2. Let Q = {[] where x ≤ 0, y ≥ 0} a. Find one vector that is in Q. N Show that Q with the standard addition and multiplication operations, is not a vector space. Find all the axioms that fail to hold. Justify your answer.
2. Let Q = {[] where x ≤ 0, y ≥ 0} a. Find one vector that is in Q. N Show that Q with the standard addition and multiplication operations, is not a vector space. Find all the axioms that fail to hold. Justify your answer.
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter1: Vectors
Section1.3: Lines And Planes
Problem 18EQ
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![Recall the axioms of a vector space: The set V, together with the operations of
addition and scalar multiplication, is said to form a vector space if the following
axioms are satisfied:
i. x+y = y + x for any x and y in V.
ii. (x + y) + z = x + (y+z) for any x, y and 'z in V.
iii. There exists an element O in V such that x + 0 = x for each x E V.
iv. For each x E V, there exists an element -x in Vsuch that x + (-x) = 0.
v. α(x + y) = ax +ay for each scalar a and any x and y in V.
vi. (a + B)x= ax + ßx for each scalar a and ß and any x E V.
vii. (aß)x= a (Bx) for each scalar a and ß and any x E V.
viii. 1x= x for all x E V.
2. Let Q = {[] where x ≤ 0, y ≥0}
a.
Find one vector that is in Q.
2
b. Show that Q with the standard addition and multiplication operations, is not a
vector space. Find all the axioms that fail to hold. Justify your answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b19024a-5609-4e35-a0a7-7058d3f26c57%2F91bf0f80-ac84-49de-8efa-779b3d5a4918%2Foaw4d.jpeg&w=3840&q=75)
Transcribed Image Text:Recall the axioms of a vector space: The set V, together with the operations of
addition and scalar multiplication, is said to form a vector space if the following
axioms are satisfied:
i. x+y = y + x for any x and y in V.
ii. (x + y) + z = x + (y+z) for any x, y and 'z in V.
iii. There exists an element O in V such that x + 0 = x for each x E V.
iv. For each x E V, there exists an element -x in Vsuch that x + (-x) = 0.
v. α(x + y) = ax +ay for each scalar a and any x and y in V.
vi. (a + B)x= ax + ßx for each scalar a and ß and any x E V.
vii. (aß)x= a (Bx) for each scalar a and ß and any x E V.
viii. 1x= x for all x E V.
2. Let Q = {[] where x ≤ 0, y ≥0}
a.
Find one vector that is in Q.
2
b. Show that Q with the standard addition and multiplication operations, is not a
vector space. Find all the axioms that fail to hold. Justify your answer.
![Step 1: Solution for (a)
To find a vector in the set Q, we need to find a
vector [x, y] that satisfies the given conditions:
1. x≤0
2. y ≥0
One possible vector that satisfies these
conditions is [-1, 2].
Here's how it fits the conditions:
1. x = -1, which is less than or equal to 0.
2. y = 2, which is greater than or equal to 0.
So, the vector [-1, 2] is in the set Q.
Step 2: Solution for (b)
To prove: Q with the standard addition and
multiplication operations is not a vector space.
To be a vector space, Q must satisfy ten axioms,
and we will check which axioms fail to hold:
1. Closure under Addition:
This axiom holds for Q because if x < 0 and y ≥0
for both u and v, then (u + v) will also have x ≤ 0
and y ≥ 0, satisfying the conditions of Q.
2. Closure under Scalar Multiplication:
This axiom need not hold for Q. If x ≤ 0 and y
0 for vector u, then for any scalar c<0, cu will
have cx ≥0 (since c can be negative) and cy ≤0.
3. Associativity of Addition:
This axiom holds for all vector spaces,
including Q.
4. Commutativity of Addition:
This axiom holds for all vector spaces,
including Q.
5. Identity Element of Addition:
This axiom holds for Q. In Q, the zero vector is
[0,0].
6. Inverse Elements of Addition:
This axiom fails to hold for Q. If u = [-1, 1], then
there is no -u in Q such that u + (-u) = [0,0] .
Thus, Q lacks inverse elements for addition.
7. Distributivity of Scalar Multiplication over
Vector Addition: This axiom doesn't hold for Q
as it fails scalar multiplication.
8. Distributivity of Scalar Multiplication over
Scalar Addition: For all vectors u in Q and all
scalars c, d, (c + d)u must be equal to cu + du.
When c=-2 and d=4, cu is not in Q but (c+d)u=
(-2+4)u=2u is in Q
Thus the axiom need not hold for Q.
9. Compatibility of Scalar Multiplication with
Field Multiplication: For all vectors u in Q and
all scalars c, d, (cd)u must be equal to c(du). If
one among c or d is negative then the axiom
fails.
This axiom need not holds for Q.
10. Identity Element of Scalar Multiplication:
The identity element must be [1,1] which is not
in Q. Therefore the axiom fails for Q.
Solution
Hence the solution.
Hope it helps.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b19024a-5609-4e35-a0a7-7058d3f26c57%2F91bf0f80-ac84-49de-8efa-779b3d5a4918%2F81pj9l3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Step 1: Solution for (a)
To find a vector in the set Q, we need to find a
vector [x, y] that satisfies the given conditions:
1. x≤0
2. y ≥0
One possible vector that satisfies these
conditions is [-1, 2].
Here's how it fits the conditions:
1. x = -1, which is less than or equal to 0.
2. y = 2, which is greater than or equal to 0.
So, the vector [-1, 2] is in the set Q.
Step 2: Solution for (b)
To prove: Q with the standard addition and
multiplication operations is not a vector space.
To be a vector space, Q must satisfy ten axioms,
and we will check which axioms fail to hold:
1. Closure under Addition:
This axiom holds for Q because if x < 0 and y ≥0
for both u and v, then (u + v) will also have x ≤ 0
and y ≥ 0, satisfying the conditions of Q.
2. Closure under Scalar Multiplication:
This axiom need not hold for Q. If x ≤ 0 and y
0 for vector u, then for any scalar c<0, cu will
have cx ≥0 (since c can be negative) and cy ≤0.
3. Associativity of Addition:
This axiom holds for all vector spaces,
including Q.
4. Commutativity of Addition:
This axiom holds for all vector spaces,
including Q.
5. Identity Element of Addition:
This axiom holds for Q. In Q, the zero vector is
[0,0].
6. Inverse Elements of Addition:
This axiom fails to hold for Q. If u = [-1, 1], then
there is no -u in Q such that u + (-u) = [0,0] .
Thus, Q lacks inverse elements for addition.
7. Distributivity of Scalar Multiplication over
Vector Addition: This axiom doesn't hold for Q
as it fails scalar multiplication.
8. Distributivity of Scalar Multiplication over
Scalar Addition: For all vectors u in Q and all
scalars c, d, (c + d)u must be equal to cu + du.
When c=-2 and d=4, cu is not in Q but (c+d)u=
(-2+4)u=2u is in Q
Thus the axiom need not hold for Q.
9. Compatibility of Scalar Multiplication with
Field Multiplication: For all vectors u in Q and
all scalars c, d, (cd)u must be equal to c(du). If
one among c or d is negative then the axiom
fails.
This axiom need not holds for Q.
10. Identity Element of Scalar Multiplication:
The identity element must be [1,1] which is not
in Q. Therefore the axiom fails for Q.
Solution
Hence the solution.
Hope it helps.
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