APSC 174: Linear Algebra for Engineers

Definition

The image (or range) of a linear transformation $T: V \rightarrow W$ is the set of all vectors in $W$ that are the result of applying $T$ to some vector in $V$ :

\text{Im}(T) = \text{range}(T) = \{T(v) : v \in V\}

This is the set of all possible outputs of the transformation.

Properties

Subspace: The image of a linear transformation $T: V \rightarrow W$ is a subspace of $W$ .
Dimension: The dimension of the image is called the rank of $T$ :

\text{rank}(T) = \dim(\text{Im}(T))

Surjectivity: A linear transformation $T$ is surjective (onto) if and only if $\text{Im}(T) = W$ , i.e., the image is the entire codomain.
- This means every vector in $W$ is the output of $T$ for some input
- By the Rank-Nullity Theorem, $T$ can only be surjective if $\dim(V) \geq \dim(W)$

Image of a Matrix

For a matrix $A \in \mathbb{R}^{m \times n}$ , the image of $A$ is the set of all possible vectors $\vec{b} \in \mathbb{R}^m$ such that $A\vec{x} = \vec{b}$ for some $\vec{x} \in \mathbb{R}^n$ :

\text{Im}(A) = \{\vec{b} \in \mathbb{R}^m : \vec{b} = A\vec{x} \text{ for some } \vec{x} \in \mathbb{R}^n\}

This corresponds to the image of the linear transformation $T_A: \mathbb{R}^n \rightarrow \mathbb{R}^m$ defined by $T_A(\vec{x}) = A\vec{x}$ .

Finding the Image of a Matrix

To find a basis for the image of a matrix $A$ :

Transform $A$ into row-reduced echelon form (RREF) using Gaussian elimination.
Identify the pivot columns in the original matrix.
The columns of the original matrix corresponding to these pivot positions form a spanning set for the image.
If you need a basis, take only the linearly independent columns from this spanning set.

Alternatively, the image of $A$ is the span of the columns of $A$ .

Example

Consider the matrix:

A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{bmatrix}

The RREF of $A$ is:

\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \end{bmatrix}

We see that only the first column is a pivot column.

Therefore, a basis for the image of $A$ is:

\left\{ \begin{bmatrix} 1 \\ 2 \end{bmatrix} \right\}

And the rank of $A$ is 1.

Relationship with the Null Space

The image and null space of a linear transformation are related through the Rank-Nullity Theorem:

\dim(V) = \text{rank}(T) + \text{nullity}(T)

This fundamental relationship tells us that:

The higher the rank (dimension of the image), the lower the nullity (dimension of the null space)
The sum of these dimensions always equals the dimension of the domain
A transformation cannot be both injective and surjective unless the domain and codomain have the same dimension

Applications

Systems of Linear Equations: The image of a coefficient matrix represents the set of right-hand sides for which the system has a solution.
Linear Transformations: Understanding the image helps identify what vectors can be reached by a transformation.
Change of Basis: When changing basis, the image gives the range of possible representations in the new basis.
Eigenspaces: The image of $(A - \lambda I)$ is complementary to the eigenspace for eigenvalue $\lambda$ .

Exercise

Find a basis for the image of the matrix:

B = \begin{bmatrix} 1 & 2 & 1 & 1 \\ 2 & 4 & 0 & 0 \\ 3 & 6 & 1 & 1 \end{bmatrix}

Determine the rank of $B$ and verify the Rank-Nullity Theorem.

Image of a Linear Transformation