Maths 2 Week 5

Linear Transformations and Their Properties

1. Introduction to Linear Transformations

Linear transformations are mappings between vector spaces that preserve vector addition and scalar multiplication. In this post, we explore transformations from ℝ² to ℝ², ℝ⁴ to ℝ⁴, and ℝ⁴ to ℝ³, examining their Nullspaces, Kernels, Range Spaces, and related concepts.

2. Nullspace and Kernel

Definition:

The Nullspace of a linear transformation T: V → W is the set of all vectors v ∈ V such that T(v) = 0. The Kernel of T is another term for this set.

Properties and Examples:

• Nullspace in ℝ² → ℝ²: For transformations like rotations or reflections, the Nullspace is often trivial (contains only 0).
• Nullspace in ℝ⁴ → ℝ⁴: Non-zero vectors may exist in the Nullspace for non-injective transformations, such as projections.
• Nullspace in ℝ⁴ → ℝ³: The Nullspace might have a higher dimension due to the transformation moving from higher to lower dimension.

3. Nullity

Definition:

Nullity is the dimension of the Nullspace of a transformation, indicating the "freedom" within the vector space that is nullified by the transformation.

Example Calculation of Nullity:

For a transformation from ℝ⁴ to ℝ³ with a Nullspace dimension of 1, the Nullity would be 1.

4. Range Space and Rank

Definition:

The Range (or Image) of a transformation T: V → W is the set of all possible values T(v) for v ∈ V. Rank is the dimension of the Range.

Examples:

• Range in ℝ² → ℝ²: Often spans ℝ² entirely if full-rank.
• Range in ℝ⁴ → ℝ⁴: May span all of ℝ⁴ if injective; otherwise, it has a lower dimension.
• Range in ℝ⁴ → ℝ³: Limited to 3 dimensions due to the codomain being ℝ³.

5. Injectivity and Surjectivity

• Injective (One-to-One): If every vector in the domain maps to a unique vector in the codomain, the transformation is injective. This implies a trivial Nullspace.
• Surjective (Onto): If the Range is the entire codomain, the transformation is surjective.

Examples:

• A transformation from ℝ² to ℝ² that rotates every vector by a fixed angle is injective and surjective.
• A projection from ℝ⁴ onto ℝ³ cannot be injective due to dimension mismatch.

6. Isomorphisms

Definition:

An Isomorphism is a bijective (both injective and surjective) linear transformation between two vector spaces, meaning there exists a perfect one-to-one mapping.

👉Examples of Linear Transformations

Example 1: Transformation from \( \mathbb{R}^2 \to \mathbb{R}^2 \)

Linear Transformation Definition

Let’s define a linear transformation \( T: \mathbb{R}^2 \to \mathbb{R}^2 \) given by the matrix:

\( T(x, y) = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \)

1. Nullspace and Kernel

To find the Nullspace, solve \( T(v) = 0 \) for \( v \in \mathbb{R}^2 \), which translates to:

\( \begin{cases} 2x + 3y = 0 \\ x + 2y = 0 \end{cases} \)

The solution is \( v = (0, 0) \), meaning the Nullspace only contains the zero vector, so the Kernel is trivial.

2. Nullity

The Nullity of this transformation is the dimension of the Nullspace, which in this case is 0.

3. Range and Rank

The Range (or Image) of \( T \) is the span of the column vectors:

\( \text{Span} \left\{ \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 3 \\ 2 \end{bmatrix} \right\} \)

Example 2

Linear Transformation Definition

Define a linear transformation \( T: \mathbb{R}^3 \to \mathbb{R}^3 \) given by the matrix:

\( T(x, y, z) = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \)

1. Nullspace and Kernel

To find the Nullspace, solve \( T(v) = 0 \), or:

\( \begin{cases} x + 2y + 3z = 0 \\ y + 4z = 0 \\ 0 = 0 \end{cases} \)

The solution is \( v = (-2, -3, 1) \), so the Nullspace is spanned by this vector, and the Kernel is non-trivial.

Example 3: Transformation from \( \mathbb{R}^4 \to \mathbb{R}^3 \)

Linear Transformation Definition

Define a linear transformation \( T: \mathbb{R}^4 \to \mathbb{R}^3 \) with the matrix:

\( T(w, x, y, z) = \begin{bmatrix} 1 & 0 & 2 & 3 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix} \)

1. Nullspace and Kernel

To find the Nullspace, solve \( T(v) = 0 \):

\( \begin{cases} w + 2y + 3z = 0 \\ x + y = 0 \\ z = 0 \end{cases} \)

We find the Nullspace as \( \text{Span} \{ (-2, -1, 1, 0) \} \).

👉Some of the questions discussed here by Lavanya Madam are good. Please use your student ID to access it:

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