Description

General Instructions

Question 1

a) NO, the vector space R2 cannot have a basis B = {b1,b2} ∈ C𝟚. Note the difference between a subset of a set and the subspace of a vector space.
R2 ⊂ C2
R2 is NOT always a subspace of C2
Here R2 is not a subspace of C2 since the field over which this vector space is defined is C.
If W is a subspace of the vector space V, it must satisfy the following property
α1x + α2y ∈ W ∀x,y ∈ W, α1,α2 ∈ F
where F is the field over which the vector space is defined.
For this question, α1,α2 ∈ C. If x,y belong to R2, then α1x + α2y must also belong to R2. But, since α1 and α2 ∈ C there exists α1,α2 such that α1x + α2y ∈ C2. This proves that R2 is not the required subspace.
Had the field been R, R2 would have indeed been a subspace of C2. However, as the field is C, it is not a subspace since the rules of linear combination fail to hold as shown above.
Just because you are able to generate all the vectors whose entries are purely real, doesn’t mean that the generated vector belongs to the real space.
b) The strict condition on α1 and α2 is that they must be complex to generate every vector in R2. Let,
b
where bij ∈ C ∀i,j ∈ 1,2. Now, we have α1b11 + α2b21 ∈ R and α1b21 + α2b22 ∈ R. For this to hold, α1,α2 ∈ C.
Remember, even if you choose b1 = [1 0]T and b2 = [0 1]T, the elements are still from the field C. The elements can be written as 1 + 0j where j is the imaginary unit.
c) An example would be the basis B = {b1,b2}, such that b1 = [1 0]T and b2 = [0 1]T. Expression for computing the coordinate vector x = [x1 x2]T is
x1
[ ] = x1b1 + x2b2 x2
d) There exists NO complex basis of R2 (following from part (a)) that constitutes an orthogonal basis. But, in order to generate vectors of dimension 2, one needs to have an orthogonal complex basis.
Question 2

a) Using the definition of the norm under consideration,

We know that:
f(x)2 > 0 ∀x ≠ 0
⟹
⟹
⟹ i.e. |f(x)|2 > 0
Now, coming to the scalability property of norm:

Proving the triangle inequality: let f(x) and g(x) be two independent functions:
Firstly we have to prove the cauchy-schwarz inequality:
⟨f(x) − ag(x),f(x) − ag(x)⟩ = ⟨f(x),f(x)⟩ − a⟨f(x),g(x)⟩ − a⟨g(x),f(x)⟩ + a2⟨g(x),g(x)⟩
= ∥f(x)∥22 − a⟨f(x),g(x)⟩ − a⟨g(x),f(x)⟩ + a2∥g(x)∥22
⟨f(x),g(x)⟩
Let a
⟨f(x),g(x)⟩ ⟨f(x),g(x)⟩ ⟨f(x),g(x)⟩2
⟨f
∥g(x)∥2 ∥g(x)∥2
= ∥f(x)∥22 − ⟨f(x),g(x)⟩2
≥ 0
⟹
⟹ ∥f(x)∥2∥g(x)∥2 ≥ ⟨f(x),g(x)⟩
Now, we can prove the triangle inequality:

dx
dx dx
⟩
≤ ∥f(x)∥22 + ∥g(x)∥22 + 2∥f(x)∥2∥g(x)∥2
= (∥f(x)∥2 + ∥g(x)∥2)2 ⟹ ∥f(x) + g(x)∥2 ≤ ∥f(x)∥2 + ∥g(x)∥2
b) Consider the inner product of the two functions:
dx
dx
because sin((m + n)x) and sin((m − n)x) are odd functions and the limits of the integral are symmetric about 0, hence the integral evaluates to 0.
c) Let f(x) = 1 ∈ [a,b]. If there exists a g(x) ∈ [a,b] that is orthogonal to f(x) then
∫ (x) dx = 0
(x) dx = 0
1. The vectors must be linearly independent.
2. The vectors must span the entire subspace (R2 in our case).
2α1 + 4α 3α1 + 5α2 = 0
v = α1b
u1 + 5u2) 1 + (3α1 + 5α2)u2
Δ =
× 4)