Exercise C25:

Define the linear transformation \(T: \mathbb{C}^3 \rightarrow \mathbb{C}^2\)

\[ T \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right) = \left[ \begin{array}{c} 2x_1 - x_2 + 5x_3 \\ -4x_1 + 2x_2 - 10x_3 \end{array} \right] \]

Verify that \(T: \mathbb{C}^3 \rightarrow \mathbb{C}^2\) is a linear transformation.

Solution:

The transformation \(T\) takes a vector from \(\mathbb{C}^3\) (which means a vector with three complex components) and maps it to \(\mathbb{C}^2\) (a vector with two complex components) according to the given rule:

\[ T \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right) = \left[ \begin{array}{c} 2x_1 - x_2 + 5x_3 \\ -4x_1 + 2x_2 - 10x_3 \end{array} \right] \]

To implement this in R, I wrote a function. This function, named T, accepts a vector x that represents our input vector from \(\mathbb{C}^3\).

Here’s how:

T <- function(x) {

  first_component <- 2 * x[1] - x[2] + 5 * x[3]
  second_component <- -4 * x[1] + 2 * x[2] - 10 * x[3]
  result <- c(first_component, second_component)
  
  return(result)
}

To use this function, I just need to provide a vector x with three elements, like this:

x <- c(1, 2, 3) # random example
transformed_x <- T(x)
print(transformed_x) # the transformed vector
## [1]  15 -30

In this example, x is a vector with elements 1, 2, and 3, and the T function applies the transformation rules to x to produce a new vector.

Transformation Linearity:

To verify that a transformation \(T: \mathbb{C}^3 \rightarrow \mathbb{C}^2\) is a linear transformation, we need to check two main properties of linear transformations:

  1. Additivity: \(T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})\) for all vectors \(\mathbf{u}, \mathbf{v} \in \mathbb{C}^3\).
  2. Homogeneity of degree 1 (Scalar Multiplication): \(T(c\mathbf{u}) = cT(\mathbf{u})\) for all vectors \(\mathbf{u} \in \mathbb{C}^3\) and all scalars \(c \in \mathbb{C}\).

Additivity

Let \(\mathbf{u} = (u_1, u_2, u_3)\) and \(\mathbf{v} = (v_1, v_2, v_3)\) be vectors in \(\mathbb{C}^3\). Then,

\[ T(\mathbf{u} + \mathbf{v}) = T \left( \begin{array}{c} u_1 + v_1 \\ u_2 + v_2 \\ u_3 + v_3 \end{array} \right) \]

Applying \(T\) gives:

\[ = \left[ \begin{array}{c} 2(u_1 + v_1) - (u_2 + v_2) + 5(u_3 + v_3) \\ -4(u_1 + v_1) + 2(u_2 + v_2) - 10(u_3 + v_3) \end{array} \right] \]

Which simplifies to:

\[ = \left[ \begin{array}{c} (2u_1 - u_2 + 5u_3) + (2v_1 - v_2 + 5v_3) \\ (-4u_1 + 2u_2 - 10u_3) + (-4v_1 + 2v_2 - 10v_3) \end{array} \right] \]

\[ = T(\mathbf{u}) + T(\mathbf{v}) \]

Homogeneity

Let \(c\) be a scalar in \(\mathbb{C}\) and \(\mathbf{u} = (u_1, u_2, u_3)\). Then,

\[ T(c\mathbf{u}) = T \left( \begin{array}{c} cu_1 \\ cu_2 \\ cu_3 \end{array} \right) \]

Applying \(T\) gives:

\[ = \left[ \begin{array}{c} 2cu_1 - cu_2 + 5cu_3 \\ -4cu_1 + 2cu_2 - 10cu_3 \end{array} \right] \]

Factor out \(c\):

\[ = c \left[ \begin{array}{c} 2u_1 - u_2 + 5u_3 \\ -4u_1 + 2u_2 - 10u_3 \end{array} \right] \]

\[ = cT(\mathbf{u}) \]

Since \(T\) satisfies both the additivity and homogeneity properties, it is a linear transformation.