Create a Vector Space
Def - Vector Space
For a set to be a vector space, it must satisfy the following rules for all vectors u, v, and w, and all scalars a and b:
- Closure under addition: u + v is always a vector in the set.
- Commutativity: u + v = v + u.
- Associativity of addition: (u + v) + w = u + (v + w).
- Additive identity: There exists a zero vector 0 such that u + 0 = u.
- Additive inverse: For every vector u, there exists a vector -u such that u + (-u) = 0.
- Closure under scalar multiplication: a * u is always a vector in the set.
- Distributivity of scalars over vectors: a * (u + v) = au + av.
- Distributivity of scalars over scalars: (a + b) * u = au + bu.
- Associativity of scalar multiplication: a * (b * u) = (a * b) * u.
- Scalar identity: 1 * u = u.
Figure 5.1 Closed under addition
Figure 5.1
It clearly shows that adding two vectors AB and BC results in another vector AC that is also a vector which is exactly what "closed under addition" means.
Figure 5.2 - Commutativity
Vector AB = Vector EF
Vector BC = Vector DE
Vector AC = Vector DF
Vector AB + Vector BC = Vector AC
Since AB = EF and BC = DE, we can substitute:
Vector DE + Vector EF = Vector BC + Vector AB
Vector DF = Vector AC
Therefore Vector AB + Vector BC = Vector AC
and Vector BC + Vector AB = Vector DF = Vector AC
This shows that changing the order of addition still gives the same resultant vector.
As this is an algebraic proof these vectors could be any vectors, there for we've proven that vector addition is commutative.
Question 5.2 - Prove Vector addition is Commutative
In your own words, describe Figure 5.2 and specifically how it can be used to demonstrate that vector addition is commutative.
Figure 5.3 - Associativity under addition
Figure 5.3 - Associativity under addition
Given (equal lengths):
E₀F₀ = E₃F₃ = E₄F₄
F₀G₀ = F₄G₄ and E₂H₂ = E₅H₅.
G₀H₀ = G₁H₁ = G₄H₄
The left side shows:
(E₀F₀ + F₀G₀) + G₀H₀
= E₁G₁ + G₁H₁ (since E₁G₁ = E₀F₀ + F₀G₀)
= E₂H₂ (adding E₁G₁ + G₁H₁)
The right side shows:
= E₄F₄ + (F₄G₄ + G₄H₄)
= E₃F₃ + F₃H₃ (since F₄G₄ + G₄H₄ = F₃H₃)
= E₅H₅
where E₂H₂ = E₅H₅.
Therefore, (u + v) + w = u + (v + w), proving that vector addition is associative.
Question 5.3
What does this figure prove about vector addition? Explain in your own words.
Figure 5.4 - Additive identity
Figure 5.4 - Additive identity: There exists a zero vector 0 such that u + 0 = u.
The figure shows:
Vector AB + Vector CC = Vector AB
Vector CC can be considered as just Point C, which has zero length and zero direction.
Therefore, any point can be used as the zero vector.
Question 5.4
What do you notice about Vector CC?
Figure 5.5 - Additive inverse
Figure 5.5 - Additive inverse: For every vector u, there exists a vector -u such that u + (-u) = 0.
The figure shows
Vector AB and Vector CD
AB + CD = 0
Therefore CD = -AB
Question 5.5
What do you notice about the length and direction of Vector AB and Vector CD?
Figure 5.6.1 - Multiplication By a Scalar Whole Number
Figure 5.6.1 - Closure under scalar multiplication: a * u for a whole number.
The figure shows Vector AB.
Point C is given on line CJ, which is parallel to AB.
Circle C has radius AB and intersects line CJ at point D.
Circle D has radius AB and intersects line CJ at point E.
Circle E has radius AB and intersects line CJ at point F.
This continues for n steps, ending at point J.
Therefore, CJ = n * AB, where n is the number of segments (CD, DE, EF, ...).
5.6.2 - Closure under scalar multiplication: a * u
Figure 5.6.2 - Closure under scalar multiplication: a * u
The Figure shows vector AB.
Vector A₀B₀ is a copy of vector AB.
Line P₁B₀ starts at B₀ and extends outward.
Points P₁, P₂, P₃, P₄, P₅ are equally spaced using the method from 5.6.1,
Therefore P₁B₀ = P₁P₂ = P₂P₃ = P₃P₄ = P₄P₅.
Parallel lines are drawn to meet A₀B₀ at points B₁, B₂, B₃, B₄, and A₀
Parallel lines are drawn from P₁, P₂, P₃, P₄, and P₅ respectively.
This forms similar triangles: B₀P₁B₁, B₀P₂B₂, B₀P₃B₃, B₀P₄B₄, and B₀P₅A₀.
Because the points on line P₁B₀ are equally spaced, A₀B₀ is cut into 5 equal pieces.
A point Pₙ is animated to travel along P₁B₀. As it moves, it shows that any rational scalar multiple of AB can be obtained in this way.
Question 5.6.2.1
Why are the triangles B₀P₁B₁, B₀P₂B₂, B₀P₃B₃, B₀P₄B₄, and B₀P₅A₀ similar to each other?
5.6.2.2
The animation shows point Pₙ moving along line P₁B₀. What does this demonstrate about scalar multiplication of a vector?
Figure 5.7 - Distributivity of scalars over vectors
Figure 5.7 - Distributivity of scalars over vectors: a * (u + v) = au + av.
The figure shows
Slider "a" where 0 ≤ a ≤ 5.
The construction Vector u + Vector v = Vector w is given.
As the slider is adjusted, Vector aw scales correspondingly.
The construction au + av = aw is shown.
As the slider is adjusted, all vectors scale accordingly.
Question 5.7
Move the slider from a = 1 to a = 3. Compare what happens to vectors u, v, and w versus vectors au, av, and aw.
5.8 Distributivity of scalars over scalars: (a + b) * u = au + bu.
The figure shows
Slider a and slider b, where 0 ≤ a ≤ 5 and 0 ≤ b ≤ 5.
Vector u is given.
Vector cu is shown, where c = a + b.
Vector au and vector bu are shown head-to-tail.
Move the sliders and observe the vectors. You will see that vector au + vector bu is always equal to vector cu.
By inspection, vector au + bu = cu.
Question 5.8
The figure shows slider a and slider b, where 0 ≤ a ≤ 5, and vector u can be manipulated to be any vector. The construction demonstrates that (a + b)u = au + bu for all values of a and b between 0 and 5, and for any vector u you choose. Does the fact that a and b are limited to values between 0 and 5 truly limit the proof? Or is it easy to assume that the property would also hold for a = 500 or a = 5,000,000? Explain your reasoning.
Figure 5.9 - Associativity of scalar multiplication: a * (b * u) = (a * b) * u.
The figure shows
Slider a and slider b, where 0 ≤ a ≤ 5 and 0 ≤ b ≤ 5.
Vector u is given.
Vector bu is shown.
Vector a(bu) is shown, this is vector u scaled by b, then by a.
Vector (ab)u is shown, this is vector u scaled by the product a*b.
Move the sliders and observe that a(bu) and (ab)*u are always equal in both length and direction.
The value of a*b is also displayed to help you verify.
Question 5.9
Since vector u can be any vector and you can move the sliders, does this prove that the property holds for all values a and b and all vectors u? Explain.
5.10 - Scalar identity
5.10 - Scalar identity: 1 * u = u.
5.10 - Scalar identity: 1 * u = u.
The figure shows
Slider a, where 0 ≤ a ≤ 5
Vector AB labelled
Vector CD = a*AB
Question 5.10
For what value of slider a are vector AB and vector CD equal?