Final answer:
To show that each of the given expressions are linear operators on R², we need to prove that they satisfy the properties of linearity. Geometrically, these linear transformations accomplish different operations such as reflection about the y-axis, reflection about the origin, swapping coordinates, scaling by a factor, and projection onto the y-axis.
Step-by-step explanation:
To show that each of the given expressions are linear operators on R², we need to prove that they satisfy the properties of linearity.
a. For L(x) = (-x₁,x₂)ᵀ, we can see that L(cu + v) = (-(cu)₁, (cu)₂)ᵀ + (-v₁, v₂)ᵀ = (-c(u₁) + (-v₁), c(u₂) + v₂)ᵀ = c(-u₁, u₂)ᵀ + (-v₁, v₂)ᵀ = cL(u) + L(v), where c is a scalar and u, v are vectors in R². This shows that the given expression is a linear operator. Geometrically, L(x) reflects the vector x about the y-axis.
b. For L(x) = -x, we have L(cu + v) = -(cu + v) = -(cu) - v = c(-u) + (-v) = cL(u) + L(v), which satisfies the properties of linearity. Geometrically, L(x) is a reflection of the vector x about the origin.
c. For L(x) = (x₂,x₁)ᵀ, we have L(cu + v) = ((cu)₂, (cu)₁)ᵀ + (v₂, v₁)ᵀ = (c(u₂) + v₂, c(u₁) + v₁)ᵀ = c(u₂, u₁)ᵀ + (v₂, v₁)ᵀ = cL(u) + L(v), which shows that the given expression is a linear operator. Geometrically, L(x) swaps the coordinates of the vector x.
d. For L(x) = 1/2x, we have L(cu + v) = 1/2(cu + v) = 1/2cu + 1/2v = c(1/2u) + 1/2v = cL(u) + L(v), satisfying the properties of linearity. Geometrically, L(x) scales the vector x by a factor of 1/2.
e. For L(x) = x₂e₂, we have L(cu + v) = (c(u₂)e₂) + (v₂)e₂ = c(u₂e₂) + v₂e₂ = cL(u) + L(v), which indicates linearity. Geometrically, L(x) projects x onto the y-axis.