# 5.5: One-to-One and Onto Transformations

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## Outcomes

1. Determine if a linear transformation is onto or one to one.

Let $$T: \mathbb{R}^n \mapsto \mathbb{R}^m$$ be a linear transformation. We define the range or image of $$T$$ as the set of vectors of $$\mathbb{R}^{m}$$ which are of the form $$T \left(\vec{x}\right)$$ (equivalently, $$A\vec{x}$$) for some $$\vec{x}\in \mathbb{R}^{n}$$. It is common to write $$T\mathbb{R}^{n}$$, $$T\left( \mathbb{R}^{n}\right)$$, or $$\mathrm{Im}\left( T\right)$$ to denote these vectors.

## Lemma $$\PageIndex{1}$$:Range of a Matrix Transformation

Let $$A$$ be an $$m\times n$$ matrix where $$A_{1},\cdots , A_{n}$$ denote the columns of $$A.$$ Then, for a vector $$\vec{x}=\left [ \begin{array}{c} x_{1} \\ \vdots \\ x_{n} \end{array} \right ]$$ in $$\mathbb{R}^n$$,

$A\vec{x}=\sum_{k=1}^{n}x_{k}A_{k}\nonumber$

Therefore, $$A \left( \mathbb{R}^n \right)$$ is the collection of all linear combinations of these products.

Proof

This follows from the definition of matrix multiplication.

Below is a video on determining if function mappings are one to one and/or onto.

This section is devoted to studying two important characterizations of linear transformations, called one to one and onto. We define them now.

## Definition $$\PageIndex{1}$$: One to One

Suppose $$\vec{x}_1$$ and $$\vec{x}_2$$ are vectors in $$\mathbb{R}^n$$. A linear transformation $$T: \mathbb{R}^n \mapsto \mathbb{R}^m$$ is called one to one (often written as $$1-1)$$ if whenever $$\vec{x}_1 \neq \vec{x}_2$$ it follows that : $T\left( \vec{x}_1 \right) \neq T \left(\vec{x}_2\right)\nonumber$

Equivalently, if $$T\left( \vec{x}_1 \right) =T\left( \vec{x}_2\right) ,$$ then $$\vec{x}_1 = \vec{x}_2$$. Thus, $$T$$ is one to one if it never takes two different vectors to the same vector.

Below is a video on one to one transformations.

The second important characterization is called onto.

## Definition $$\PageIndex{2}$$: Onto

Let $$T: \mathbb{R}^n \mapsto \mathbb{R}^m$$ be a linear transformation. Then $$T$$ is called onto if whenever $$\vec{x}_2 \in \mathbb{R}^{m}$$ there exists $$\vec{x}_1 \in \mathbb{R}^{n}$$ such that $$T\left( \vec{x}_1\right) = \vec{x}_2.$$

We often call a linear transformation which is one-to-one an injection. Similarly, a linear transformation which is onto is often called a surjection.

Below is a video on one to one and onto transformation definitions.

Below is a video on onto transformations.

The following proposition is an important result.

## Proposition $$\PageIndex{1}$$:One to One

Let $$T:\mathbb{R}^n \mapsto \mathbb{R}^m$$ be a linear transformation. Then $$T$$ is one to one if and only if $$T(\vec{x}) = \vec{0}$$ implies $$\vec{x}=\vec{0}$$.

Proof

We need to prove two things here. First, we will prove that if $$T$$ is one to one, then $$T(\vec{x}) = \vec{0}$$ implies that $$\vec{x}=\vec{0}$$. Second, we will show that if $$T(\vec{x})=\vec{0}$$ implies that $$\vec{x}=\vec{0}$$, then it follows that $$T$$ is one to one. Recall that a linear transformation has the property that $$T(\vec{0}) = \vec{0}$$.

Suppose first that $$T$$ is one to one and consider $$T(\vec{0})$$. $T(\vec{0})=T\left( \vec{0}+\vec{0}\right) =T(\vec{0})+T(\vec{0})\nonumber$ and so, adding the additive inverse of $$T(\vec{0})$$ to both sides, one sees that $$T(\vec{0})=\vec{0}$$. If $$T(\vec{x})=\vec{0}$$ it must be the case that $$\vec{x}=\vec{0}$$ because it was just shown that $$T(\vec{0})=\vec{0}$$ and $$T$$ is assumed to be one to one.

Now assume that if $$T(\vec{x})=\vec{0},$$ then it follows that $$\vec{x}=\vec{0}.$$ If $$T(\vec{v})=T(\vec{u}),$$ then $T(\vec{v})-T(\vec{u})=T\left( \vec{v}-\vec{u}\right) =\vec{0}\nonumber$ which shows that $$\vec{v}-\vec{u}=0$$. In other words, $$\vec{v}=\vec{u}$$, and $$T$$ is one to one.

Note that this proposition says that if $$A=\left [ \begin{array}{ccc} A_{1} & \cdots & A_{n} \end{array} \right ]$$ then $$A$$ is one to one if and only if whenever $0 = \sum_{k=1}^{n}c_{k}A_{k}\nonumber$ it follows that each scalar $$c_{k}=0$$.

Below is a video on comparing one to one and onto matrix transformations.

Below is a video on determining if transformation mappings are one to one and/or onto.

We will now take a look at an example of a one to one and onto linear transformation.

## Example $$\PageIndex{1}$$: A One to One and Onto Linear Transformation

Suppose $T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array} \right ] \left [ \begin{array}{r} x \\ y \end{array} \right ]\nonumber$ Then, $$T:\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$$ is a linear transformation. Is $$T$$ onto? Is it one to one?

Solution

Recall that because $$T$$ can be expressed as matrix multiplication, we know that $$T$$ is a linear transformation. We will start by looking at onto. So suppose $$\left [ \begin{array}{c} a \\ b \end{array} \right ] \in \mathbb{R}^{2}.$$ Does there exist $$\left [ \begin{array}{c} x \\ y \end{array} \right ] \in \mathbb{R}^2$$ such that $$T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ] ?$$ If so, then since $$\left [ \begin{array}{c} a \\ b \end{array} \right ]$$ is an arbitrary vector in $$\mathbb{R}^{2},$$ it will follow that $$T$$ is onto.

This question is familiar to you. It is asking whether there is a solution to the equation $\left [ \begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]\nonumber$ This is the same thing as asking for a solution to the following system of equations. $\begin{array}{c} x+y=a \\ x+2y=b \end{array}\nonumber$ Set up the augmented matrix and row reduce. $\left [ \begin{array}{rr|r} 1 & 1 & a \\ 1 & 2 & b \end{array} \right ] \rightarrow \left [ \begin{array}{rr|r} 1 & 0 & 2a-b \\ 0 & 1 & b-a \end{array} \right ] \label{ontomatrix}$ You can see from this point that the system has a solution. Therefore, we have shown that for any $$a, b$$, there is a $$\left [ \begin{array}{c} x \\ y \end{array} \right ]$$ such that $$T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]$$. Thus $$T$$ is onto.

Now we want to know if $$T$$ is one to one. By Proposition $$\PageIndex{1}$$ it is enough to show that $$A\vec{x}=0$$ implies $$\vec{x}=0$$. Consider the system $$A\vec{x}=0$$ given by: $\left [ \begin{array}{cc} 1 & 1 \\ 1 & 2\\ \end{array} \right ] \left [ \begin{array}{c} x\\ y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber$

This is the same as the system given by

$\begin{array}{c} x + y = 0 \\ x + 2y = 0 \end{array}\nonumber$

We need to show that the solution to this system is $$x = 0$$ and $$y = 0$$. By setting up the augmented matrix and row reducing, we end up with $\left [ \begin{array}{rr|r} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right ]\nonumber$

This tells us that $$x = 0$$ and $$y = 0$$. Returning to the original system, this says that if

$\left [ \begin{array}{cc} 1 & 1 \\ 1 & 2\\ \end{array} \right ] \left [ \begin{array}{c} x\\ y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber$

then $\left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber$

In other words, $$A\vec{x}=0$$ implies that $$\vec{x}=0$$. By Proposition $$\PageIndex{1}$$, $$A$$ is one to one, and so $$T$$ is also one to one.

We also could have seen that $$T$$ is one to one from our above solution for onto. By looking at the matrix given by $$\eqref{ontomatrix}$$, you can see that there is a unique solution given by $$x=2a-b$$ and $$y=b-a$$. Therefore, there is only one vector, specifically $$\left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} 2a-b\\ b-a \end{array} \right ]$$ such that $$T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]$$. Hence by Definition $$\PageIndex{1}$$, $$T$$ is one to one.

## Example $$\PageIndex{2}$$:An Onto Transformation

Let $$T: \mathbb{R}^4 \mapsto \mathbb{R}^2$$ be a linear transformation defined by $T \left [ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right ] = \left [ \begin{array}{c} a + d \\ b + c \end{array} \right ] \mbox{ for all } \left [ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right ] \in \mathbb{R}^4\nonumber$ Prove that $$T$$ is onto but not one to one.

Solution

You can prove that $$T$$ is in fact linear.

To show that $$T$$ is onto, let $$\left [ \begin{array}{c} x \\ y \end{array} \right ]$$ be an arbitrary vector in $$\mathbb{R}^2$$. Taking the vector $$\left [ \begin{array}{c} x \\ y \\ 0 \\ 0 \end{array} \right ] \in \mathbb{R}^4$$ we have $T \left [ \begin{array}{c} x \\ y \\ 0 \\ 0 \end{array} \right ] = \left [ \begin{array}{c} x + 0 \\ y + 0 \end{array} \right ] = \left [ \begin{array}{c} x \\ y \end{array} \right ]\nonumber$ This shows that $$T$$ is onto.

By Proposition $$\PageIndex{1}$$ $$T$$ is one to one if and only if $$T(\vec{x}) = \vec{0}$$ implies that $$\vec{x} = \vec{0}$$. Observe that $T \left [ \begin{array}{r} 1 \\ 0 \\ 0 \\ -1 \end{array} \right ] = \left [ \begin{array}{c} 1 + -1 \\ 0 + 0 \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber$ There exists a nonzero vector $$\vec{x}$$ in $$\mathbb{R}^4$$ such that $$T(\vec{x}) = \vec{0}$$. It follows that $$T$$ is not one to one.

Below is a video on determining if a linear transformation is one to one and/or onto.

Below is another video on determining if a linear transformation is one to one and/or onto.

The above examples demonstrate a method to determine if a linear transformation $$T$$ is one to one or onto. It turns out that the matrix $$A$$ of $$T$$ can provide this information.

## Theorem $$\PageIndex{1}$$:Matrix of a One to One or Onto Transformation

Let $$T: \mathbb{R}^n \mapsto \mathbb{R}^m$$ be a linear transformation induced by the $$m \times n$$ matrix $$A$$. Then $$T$$ is one to one if and only if the rank of $$A$$ is $$n$$. $$T$$ is onto if and only if the rank of $$A$$ is $$m$$.

Consider Example $$\PageIndex{2}$$. Above we showed that $$T$$ was onto but not one to one. We can now use this theorem to determine this fact about $$T$$.

## Example $$\PageIndex{3}$$:An Onto Transformation

Let $$T: \mathbb{R}^4 \mapsto \mathbb{R}^2$$ be a linear transformation defined by $T \left [ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right ] = \left [ \begin{array}{c} a + d \\ b + c \end{array} \right ] \mbox{ for all } \left [ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right ] \in \mathbb{R}^4\nonumber$ Prove that $$T$$ is onto but not one to one.

Solution

Using Theorem $$\PageIndex{1}$$ we can show that $$T$$ is onto but not one to one from the matrix of $$T$$. Recall that to find the matrix $$A$$ of $$T$$, we apply $$T$$ to each of the standard basis vectors $$\vec{e}_i$$ of $$\mathbb{R}^4$$. The result is the $$2 \times 4$$ matrix A given by $A = \left [ \begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{array} \right ]\nonumber$ Fortunately, this matrix is already in reduced row-echelon form. The rank of $$A$$ is $$2$$. Therefore by the above theorem $$T$$ is onto but not one to one.

Below is a video on classifying matrix multiplication:  onto and one to one.

Below is a video on determining if a 4x4 matrix represents a one to one and/or onto transformation.

Recall that if $$S$$ and $$T$$ are linear transformations, we can discuss their composite denoted $$S \circ T$$. The following examines what happens if both $$S$$ and $$T$$ are onto.

## Example $$\PageIndex{4}$$:Composite of Onto Transformations

Let $$T: \mathbb{R}^k \mapsto \mathbb{R}^n$$ and $$S: \mathbb{R}^n \mapsto \mathbb{R}^m$$ be linear transformations. If $$T$$ and $$S$$ are onto, then $$S \circ T$$ is onto.

Solution

Let $$\vec{z}\in \mathbb{R}^m$$. Since $$S$$ is onto, there exists a vector $$\vec{y}\in \mathbb{R}^n$$ such that $$S(\vec{y})=\vec{z}$$. Furthermore, since $$T$$ is onto, there exists a vector $$\vec{x}\in \mathbb{R}^k$$ such that $$T(\vec{x})=\vec{y}$$. Thus $\vec{z} = S(\vec{y}) = S(T(\vec{x})) = (ST)(\vec{x}),\nonumber$ showing that for each $$\vec{z}\in \mathbb{R}^m$$ there exists and $$\vec{x}\in \mathbb{R}^k$$ such that $$(ST)(\vec{x})=\vec{z}$$. Therefore, $$S \circ T$$ is onto.

The next example shows the same concept with regards to one-to-one transformations.

## Example $$\PageIndex{5}$$:Composite of One to One Transformations

Let $$T: \mathbb{R}^k \mapsto \mathbb{R}^n$$ and $$S: \mathbb{R}^n \mapsto \mathbb{R}^m$$ be linear transformations. Prove that if $$T$$ and $$S$$ are one to one, then $$S \circ T$$ is one-to-one.

Solution

To prove that $$S \circ T$$ is one to one, we need to show that if $$S(T (\vec{v})) = \vec{0}$$ it follows that $$\vec{v} = \vec{0}$$. Suppose that $$S(T (\vec{v})) = \vec{0}$$. Since $$S$$ is one to one, it follows that $$T (\vec{v}) = \vec{0}$$. Similarly, since $$T$$ is one to one, it follows that $$\vec{v} = \vec{0}$$. Hence $$S \circ T$$ is one to one.

This page titled 5.5: One-to-One and Onto Transformations is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) .