# Linear Algebra for Team-Based Inquiry Learning

## 2022 Edition

Steven Clontz | Drew Lewis |
---|---|

University of South Alabama | University of South Alabama |

#### August 2, 2022

# Section 3.4: Injective and Surjective Linear Maps (AT4)

### Definition 3.4.1

Let \(T: V \rightarrow W\) be a linear transformation. \(T\) is called injective or one-to-one if \(T\) does not map two distinct vectors to the same place. More precisely, \(T\) is injective if \(T(\vec{v}) \neq T(\vec{w})\) whenever \(\vec{v} \neq \vec{w}\text{.}\)

### Activity 3.4.2 (~3 min)

Let \(T: \IR^3 \rightarrow \IR^2\) be given by

Yes, because \(T(\vec v)=T(\vec w)\) whenever \(\vec v=\vec w\text{.}\)

Yes, because \(T(\vec v)\not=T(\vec w)\) whenever \(\vec v\not=\vec w\text{.}\)

No, because \(T\left(\left[\begin{array}{c}0\\0\\1\end{array}\right]\right) \not= T\left(\left[\begin{array}{c}0\\0\\2\end{array}\right]\right) \text{.}\)

No, because \(T\left(\left[\begin{array}{c}0\\0\\1\end{array}\right]\right) = T\left(\left[\begin{array}{c}0\\0\\2\end{array}\right]\right) \text{.}\)

### Activity 3.4.3 (~2 min)

Let \(T: \IR^2 \rightarrow \IR^3\) be given by

Yes, because \(T(\vec v)=T(\vec w)\) whenever \(\vec v=\vec w\text{.}\)

Yes, because \(T(\vec v)\not=T(\vec w)\) whenever \(\vec v\not=\vec w\text{.}\)

No, because \(T\left(\left[\begin{array}{c}1\\2\end{array}\right]\right) \not= T\left(\left[\begin{array}{c}3\\4\end{array}\right]\right) \text{.}\)

No, because \(T\left(\left[\begin{array}{c}1\\2\end{array}\right]\right) = T\left(\left[\begin{array}{c}3\\4\end{array}\right]\right) \text{.}\)

### Definition 3.4.4

Let \(T: V \rightarrow W\) be a linear transformation. \(T\) is called surjective or onto if every element of \(W\) is mapped to by an element of \(V\text{.}\) More precisely, for every \(\vec{w} \in W\text{,}\) there is some \(\vec{v} \in V\) with \(T(\vec{v})=\vec{w}\text{.}\)

### Activity 3.4.5 (~3 min)

Let \(T: \IR^2 \rightarrow \IR^3\) be given by

Yes, because for every \(\vec w=\left[\begin{array}{c}x\\y\\z\end{array}\right]\in\IR^3\text{,}\) there exists \(\vec v=\left[\begin{array}{c}x\\y\end{array}\right]\in\IR^2\) such that \(T(\vec v)=\vec w\text{.}\)

No, because \(T\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)\) can never equal \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] \text{.}\)

No, because \(T\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)\) can never equal \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right] \text{.}\)

### Activity 3.4.6 (~2 min)

Let \(T: \IR^3 \rightarrow \IR^2\) be given by

Yes, because for every \(\vec w=\left[\begin{array}{c}x\\y\end{array}\right]\in\IR^2\text{,}\) there exists \(\vec v=\left[\begin{array}{c}x\\y\\42\end{array}\right]\in\IR^3\) such that \(T(\vec v)=\vec w\text{.}\)

Yes, because for every \(\vec w=\left[\begin{array}{c}x\\y\end{array}\right]\in\IR^2\text{,}\) there exists \(\vec v=\left[\begin{array}{c}0\\0\\z\end{array}\right]\in\IR^3\) such that \(T(\vec v)=\vec w\text{.}\)

No, because \(T\left(\left[\begin{array}{c}x\\y\\z\end{array}\right]\right)\) can never equal \(\left[\begin{array}{c} 3\\-2 \end{array}\right] \text{.}\)

### Observation 3.4.7

As we will see, it's no coincidence that the \(\RREF\) of the injective map's standard matrix

### Activity 3.4.8 (~5 min)

Let \(T: V \rightarrow W\) be a linear transformation where \(\ker T\) contains multiple vectors. What can you conclude?

\(T\) is injective

\(T\) is not injective

\(T\) is surjective

\(T\) is not surjective

### Fact 3.4.9

A linear transformation \(T\) is injective *if and only if* \(\ker T = \{\vec{0}\}\text{.}\) Put another way, an injective linear transformation may be recognized by its trivial kernel.

### Activity 3.4.10 (~5 min)

Let \(T: V \rightarrow \IR^5\) be a linear transformation where \(\Im T\) is spanned by four vectors. What can you conclude?

\(T\) is injective

\(T\) is not injective

\(T\) is surjective

\(T\) is not surjective

### Fact 3.4.11

A linear transformation \(T:V \rightarrow W\) is surjective *if and only if* \(\Im T = W\text{.}\) Put another way, a surjective linear transformation may be recognized by its identical codomain and image.

### Activity 3.4.12 (~15 min)

Let \(T: \IR^n \rightarrow \IR^m\) be a linear map with standard matrix \(A\text{.}\) Sort the following claims into two groups of *equivalent* statements: one group that means \(T\) is *injective*, and one group that means \(T\) is *surjective*.

The kernel of \(T\) is trivial, i.e. \(\ker T=\{\vec 0\}\text{.}\)

The columns of \(A\) span \(\IR^m\text{.}\)

The columns of \(A\) are linearly independent.

Every column of \(\RREF(A)\) has a pivot.

Every row of \(\RREF(A)\) has a pivot.

The image of \(T\) equals its codomain, i.e. \(\Im T=\IR^m\text{.}\)

The system of linear equations given by the augmented matrix \(\left[\begin{array}{c|c}A & \vec{b} \end{array}\right]\) has a solution for all \(\vec{b} \in \IR^m\text{.}\)

The system of linear equations given by the augmented matrix \(\left[\begin{array}{c|c} A & \vec{0} \end{array}\right]\) has exactly one solution.

### Observation 3.4.13

The easiest way to determine if the linear map with standard matrix \(A\) is injective is to see if \(\RREF(A)\) has a pivot in each column.

The easiest way to determine if the linear map with standard matrix \(A\) is surjective is to see if \(\RREF(A)\) has a pivot in each row.

### Activity 3.4.14 (~3 min)

What can you conclude about the linear map \(T:\IR^2\to\IR^3\) with standard matrix \(\left[\begin{array}{cc} a & b \\ c & d \\ e & f \end{array}\right]\text{?}\)

Its standard matrix has more columns than rows, so \(T\) is not injective.

Its standard matrix has more columns than rows, so \(T\) is injective.

Its standard matrix has more rows than columns, so \(T\) is not surjective.

Its standard matrix has more rows than columns, so \(T\) is surjective.

### Activity 3.4.15 (~2 min)

What can you conclude about the linear map \(T:\IR^3\to\IR^2\) with standard matrix \(\left[\begin{array}{ccc} a & b & c \\ d & e & f \end{array}\right]\text{?}\)

Its standard matrix has more columns than rows, so \(T\) is not injective.

Its standard matrix has more columns than rows, so \(T\) is injective.

Its standard matrix has more rows than columns, so \(T\) is not surjective.

Its standard matrix has more rows than columns, so \(T\) is surjective.

### Fact 3.4.16

The following are true for any linear map \(T:V\to W\text{:}\)

If \(\dim(V)>\dim(W)\text{,}\) then \(T\) is not injective.

If \(\dim(V)<\dim(W)\text{,}\) then \(T\) is not surjective.

Basically, a linear transformation cannot reduce dimension without collapsing vectors into each other, and a linear transformation cannot increase dimension from its domain to its image.

But dimension arguments *cannot* be used to prove a map *is* injective or surjective.

### Activity 3.4.17 (~5 min)

Suppose \(T: \IR^n \rightarrow \IR^4\) with standard matrix \(A=\left[\begin{array}{cccc} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ a_{31}&a_{32}&\cdots&a_{3n}\\ a_{41}&a_{42}&\cdots&a_{4n}\\ \end{array}\right]\) is both injective and surjective (we call such maps bijective).

*Part 1.*

How many pivot rows must \(\RREF A\) have?

### Activity 3.4.17 (~5 min)

Suppose \(T: \IR^n \rightarrow \IR^4\) with standard matrix \(A=\left[\begin{array}{cccc} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ a_{31}&a_{32}&\cdots&a_{3n}\\ a_{41}&a_{42}&\cdots&a_{4n}\\ \end{array}\right]\) is both injective and surjective (we call such maps bijective).

*Part 2.*

How many pivot columns must \(\RREF A\) have?

### Activity 3.4.17 (~5 min)

Suppose \(T: \IR^n \rightarrow \IR^4\) with standard matrix \(A=\left[\begin{array}{cccc} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ a_{31}&a_{32}&\cdots&a_{3n}\\ a_{41}&a_{42}&\cdots&a_{4n}\\ \end{array}\right]\) is both injective and surjective (we call such maps bijective).

*Part 3.*

What is \(\RREF A\text{?}\)

### Activity 3.4.18 (~5 min)

Let \(T: \IR^n \rightarrow \IR^n\) be a bijective linear map with standard matrix \(A\text{.}\) Label each of the following as true or false.

\(\RREF(A)\) is the identity matrix.

The columns of \(A\) form a basis for \(\IR^n\)

The system of linear equations given by the augmented matrix \(\left[\begin{array}{c|c} A & \vec{b} \end{array}\right]\) has exactly one solution for each \(\vec b \in \IR^n\text{.}\)

### Observation 3.4.19

The easiest way to show that the linear map with standard matrix \(A\) is bijective is to show that \(\RREF(A)\) is the identity matrix.

### Activity 3.4.20 (~3 min)

Let \(T: \IR^3 \rightarrow \IR^3\) be given by the standard matrix

\(T\) is neither injective nor surjective

\(T\) is injective but not surjective

\(T\) is surjective but not injective

\(T\) is bijective.

### Activity 3.4.21 (~3 min)

Let \(T: \IR^3 \rightarrow \IR^3\) be given by

\(T\) is neither injective nor surjective

\(T\) is injective but not surjective

\(T\) is surjective but not injective

\(T\) is bijective.

### Activity 3.4.22 (~3 min)

Let \(T: \IR^2 \rightarrow \IR^3\) be given by

\(T\) is neither injective nor surjective

\(T\) is injective but not surjective

\(T\) is surjective but not injective

\(T\) is bijective.

### Activity 3.4.23 (~3 min)

Let \(T: \IR^3 \rightarrow \IR^2\) be given by

\(T\) is neither injective nor surjective

\(T\) is injective but not surjective

\(T\) is surjective but not injective

\(T\) is bijective.