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

### Learning Outcomes

Determine if a given linear map is injective and/or surjective.

### Subsection 3.4.1 Class Activities

#### 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.

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

Is \(T\) injective?

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.

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

Is \(T\) injective?

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.

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

Is \(T\) surjective?

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.

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

Is \(T\) surjective?

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

has all pivot columns. Similarly, the \(\RREF\) of the surjective map's standard matrix

has a pivot in each row.

#### Activity 3.4.8.

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.

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.

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.

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.

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.

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).

##### (a)

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

##### (b)

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

##### (c)

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

#### Activity 3.4.18.

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.

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

Which of the following must be true?

\(T\) is neither injective nor surjective

\(T\) is injective but not surjective

\(T\) is surjective but not injective

\(T\) is bijective.

#### Activity 3.4.21.

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

Which of the following must be true?

\(T\) is neither injective nor surjective

\(T\) is injective but not surjective

\(T\) is surjective but not injective

\(T\) is bijective.

#### Activity 3.4.22.

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

Which of the following must be true?

\(T\) is neither injective nor surjective

\(T\) is injective but not surjective

\(T\) is surjective but not injective

\(T\) is bijective.

#### Activity 3.4.23.

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

Which of the following must be true?

\(T\) is neither injective nor surjective

\(T\) is injective but not surjective

\(T\) is surjective but not injective

\(T\) is bijective.

### Subsection 3.4.2 Videos

### Subsection 3.4.3 Slideshow

Slideshow of activities available at `https://teambasedinquirylearning.github.io/linear-algebra/2022/AT4.slides.html`

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### Exercises 3.4.4 Exercises

Exercises available at `https://teambasedinquirylearning.github.io/linear-algebra/2022/exercises/#/bank/AT4/`

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### Subsection 3.4.5 Mathematical Writing Explorations

#### Exploration 3.4.24.

Suppose that \(f:V \rightarrow W\) is a linear transformation between two vector spaces \(V\) and \(W\text{.}\) State carefully what conditions \(f\) must satisfy. Let \(\vec{0_V}\) and \(\vec{0_W}\) be the zero vectors in \(V\) and \(W\) respectively.Prove that \(f\) is one-to-one if and only if \(f(\vec{0_V}) = \vec{0_W}\text{,}\) and that \(\vec{0_V}\) is the unique element of \(V\) which is mapped to \(\vec{0_W}\text{.}\) Remember that this needs to be done in both directions. First prove the if and only if statement, and then show the uniqueness.

Do not use subtraction in your proof. The only vector space operation we have is addition, and a structure preserving function only preserves addition. If you are writing \(\vec{v} - \vec{v} = \vec{0}_V\text{,}\) what you really mean is that \(\vec{v} \oplus \vec{v}^{-1} = \vec{0}_V\text{,}\) where \(\vec{v}^{-1}\) is the additive inverse of \(\vec{v}\text{.}\)

#### Exploration 3.4.25.

Start with an \(n\)-dimensional vector space \(V\text{.}\) We can define the dual of \(V\text{,}\) denoted \(V^*\text{,}\) by

Prove that \(V\) is isomorphic to\(V^*\text{.}\) Here are some things to think about as you work through this.

Start by assuming you have a basis for \(V\text{.}\) How many basis vectors should you have?

For each basis vector in \(V\text{,}\) define a function that returns 1 if it's given that basis vector, and returns 0 if it's given any other basis vector. For example, if \(\vec{b_i}\) and \(\vec{b_j}\) are each members of the basis for \(V\text{,}\) and you'll need a function \(f_i:V \rightarrow \{0,1\}\text{,}\) where \(f_i(b_i) = 1\) and \(f_i(b_j)= 0\) for all \(j \neq i\text{.}\)

How many of these functions will you need? Show that each of them is in \(V^*\text{.}\)

Show that the functions you found in the last part are a basis for \(V^*\text{?}\) To do this, take an arbitrary function \(h \in V^*\) and some vector \(\vec{v} \in V\text{.}\) Write \(\vec{v}\) in terms of the basis you chose earlier. How can you write \(h(\vec{v})\text{,}\) with respect to that basis? Pay attention to the fact that all functions in \(V^*\) are linear.

Now that you've got a basis for \(V\) and a basis for \(V^*\text{,}\) can you find an isomorphism?

### Subsection 3.4.6 Sample Problem and Solution

Sample problem Example B.1.17.