5: Matrix Transformations as Functions

Matrix Transformations

(Linear Transformation)

A matrix transformation (or linear transformation) is a function $T: \mathbb{R}^n \to \mathbb{R}^m$ defined by:

where $A$ is an $m \times n$ matrix.

Key Properties:

1. $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ (preserves addition)
2. $T(c\mathbf{v}) = cT(\mathbf{v})$ (preserves scalar multiplication)

These properties mean $T(c_1\mathbf{v}_1 + c_2\mathbf{v}_2) = c_1T(\mathbf{v}_1) + c_2T(\mathbf{v}_2)$ for any scalars and vectors.

Viewing Transformations as Functions

When we write $T(\mathbf{x}) = A\mathbf{x}$, we’re treating matrix multiplication as a function:

• Domain: $\mathbb{R}^n$ (all possible input vectors)
• Codomain: $\mathbb{R}^m$ (the space where outputs live)
• Range (Image): $\{\mathbf{y} \in \mathbb{R}^m \mid \mathbf{y} = A\mathbf{x} \text{ for some } \mathbf{x}\}$

The range is the set of all possible outputs,it’s the column space of $A$.

Injectivity (One-to-One)

(Injective / One-to-One)

A transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is injective (one-to-one) if:

Equivalently: Different inputs produce different outputs.

Equivalently: $T(\mathbf{x}) = \mathbf{0}$ has only the trivial solution $\mathbf{x} = \mathbf{0}$.

(Testing for Injectivity)

For $T(\mathbf{x}) = A\mathbf{x}$ where $A$ is $m \times n$:

Equivalently:

• The columns of $A$ are linearly independent
• $\text{Nul}(A) = \{\mathbf{0}\}$ (null space contains only zero vector)
• $\text{rref}(A)$ has a pivot in every column
• $\text{rank}(A) = n$ (number of columns)

Geometric Intuition: Injective transformations don’t “collapse” dimensions,they preserve distinctness.

Example: Testing Injectivity

Is $T(\mathbf{x}) = \begin{bmatrix} 1 & 2 \\ 3 & 6 \\ 0 & 1 \end{bmatrix}\mathbf{x}$ injective?

Solution: Check if $A\mathbf{x} = \mathbf{0}$ has only the trivial solution:

Pivot in every column → Yes, injective.

Surjectivity (Onto)

(Surjective / Onto)

A transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is surjective (onto) if:

Equivalently: For every $\mathbf{b} \in \mathbb{R}^m$, there exists some $\mathbf{x} \in \mathbb{R}^n$ such that $T(\mathbf{x}) = \mathbf{b}$.

Equivalently: Every vector in the codomain is “hit” by some input.

(Testing for Surjectivity)

For $T(\mathbf{x}) = A\mathbf{x}$ where $A$ is $m \times n$:

Equivalently:

• The columns of $A$ span $\mathbb{R}^m$
• $\text{Col}(A) = \mathbb{R}^m$
• $\text{rref}(A)$ has a pivot in every row
• $\text{rank}(A) = m$ (number of rows)

Geometric Intuition: Surjective transformations “cover” the entire codomain,no gaps.

Example: Testing Surjectivity

Is $T(\mathbf{x}) = \begin{bmatrix} 1 & 2 & 0 \\ 3 & 6 & 1 \end{bmatrix}\mathbf{x}$ surjective (as a map $\mathbb{R}^3 \to \mathbb{R}^2$)?

Solution: Check if rref has a pivot in every row:

Pivot in every row → Yes, surjective.

Bijectivity (One-to-One and Onto)

(Bijective)

A transformation is bijective if it is both injective and surjective.

Properties of Bijections:

1. Every output has exactly one input that produces it
2. The transformation is invertible (has an inverse function $T^{-1}$)
3. $T$ establishes a perfect “pairing” between domain and codomain

(When is a Matrix Transformation Bijective?)

For $T(\mathbf{x}) = A\mathbf{x}$:

Equivalently:

• $A$ is an $n \times n$ matrix with $\text{rank}(A) = n$
• $\text{rref}(A) = I_n$ (the identity matrix)
• $\det(A) \neq 0$
• Columns of $A$ form a basis for $\mathbb{R}^n$

Note: For non-square matrices:

• If $m < n$ (more columns than rows), $T$ cannot be injective
• If $m > n$ (more rows than columns), $T$ cannot be surjective

Example: Bijective Transformation

Is $T(\mathbf{x}) = \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}\mathbf{x}$ bijective ($\mathbb{R}^2 \to \mathbb{R}^2$)?

Solution: Check if $A$ is invertible:

Identity matrix → Yes, bijective (and $A$ has an inverse).

Summary Table

Connecting to Linear Systems

Given $T(\mathbf{x}) = A\mathbf{x}$:

1. Injective ⇔ $A\mathbf{x} = \mathbf{b}$ has at most one solution for any $\mathbf{b}$
2. Surjective ⇔ $A\mathbf{x} = \mathbf{b}$ has at least one solution for any $\mathbf{b}$
3. Bijective ⇔ $A\mathbf{x} = \mathbf{b}$ has exactly one solution for any $\mathbf{b}$

Visual Intuition

Injective but not Surjective

Imagine embedding a plane into 3D space,points don’t overlap (injective), but not all of 3D is covered (not surjective).

Surjective but not Injective

Imagine projecting 3D onto a plane,every point on the plane is hit (surjective), but many 3D points map to the same plane point (not injective).

Bijective

A rotation, reflection, or scaling in $\mathbb{R}^n$,every point has a unique pre-image and every point is reached.