One-to-One Linear Transformations

One-to-One Linear Transformations

Definition: A linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ that maps distinct points/vectors from $\mathbb{R}^n$ into distinct points/vectors in $\mathbb{R}^m$ is said to be a one-to-one transformation or an injective transformation. Thus for every vector $\vec{w} \in \mathbb{R}^m$, there exists exactly one vector $\vec{x} \in \mathbb{R}^n$ such that $T(\vec{x}) = \vec{w}$.

One prime example of a linear transformation that is one-to-one is the linear operator $T: \mathbb{R}^2 \to \mathbb{R}^2$ that takes any vector $\vec{x}$ and rotates it by a fixed angle $\theta$. We note that all vectors $\vec{x}$ will be rotated proportionally by $\theta$, and thus no image vector $\vec{w}$ can be obtained by more than one input vector $\vec{x}$:


On the otherhand, the linear operation $T: \mathbb{R}^2 \to \mathbb{R}^2$ that takes any vector $\vec{x}$ and orthogonally projects it onto the $x$-axis is not a one-to-one transformation. Suppose that we have two vectors $\vec{x_1}$ and $\vec{x_2}$ that have the same x-coordinate. Since the y-coordinate of both vectors become zero, both vectors will be mapped to the same image under the transformation:


The following theorem will give us an important relationship between one-to-one transformations and matrices.

Theorem 1: If $T_A: \mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation from multiplication by the $n \times n$ matrix $A$, then the following statements are equivalent: 1. $A$ is invertible $\Leftrightarrow$ 2. The range of $T_A$ is $\mathbb{R}^n$ $\Leftrightarrow$ 3. $T_A$ is a one-to-one transformation.
  • Proof of Theorem 1:
  • $1 \Leftrightarrow 2$: Let $A$ be an $n \times n$ invertible matrix. We know that the equation $w = Ax$ is consistent for every $n \times 1$ matrix $w$ as there is exactly one solution $x$ that satisfies the equation. The range of $T_A$ is thus all of $\mathbb{R}^n$, and this only happens when $A$ is invertible. We note that if $A$ is not invertible then for some $w$, $w = Ax$ is inconsistent so the range of $T_A$ can't be all of $\mathbb{R}^n$.
  • $2 \Leftrightarrow 3$: Suppose that the range of $T_A$ is all of $\mathbb{R}^n$. Thus every image $\vec{w}$ is associated to precisely one vector $\vec{x} \in \mathbb{R}^n$ as the domain is $\mathbb{R}^n$.
  • $3 \Leftrightarrow 1$: Suppose that $T_A$ is one-to-one. Then every vector $\vec{w}$ in the range is associated to exactly one vector $\vec{x}$. If $A$ is not invertible, then this cannot be true as $w = Ax$ cannot be consistent for every $w$ otherwise. $\blacksquare$

Example 1

Let the linear operator $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a rotational operator of any vector $\vec{x}$ by a fixed angle $\theta$. Prove this transformation is one-to-one.

We can prove this transformation is one-to-one by showing that the standard matrix for this transformation is invertible. Recall that the standard matrix for a rotation in 2-space by a fixed angle $\theta$ is $A = \begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta \end{bmatrix}$. The determinant of this matrix is thus:

\begin{align} \det (A) = \cos \theta \cdot \cos \theta - (-\sin \theta) \cdot \sin \theta \\ \det (A) = \cos ^2 \theta + \sin ^2 \theta \\ \det (A) = 1 \end{align}

Note that we made the substitution that $\cos ^2 \theta + \sin ^2 \theta = 1$ from the unit circle. Since $\det (A) ≠ 0$, it goes that $A$ is invertible and that this linear operator is hence one-to-one.

Example 2

Let the linear operator $T: \mathbb{R}^2 \to \mathbb{R}^2$ be defined by the following equations $w_1 = 2x - 5y$ and $w_2 = 4x + 3y$. Is this transformation one-to-one?

The standard matrix for this operation is $A = \begin{bmatrix} 2 & -5\\ 4 & 3 \end{bmatrix}$, and $\det(A) = 26 ≠ 0$, so this transformation is one-to-one.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License