition but with scalar multiplication defined by
𝒌(𝒙, 𝒚, 𝒛) = (𝒌𝟐 𝒙, 𝒌𝟐 𝒚, 𝒌𝟐 𝒛)
is a vector space or not?
.
Axiom 8 fails.
Example # 8: Determine whether the set of all pairs of real numbers of the form
(𝟏, 𝒙) with the operations
(𝟏, 𝒚) + (𝟏, 𝒚′ ) = (𝟏, 𝒚 + 𝒚′ )
and
𝒌(𝟏, 𝒚) = (𝟏, 𝒌𝒚)
is a vector space or not?
Example # 9: Determine whether 𝑉 is a vector space or not.
V= the set of all triples of the form (𝑥, 𝑦, 𝑧) with the operations
(𝑥, 𝑦, 𝑧) + (𝑥 ′ , 𝑦 ′ , 𝑧 ′ ) = (𝑥 + 𝑥 ′ , 𝑦 + 𝑦 ′ , 𝑧 + 𝑧 ′ )
& 𝑘(𝑥, 𝑦, 𝑧) = (𝑘𝑥, 𝑦, 𝑧)
Example # 10: Let 𝑽 be the set of all 𝟐 × 𝟐 matrices with real entries and take the
vector space operations on 𝑽 to be usual operations of matrix addition and scalar
multiplication i.e.
𝒖𝟏𝟏 𝒖𝟏𝟐 𝒗𝟏𝟏 𝒗𝟏𝟐 𝒖𝟏𝟏 + 𝒗𝟏𝟏 𝒖𝟏𝟐 + 𝒗𝟏𝟐
⃗ +𝒗
𝒖 ⃗ = [𝒖 ] + [ ] = [
𝟐𝟏 𝒖𝟐𝟐 𝒗𝟐𝟏 𝒗𝟐𝟐 𝒖𝟐𝟏 + 𝒗𝟐𝟏 𝒖𝟐𝟐 + 𝒗𝟐𝟐 ] − − − (𝟏)
𝒖𝟏𝟏 𝒖𝟏𝟐 𝒌𝒖𝟏𝟏 𝒌𝒖𝟏𝟐
⃗ = 𝒌 [𝒖
𝒌𝒖 𝒖𝟐𝟐 ] = [ ] − − − − − − − (𝟐)
𝟐𝟏 𝒌𝒖𝟐𝟏 𝒌𝒖𝟐𝟐
Determine whether 𝑽 is a vector space or not?
Solution:
1. From equation (1), it is clear that 𝑉 is closed under addition.
2. ⃗ + (𝑣 + 𝑤
𝑢 ⃗⃗ ) = (𝑢
⃗ + 𝑣) + 𝑤
⃗⃗
𝑢11 𝑢12 0 0 𝑢11 𝑢12
3. 𝑢
⃗ + 0 = [𝑢 ] + [ ] = [𝑢21 𝑢22 ] = 𝑢
⃗
21 𝑢22 0 0
𝑢11 𝑢12 −𝑢11 −𝑢12
⃗ + (−𝑢
4. 𝑢 ⃗ ) = [𝑢 ] + [ −𝑢21 −𝑢22 ]
21 𝑢22
𝑢 + (−𝑢11 ) 𝑢12 + (−𝑢12 )
= [ 11 ]
𝑢21 + (−𝑢21 ) 𝑢22 + (−𝑢22 )
0 0 ⃗
=[ ]=0
0 0
𝑢11 𝑢12 𝑣11 𝑣12
5. 𝑢
⃗ + 𝑣 = [𝑢 𝑢22 ] + [ 𝑣21 𝑣22 ]
21
𝑢 +𝑣 𝑢12 + 𝑣12
= [𝑢11 + 𝑣11 𝑢22 + 𝑣22 ]
21 21
𝑣 +𝑢 𝑣12 + 𝑢12
= [𝑣11 + 𝑢11 𝑣22 + 𝑢22 ]
21 21
=𝑣+𝑢
⃗
Similarly, you can prove all the properties of scalar multiplication.
(Prove it by yourself).
Since all properties of vector space with usual operations of matrix addition and
scalar multiplication satisfied by the given set 𝑉, therefore, 𝑉 is a vector space.
Example # 11: Let 𝑽 = 𝑹𝒏 and define operations on 𝑽 to be the usual operations of
addition given by
⃗ + 𝑣 = (𝑢1 , 𝑢2 , 𝑢3 , … , 𝑢𝑛 ) + (𝑣1 , 𝑣2 , 𝑣3 , … , 𝑣𝑛 ) = (𝑢1 + 𝑣1 , 𝑢2 + 𝑣2 , … , 𝑢𝑛 + 𝑣𝑛 )
𝑢
and scalar multiplication given by
⃗ = 𝑘(𝑢1 , 𝑢2 , 𝑢3 , … , 𝑢𝑛 ) = (𝑘𝑢1 , 𝑘𝑢2 , 𝑘𝑢3 , … , 𝑘𝑢𝑛 )
𝑘𝑢
Show that V is a vector space.
Example # 12: Let 𝑽 be the set of polynomials of the form
𝑃(𝑡) = 𝑎𝑛 𝑡 𝑛 + 𝑎𝑛−1 𝑡 𝑛−1 + ⋯ + 𝑎1 𝑡 + 𝑎0 .
Determine whether 𝑽 is a vector space under the usual operations of addition and
scalar multiplication or not?
THEOREM:
Let 𝑉 be a vector space, 𝒖 be a vector in 𝑉, and 𝑘 be a scalar; then:
a) 0 𝒖 = 0
b) 𝑘 𝟎 = 0
c) (−1)𝒖 = −𝒖
d) If 𝑘 𝒖 = 0, then 𝑘 = 0 or 𝒖 = 𝟎.
Practice Questions for Students
Q # 1: Let 𝑉 be the set of all ordered pairs of real numbers, and consider the following addition
⃗ = (𝒖𝟏 , 𝒖𝟐 ) and 𝒗
and scalar multiplication operations on 𝒖 ⃗ = (𝒗𝟏 , 𝒗𝟐 ):
⃗ + ⃗𝒗 = (𝒖𝟏 , 𝒖𝟐 ) + (𝒗𝟏 , 𝒗𝟐 ) = (𝒖𝟏 + 𝒗𝟏 , 𝒖𝟐 + 𝒗𝟐 ) − − − (𝟏)
𝒖
⃗ = 𝒌(𝒖𝟏 , 𝒖𝟐 ) = (𝟎, 𝒌𝒖𝟐 ) − − − (𝟐)
𝒌𝒖
⃗ +𝒗
a) Compute 𝒖 ⃗ and 𝒌𝒖 ⃗ = (−𝟏, 𝟐), 𝒗
⃗ for 𝒖 ⃗ = (𝟑, 𝟒) and 𝒌 = 𝟑.
b) Since addition of 𝑽 is standard operations on 𝑹𝟐 , certain vector space axioms hold for 𝑽
because they are known to hold for 𝑹𝟐 . Which axioms are they?
⃗ = (−𝟏, 𝟐), then find the addition inverse of 𝒖
c) If 𝒖 ⃗.
d) Show that axiom 7, 8, & 9 hold, where 𝒎 = 𝟐.
e) Show that axiom 10 fails and hence 𝑽 is not a vector space under the given operations.
Answer: Do yourself
⃗ +𝒗
(a) 𝒖 ⃗ = (𝟐, 𝟔) and 𝒌𝒖
⃗ = (𝟎, 𝟔).
c) Axiom 1-5
Q# 2: Let 𝑉 be the set of all ordered pairs of real numbers, and consider the following addition
⃗ = (𝒖𝟏 , 𝒖𝟐 ) and 𝒗
and scalar multiplication operations on 𝒖 ⃗ = (𝒗𝟏 , 𝒗𝟐 ):
⃗ = (𝒖𝟏 , 𝒖𝟐 ) + (𝒗𝟏 , 𝒗𝟐 ) = (𝒖𝟏 + 𝒗𝟏 + 𝟏, 𝒖𝟐 + 𝒗𝟐 + 𝟏) − − − (𝟏)
⃗ +𝒗
𝒖
⃗ = 𝒌(𝒖𝟏 , 𝒖𝟐 ) = (𝒌𝒖𝟏 , 𝒌𝒖𝟐 ) − − − (𝟐)
𝒌𝒖
⃗ +𝒗
a) Compute 𝒖 ⃗ and 𝒌𝒖 ⃗ = (𝟎, 𝟒), 𝒗
⃗ for 𝒖 ⃗ = (𝟏, −𝟑) and 𝒌 = 𝟐.
⃗ = (−𝟏, 𝟐), then find the addition inverse of 𝒖
b) If 𝒖 ⃗.
⃗ + (−𝒖
⃗ such that 𝒖
c) Show that axiom 5 holds by producing an ordered pair −𝒖 ⃗ ) = 𝟎 for
⃗ = (𝒖𝟏 , 𝒖𝟐 )
𝒖
d) Find two vector space axioms that fail to hold.
