Question

show that T is a linear transformation by finding a matrix that implements the mapping. Note that X₁, X₂, ... are not vectors but are entries in vectors. 17. T(X₁, X₂, X₃, x₄) = (0, X₁ + X₂, X₂ + x₃, x₃ + x₄)

Answer 1

To show that T is a linear transformation, we need to show that it satisfies the properties of linearity: additivity and scalar multiplication. By performing the necessary calculations and comparisons, we can conclude that T(X₁, X₂, X₃, x₄) = (0, X₁ + X₂, X₂ + X₃, x₃ + x₄) is indeed a linear transformation.

Step-by-step explanation:

To show that T is a linear transformation, we need to show that it satisfies the two properties of linearity: additivity and scalar multiplication.

1. Additivity: T(u + v) = T(u) + T(v)

T(X₁, X₂, X₃, x₄) + T(Y₁, Y₂, Y₃, y₄) = (0, X₁ + X₂, X₂ + X₃, x₃ + x₄) + (0, Y₁ + Y₂, Y₂ + Y₃, y₃ + y₄) = (0, X₁ + X₂ + Y₁ + Y₂, X₂ + X₃ + Y₂ + Y₃, x₃ + x₄ + y₃ + y₄)

T(X₁ + Y₁, X₂ + Y₂, X₃ + Y₃, x₄ + y₄) = (0, X₁ + Y₁ + X₂ + Y₂, X₂ + Y₂ + X₃ + Y₃, x₃ + y₃ + x₄ + y₄)

Since (0, X₁ + X₂ + Y₁ + Y₂, X₂ + X₃ + Y₂ + Y₃, x₃ + x₄ + y₃ + y₄) = (0, X₁ + Y₁ + X₂ + Y₂, X₂ + Y₂ + X₃ + Y₃, x₃ + y₃ + x₄ + y₄), we can conclude that the transformation T is additive.

2. Scalar Multiplication: T(cu) = cT(u)

T(cX₁, cX₂, cX₃, cx₄) = (0, cX₁ + cX₂, cX₂ + cX₃, cx₃ + cx₄) = c(0, X₁ + X₂, X₂ + X₃, x₃ + x₄) = cT(X₁, X₂, X₃, x₄)

Therefore, T(X₁, X₂, X₃, x₄) = (0, X₁ + X₂, X₂ + X₃, x₃ + x₄) is a linear transformation.