HERE IS MY QUESTION:

The set of elements belonging to R2 is usually denoted as {(a, b) | a, b∈R}. Combining elements within this set under the operations of addition and scalar multiplication should use the following notation:

Addition Example: (–2,10)+(–5,0)=(–2–5,10+0)=(–7,10)

Scalar Multiplication Example: –10 × (1, –7) = (–10 × 1, –10 × –7) = (–10, 70), where –10 is a scalar.

Provide a written explanation (suggested length of 3–5 pages) of vector space in which you do the following:

A. Provide a mathematical definition for a vector space.

B. Indicate whether R2 is a vector space.

1. Justify your assertions by applying the provided mathematical definition for a vector space.

C. Determine whether R2 is spanned by (1,1) and (3,2), showing all work.

D. Define a nontrivial subspace of R2, showing all work.

HERE IS MY ANSWER:

  1. Vector Spaces – Definition

A vector space on the set of real numbers ℝis a set V – whose elements are called vectors - equipped with two operations, called addition of two vectors (denoted u + v, or (u1,u2,u3,…un) + (v1,v2,v3,…vn)) and scalar multiplication,(denoted au, or a x (u1,u2,u3,…un)). The addition operation is a function which associates to each pair of vectors u, v ∈V the sum u + v ∈V . The scalar multiplication is a function associating to each a ∈ℝand each u ∈V the product au ∈V .

Let u, v, and w ∈V, and let a and b ∈ℝ.

The addition and scalar multiplication operations satisfy the ten following properties:

  1. u + v ∈V (closure under addition of vectors)
  2. u + v = v + u for any u,v∈V (commutative property of addition)
  3. (u + v) + w = u + (v + w) for any u, v, w ∈V (associative property of addition)
  4. There exists a vector 0 ∈V with the property that u + 0 = u for any u ∈V (zero element or additive identity)
  5. For any u ∈V there exists a vector −u ∈V such that u + (−u) = 0 (additive inverse)
    Furthermore, scalar multiplication satisfies the following properties:
  6. au ∈V (closure under scalar multiplication)
  7. (ab)u = a(bu) for any a, b∈ℝ, u ∈V (associative property of scalar multiplication)
  8. (a + b)u = au + bufor any a, b ∈ℝ, u ∈V (distributive property of scalar multiplication with regards to addition of scalars)
  9. a(u + v) = au + avfor any a ∈ℝ, u,v∈V (distributive property of scalar multiplication with regards to addition of vectors)
  10. 1v = v for any v ∈V (preservation of scale or multiplicative identity property)
  1. ℝ2 is a vector space.

Proof:

By definition, ℝ2is the set of ordered pairs of real numbers, i.e.

ℝ2:= {(a1, a2 ) : a1,a2 ∈R}.

The elements of ℝ2are called vectors. Given a vector x = (x1, x2), the numbers x1,x2are called the components of x. Similarly, given a vector u = (u1, u2), the numbers u1,u2are called the components of u.

R1 = R may be interpreted as the number line and R2 as the Cartesian plane.

The set of elements belonging toℝ 2is often denoted as{(a,b) |a,b∈ℝ }.

We need to prove that all elements of ℝ2 satisfy the ten properties of vector spaces.

Let u,v,&w be three vectors from ℝ 2, ie. u,v,w∈ℝ2 . u,v,w can be denoted or represented as (u1,u2), (v1,v2), (w1,w2), where u1,u2,v1,v2,w1,w2 are real numbers, ie.u1,u2,v1,v2,w1,w2∈ℝ.

Also, let a,b be scalars, ie. a,b, ∈ℝ.

  1. We have u+v = (u1,u2)+ (v1,v2) = (u1+ v1, u2+ v2). Because u1+ v1, u2+ v2 ∈ℝ, u+v ∈ℝ2 and there is closure under addition of vectors;therefore the first property is satisfied.
  2. Then u + v= (u1,u2)+ (v1,v2) = (u1+ v1, u2+ v2) = (v1+ u1, v2+ u2) = v+u. The addition is commutative; therefore the second property is satisfied.
  3. We have (u + v) + w=(u1+ v1, u2+ v2)+(w1,w2)=(u1,u2)+(v1+w1,v2+w2)= u+(v+w). The addition is associative; therefore the third property is satisfied.
  4. And u+0 = (u1,u2)+ (0,0) =(u1+0,u2+0) = (u1,u2) = u. There is a zero element:therefore the fourth property is satisfied.
  5. We have u + (−u) =(u1,u2)+ (-u1,-u2) = (u1-u1,u2-u2)= (0,0). There is an additive inverse vector; therefore the fifth property is satisfied.
  6. We have au = a(u1,u2) = (au1,au2) . Because au1,au2 are real numbers, ie.au1,au2∈V there is closure under scalar multiplication: therefore the sixth property is satisfied.
  7. We have (ab)u= ab(u1,u2) = (abu1,abu2) = a(bu1,bu2) =a(bu). The scalar multiplication is associative: therefore the seventh property is satisfied.
  8. We have (a + b)u= ((a+b)u1, (a+b)u2) = ((au1+b u1), (au2+b u2)) = (au1, au2) +(bu1, bu2) =au + bu. The scalar multiplication is distributive with respect to addition of scalars; therefore the eighth property issatisfied.
  9. We have a(u + v) = a(u1+ v1, u2+ v2) = (au1+ av1, au2+ av2) = (au1,au2)+ (av1,av2)) = a(u1,u2) + a(v1,v2) = au + av. The scalar multiplication is distributive with respect to addition of vectors; therefore the ninth property is satisfied.
  10. We have 1v= 1(v1,v2) = (1v1,1v2) = (v1,v2) = v. There is preservation of scale; therefore the tenth property is satisfied.
  1. R2is spanned by (1,1) and (3,2).

Proof:

  1. First we show that R2is spanned by (1,0) and (0,1).
    v = (v1,v2) = (v1+0,0+v2) = (v1,0) + (0,v2) = v1(1,0) + v2 (0,1). Ie. R2is spanned by (1,0) and (0,1), or any vector in R2is a linear combination of (1,0) and (0,1).
  2. Second, we show that (1,0) and (0,1) are each a linear combination of (1,1) and (3,2).
    Let a,b,c,d be real numbers, ie. a,b,c,d∈ℝ.
    a(1,1) + b(3,2) = (1,0) ie.
    a+3b=1anda+2b=0. Solving these simultaneous equations yields a = -2, b = 1 ie. (1,0) = -2(1,1) + 1(3,2)
    c(1,1) + d(3,2) = (0,1) ie.
    c+3d=0andc+2d=1. Solving these simultaneous equations yields c=3, d=-1
    ie. (0,1) = 3(1,1) + -1(3,2)
  3. From paragraph 1., we have any vector v = (v1,v2) = v1(1,0) + v2 (0,1)
    Substituting the results obtained in 2., for (1,0) and (0,1), we have,
    (v1,v2) = v1(1,0) + v2 (0,1) = v1(-2(1,1) + 1(3,2)) + v2(3(1,1) + -1(3,2))
    = -2v1(1,1) + v1(3,2)) + 3v2((1,1) - v2(3,2)) = -2v1(1,1) + 3v2(1,1) + v1(3,2)) - v2(3,2) = (2v1 +3 v2) (1,1) + (v1 - v2)(3,2).
    ie. any vector (v1,v2) ∈ ℝ2 can be expressed as a linear combination of (1,1) and (3,2); ie. R2is spanned by (1,1) and (3,2).
  1. An example of a non-trivial subspace of ℝ2 is {(a,3a) |a∈R}.
    This can be described as all points on a straight line with a slope of 3, passing through the origin, in the xy two-dimensional plane. It can be shown that all points satisfy the ten properties of a vector space.

HERE’S WHAT I needfixed:

An appropriatenontrivialsubspaceisgiven as {(a, 3a) | a in R}. Closureunder addition and scalar multiplication has not been established.

Most steps are shown to support thatmost of the propertieshold. One criticalstepismissing to establish the associative property, and justification for criticalstepsusing the laws of real numbers are not evident.