Matrices: Surjective, Injective, and Bijective Linear Transformations

IceT Thaewanarumitkul

Transformation is another synonym for “function” but its terminology implies the heavy focus on “movement” of geometric components such as lines, curves, and regions. Today, we will explore the properties of linear transformation which is one of the central topics in linear algebra.

Linear transformations

A linear transformation, in the world of matrices, is a transformation satisfying two conditions.

The transformation preserve the properties of lines and curves.
The origin is fixed.

These two conditions seems complicated to imagine geometrically and we will talk about them numerically instead.

Consider the equation $b = Ax$ when $x$ is a vector in $\mathbb{R}^n$ , $A$ is an $m \times n$ matrix, and $b$ is the output vector in $\mathbb{R}^n$ represents a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ . We also can write the conditions of a linear transformation in terms of function.

Let $T \colon V \to W$ be a linear transformation from $V \in \mathbb{R}^n$ to $W \in \mathbb{R}^m$ . Here, $V$ is called the domain of $T$ and $W$ is called the codomain of $T$ . The range of $T$ , denoted by $\text{range}(T)$ or $\text{Im}(T)$ is the set of all possible outputs, which is a subset of $W$ , defined by

$\text{range}(T)$ = $\{ T(x) \mid x \in V \}$

Since $T$ is a linear transformation, the two conditions of linear transformation still hold and can be described as

$T(kx+ly)$ = $kT(x) + lT(y)$ for all $x, y \in V$ and all $k,l \in \mathbb R$ .
$T(0) = 0$ (The origin).

Looking back to the equation $b = Ax$ , we realize that $b$ is a linear combination of the columns of $A$ ; therefore, $\text{range}(A)$ is the same as the column space of $A$ .

Surjectivity of linear transformations

A linear transformation $T \colon V \to W$ is said to be surjective if $\text{range}(T) = W$ . Descriptively, it means all vectors in $W$ can be transformed from $V$ via $T$ .

Example. Let $T \colon \mathbb{R}^2 \to \mathbb{R}^2$ be given by $T(\begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix})$ = $\begin{bmatrix} x_1 - x_2 \\ -2x_1 + 2x_2 \\ \end{bmatrix}$ .

We see that every vector in $\text{range}(T)$ is of the form $\begin{bmatrix} k \\ -2k \\ \end{bmatrix}$ , so $\text{range}(T)$ is $\text{span}\{\begin{bmatrix} 1 \\ -2 \\ \end{bmatrix}\}$ . Thus, $T$ is not surjective because $\text{range}(T)$ is not the same as $\mathbb{R}^2$ .

Injectivity of linear transformations

A linear transformation $T \colon V \to W$ is considered injective if for all distinct $x,y \in V$ , we have $T(x) \neq T(y)$ . Basically, this means the different inputs give the different outputs.

Example. Let $T \colon \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be given by $T(\begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix})$ = $\begin{bmatrix} x_1 - x_2 \\ -x_1 + 2x_2 \\ \end{bmatrix}$ . We can prove that $T$ is injective.

Proof. Let $u = \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix}$ and $v = \begin{bmatrix} x_3 \\ x_4 \\ \end{bmatrix}$ . Suppose that $T(u) = T(v)$ . Then $\begin{bmatrix} x_1 - x_2 \\ -x_1 + 2x_2 \\ \end{bmatrix} = \begin{bmatrix} x_3 - x_4 \\ -x_3 + 2x_4 \\ \end{bmatrix}$ . Thus, $x_1 - x_2 = x_3 - x_4$ and $-x_1 + 2x_2 = -x_3 + 2x_4$ . It follows that $x_1 - x_2 + (-x_1 + 2x_2) = x_3 - x_4 + (-x_3 + 2x_4)$ . Then $x_2 = x_4$ . Since $x_2 = x_4$ , we also have $x_1 = x_3$ . Therefore, $u = v$ . Hence, $T$ is injective.

Bijectivity of linear transformations

A linear transformation $T \colon V \to W$ is bijective if $T$ is both injective and surjective.

Example. Let $T \colon \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be given by $T(\begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix})$ = $\begin{bmatrix} x_1 - x_2 \\ -x_1 +2x_2 \\ \end{bmatrix}$ . We can prove that $T$ is bijective.

Proof. Let $z$ = $\begin{bmatrix} m \\ n \\ \end{bmatrix} \in \mathbb{R}^2$ be arbitrary. There exists $y = \begin{bmatrix} 2m + n \\ m + n \\ \end{bmatrix} \in \mathbb{R}^2$ such that

$T(y) = \begin{bmatrix} (2m+n) - (m+n) \\ -(2m+n) +2(m+n) \\ \end{bmatrix} = \begin{bmatrix} m \\ n \\ \end{bmatrix}= z$ .

Thus, any vector $z \in \mathbb{R}^2$ can be transformed from a vector $y \in \mathbb{R}^2$ . Thus, T is surjective. Since we already proved that this linear transformation $T$ is injective, then $T$ is bijective.

Bijectivity of composite linear transformations

In a composite transformation, we have linear transformation $S \colon V \rightarrow W$ and $T \colon W \rightarrow Z$ defined by $b$ = $[S]x$ and $c$ = $[T]x$ respectively when $[S]$ and $[T]$ are matrices of linear transformation.

We see that $T \circ S$ is a linear transformation from $V$ to $Z$ defined by $d$ = $[T][S]x$ . If $T \circ S$ is bijective, from the lemma we proved in Homework 4, $T$ will be surjective, and $S$ will be injective. However, the converse of the statement is not necessarily true. If $T$ is surjective and $S$ is injective, $T \circ S$ may not be bijective.

Example. Let $S \colon \mathbb{R}^2 \to\mathbb{R}^3$ be a linear transformation defined by $S(x) = \begin{bmatrix} 1 &2 \\ 1 & 1 \\ 1 & 0 \\ \end{bmatrix} x$ . Let $T \colon \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation given by $T(x) = \begin{bmatrix} -1 & 2 & 0 \\ -1 & 2 & 1 \\ \end{bmatrix} x$ . We can show that $S$ and $T$ are injective and surjective, respectively; however, $T \circ S$ is not bijective.

Let $x_1, x_2 \in \mathbb{R}^2$ such that $S(x_1) = S(x_2)$ . Because $S$ is a linear transformation, $S(x_1 - x_2) = S(x_1) - S(x_2) = 0$ . We see that the columns of $\begin{bmatrix} 1 & 2 \\ 1 & 1 \\ 1 & 0 \\ \end{bmatrix}$ are linearly independent. Then $x_1-x_2=0$ . Thus, $x_1 = x_2$ . Therefore, $S$ is injective.

Let $y =\begin{bmatrix} y_1 \\ y_2 \\ \end{bmatrix} \in \mathbb{R}^2$ be arbitrary. Then there exists $x = \begin{bmatrix} y_1 \\ 0 \\ y_1 - y_2 \\ \end{bmatrix} \in \mathbb{R}^3$ such that $T(x) = y$ . Thus, $T$ is surjective.

Now consider $[T][S] = \begin{bmatrix} -1 & 2 & 0 \\ -1 & 2 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 &2 \\ 1 & 1 \\ 1 & 0 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 0 \\ \end{bmatrix}$ . We see that every vector in $\text{range}(T \circ S)$ is of the form $\begin{bmatrix} k \\ 2k \\ \end{bmatrix}$ , so $\text{range}(T)$ is $\text{span}\{\begin{bmatrix} 1 \\ -2 \\ \end{bmatrix}\}$ . Thus, $T$ is not surjective because $\text{range}(T)$ is not the same as $\mathbb{R}^2$ .

Citation

Taboga, Marco (2021). “Surjective, injective and bijective linear maps”, Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/surjective-injective-bijective-linear-maps.
Cheung, Kevin (2017). “Linear Transformations”, Carleton University.

One Comment


June 3, 2023
uno

Hey IceT! Thanks for the really insightful post on how linear transformations relate in the domain of linear algebra! I was really able to understand how matrices and vectors work alongside linear transformations after reading your post. While this is one of the topics that I am not the best at, your clear explanations were easy to follow and was very interesting. A followup question I have regarding your post would be to ask if there’s any real world applications for this concept – physically or intangibly?