In linear algebra, a *transformation* between two vector spaces is a rule that assigns a vector in one space to a vector in the other space. Linear transformations are transformations that satisfy a particular property around addition and scalar multiplication. In this lesson, we will look at the basic notation of transformations, what is meant by “image” and “range”, as well as what makes a linear transformation different from other transformations.

Table of Contents

- The idea of a mapping
- Terminology: domain, codomain, image, and range for linear and other transformations
- What makes a transformation linear?

## The idea of a mapping

In mathematics, sometimes we use the word mapping to describe the same idea of a transformation. You are already familiar with mappings. For example, we could make up a rule that maps the real number to the real numbers. One such rule could be “multiply by 10”. Then 8 would be mapped to 80, and 3 would be mapped to 30, and so on.

Transformations in linear algebra are mappings as well, but they map vectors to vectors. This can be done with a rule described using a formula, or in the case of mappings between \(R^n\) and \(R^m\), maybe a matrix.

Remember that not all transformations are linear, but many that you study in linear algebra will be, and that yields a lot of useful theorems and problem solving techniques.

In this lesson, we will only consider transformations between the vector spaces\(R^n\) and \(R^m\) (for some m and n). See: Euclidean space.

## Terminology: domain, codomain, image, and range for linear and other transformations

When a transformation T “maps” vectors in \(R^n\) to vectors in \(R^m\), we write:

\(T: R^n \rightarrow R^m\)

We then call \(R^n\) the *domain* and \(R^m\) the *codomain*. That is, T maps vectors in the domain to vectors in the codomain.

## The image of a vector under a transformation and the range of a transformation

Suppose that our rule assigns the vector \(\vec{x}\) to the vector \(\vec{y}\). Then, just like in our algebra and calculus classes, we can write:

\(T(\vec{x}) = \vec{y}\)

We would then say that \(\vec{y}\) is the *image* of \(\vec{x}\) under T. The set of all images under T is called the *range* of T, denoted range(T).

Note that the range of T is a subset, or part of, the codomain. These two sets of vectors may or may not equal. This is something we will study more when we look at 1-1 and onto transformations.

### Example

Let \(T\left(\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}\right) = \begin{bmatrix} 2x \\ 5y \\ 3z \\ \end{bmatrix}\)

Find the image of \(\vec{v} = \begin{bmatrix} -1 \\ 1 \\ 4 \\ \end{bmatrix}\)

### Solution

This is just asking us to find \(T(\vec{v})\). Using the rule:

\(\begin{align}T\left(\begin{bmatrix} -1 \\ 1 \\ 4 \\ \end{bmatrix}\right) &= \begin{bmatrix} 2(-1) \\ 5(1) \\ 3(4) \\ \end{bmatrix} \\ &= \boxed{\begin{bmatrix} -2 \\ 5 \\ 12 \\ \end{bmatrix}}\end{align}\)

This shows us that T maps the vector \(\begin{bmatrix} -1 \\ 1 \\ 4 \\ \end{bmatrix}\) to the vector \(\begin{bmatrix} -2 \\ 5 \\ 12 \\ \end{bmatrix}\). Thus, we also know that \(\begin{bmatrix} -2 \\ 5 \\ 12 \\ \end{bmatrix}\) is in the range(T).

## What makes a transformation linear?

For a transformation to be linear, the following must hold for all vectors \(\vec{u}\) and \(\vec{v}\) in the domain and all scalars, \(c\) and \(d\).

\(T(c\vec{v}) = cT(\vec{v})\)

\(T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})\)

These two rules can be combined to a single rule that must hold.

\(T(c\vec{u} + d\vec{v}) = cT(\vec{u}) + dT(\vec{v})\)

This means that we can factor out scalars before applying T and break T up over addition or subtraction. Linear transformations make up a whole class of transformations that are studied in linear algebra. For a more in depth look at this rule, you can read the following article:

How to show a transformation is linear using the definition.

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