(a) Let [r s] ∈R^2 and T: R2→R2 be defined by T(⃗u)=⃗u+ [r s].
Show that if r /= 0 or s/=0 thenT is not a linear transformation of R^2
(b) i. Let Tr,s : R3 → R3 be the linear transformation defined by
Tr,s ([ x y 1]) = [x + r, y +s, 1])
Write down the matrix of Tr,s. (Note: it is a 3 × 3 matrix.)
ii. Show that Tr,s is an isomorphism of R3.
(a) To show that T is not a linear transformation if r ≠ 0 or s ≠ 0, we need to demonstrate that T violates at least one of the properties of a linear transformation.
Property 1: T(u + v) = T(u) + T(v)
To check this property, let's consider two vectors u = [a b] and v = [c d].
T(u + v) = T([a b] + [c d])
= T([a+c, b+d])
= [a+c, b+d] + [r s]
= [a+c+r, b+d+s]
T(u) + T(v) = T([a b]) + T([c d])
= [a, b] + [r s] + [c, d] + [r s]
= [a+r, b+s] + [c+r, d+s]
= [a+c+2r, b+d+2s]
If T were a linear transformation, T(u + v) would be equal to T(u) + T(v). However, as we can see from the calculations above, [a+c+r, b+d+s] is not equal to [a+c+2r, b+d+2s]. Hence, Property 1 is violated.
Therefore, T is not a linear transformation if r ≠ 0 or s ≠ 0.
(b)
i. To write down the matrix of Tr,s, we need to express the linear transformation in terms of a matrix.
Tr,s ([x y 1]) = [x + r, y + s, 1]
To represent this in matrix form, we write the coefficients of x, y, and 1 in a matrix:
[1 0 r]
[0 1 s]
[0 0 1]
Thus, the matrix of Tr,s is:
[1 0 r]
[0 1 s]
[0 0 1]
ii. To show that Tr,s is an isomorphism of R3, we need to demonstrate that it is a linear transformation and that it is bijective (one-to-one and onto).
1. Linear transformation:
To show that Tr,s is a linear transformation, we need to verify the properties of a linear transformation.
- Property 1: T(u + v) = T(u) + T(v)
- Property 2: T(ku) = kT(u), where k is a scalar
From the definition of Tr,s, we can see that it satisfies both properties:
- Property 1: Tr,s(u + v) = (u + v) + [r, s, 0] = u + v + [r, s, 0] = Tr,s(u) + Tr,s(v)
- Property 2: Tr,s(ku) = ku + [r, s, 0] = k(u + [r, s, 0]) = kTr,s(u)
Therefore, Tr,s is a linear transformation.
2. Bijective (one-to-one and onto):
To show that Tr,s is bijective, we need to demonstrate that it is both one-to-one and onto.
- One-to-one: If Tr,s is one-to-one, then for any two distinct vectors u and v in R3, Tr,s(u) is not equal to Tr,s(v). Since Tr,s only adds r and s to the first two components of the vector, it preserves the uniqueness of the vectors. Therefore, Tr,s is one-to-one.
- Onto: To show that Tr,s is onto, we need to demonstrate that for every vector w in R3, there exists a vector u in R3 such that Tr,s(u) = w. Since Tr,s adds r and s to the first two components of the vector, we can simply subtract r and s from w to obtain the original vector u. Hence, Tr,s is onto.
Therefore, Tr,s is both a linear transformation and an isomorphism of R3.