Question

Which of the following forms a linear transformation:

• A. \( T: \mathbb{R}^2 \to \mathbb{R}, T(x, y) = xy \)
• B. \( T: \mathbb{R}^2 \to \mathbb{R}^3, T(x, y) = (x+1, 2y, x+y) \)
• C. \( T: \mathbb{R}^3 \to \mathbb{R}^2, T(x, y, z) = (|x|, 0) \)
• D. \( T: \mathbb{R}^2 \to \mathbb{R}^2, T(x, y) = (x+y, x) \)

Hint:

A transformation \(T\) from \(\mathbb{R}^n\) to \(\mathbb{R}^m\) is linear if and only if each component of the output vector is a linear combination of the input variables (e.g., \(T(x,y) = (ax+by, cx+dy)\)). Any constant terms (like the '+1' in option B), non-linear terms (\(xy\)), or functions like absolute value (\(|x|\)) will make the transformation non-linear.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A transformation \( T: V \to W \) between two vector spaces V and W is linear if it satisfies two conditions for all vectors \(\mathbf{u}, \mathbf{v} \in V\) and any scalar \(c\): 1. Additivity: \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \) 2. Homogeneity: \( T(c\mathbf{u}) = cT(\mathbf{u}) \) A quick check is to see if \( T(\mathbf{0}) = \mathbf{0} \). If not, the transformation cannot be linear.

Step 2: Detailed Explanation:
Let's check each transformation:
A. \( T(x, y) = xy \): Let's check homogeneity. Let \(\mathbf{u}=(x,y)\) and \(c\) be a scalar. \( T(c\mathbf{u}) = T(cx, cy) = (cx)(cy) = c^2xy \). \( cT(\mathbf{u}) = c(xy) = cxy \). Since \( c^2xy \neq cxy \) in general, this is not a linear transformation.
B. \( T(x, y) = (x+1, 2y, x+y) \): Let's check the zero vector condition. \( T(0,0) = (0+1, 2(0), 0+0) = (1, 0, 0) \). Since \( T(\mathbf{0}) \neq \mathbf{0} \), this is not a linear transformation.
C. \( T(x, y, z) = (|x|, 0) \): Let's check homogeneity. Let \(\mathbf{u}=(x,y,z)\) and \(c=-1\). \( T(-\mathbf{u}) = T(-x, -y, -z) = (|-x|, 0) = (|x|, 0) \). \( -T(\mathbf{u}) = -( |x|, 0) = (-|x|, 0) \). Since \( (|x|, 0) \neq (-|x|, 0) \) for \(x \neq 0\), this is not a linear transformation.
D. \( T(x, y) = (x+y, x) \): 1. Additivity: Let \(\mathbf{u}=(x_1, y_1)\) and \(\mathbf{v}=(x_2, y_2)\). \( T(\mathbf{u}+\mathbf{v}) = T(x_1+x_2, y_1+y_2) = ((x_1+x_2)+(y_1+y_2), x_1+x_2) \). \( T(\mathbf{u})+T(\mathbf{v}) = (x_1+y_1, x_1) + (x_2+y_2, x_2) = (x_1+y_1+x_2+y_2, x_1+x_2) \). They are equal. Additivity holds. 2. Homogeneity: Let \(\mathbf{u}=(x,y)\) and \(c\) be a scalar. \( T(c\mathbf{u}) = T(cx, cy) = (cx+cy, cx) = c(x+y, x) \). \( cT(\mathbf{u}) = c(x+y, x) \). They are equal. Homogeneity holds. Since both conditions are met, this is a linear transformation.
Step 3: Final Answer:
The transformation \( T(x, y) = (x+y, x) \) is the only linear transformation among the options.