## Section3.4Injective and Surjective Linear Maps (AT4)

### Subsection3.4.1Class Activities

#### Definition3.4.1.

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 $$\vec{v} \neq \vec{w}\text{.}$$

#### Activity3.4.2.

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

\begin{equation*} T\left(\left[\begin{array}{c}x \\ y\\z \end{array}\right] \right) = \left[\begin{array}{c} x \\ y \end{array}\right] \hspace{3em} \text{with standard matrix } \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] \end{equation*}

Is $$T$$ injective?

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

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

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

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

#### Activity3.4.3.

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

\begin{equation*} T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right) = \left[\begin{array}{c} x \\ y \\ 0 \end{array}\right] \hspace{3em} \text{with standard matrix } \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] \end{equation*}

Is $$T$$ injective?

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

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

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

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

#### Definition3.4.4.

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 $$T(\vec{v})=\vec{w}\text{.}$$

#### Activity3.4.5.

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

\begin{equation*} T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right) = \left[\begin{array}{c} x \\ y \\ 0 \end{array}\right] \hspace{3em} \text{with standard matrix } \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] \end{equation*}

Is $$T$$ surjective?

1. 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 $$T(\vec v)=\vec w\text{.}$$

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

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

#### Activity3.4.6.

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

\begin{equation*} T\left(\left[\begin{array}{c}x \\ y\\z \end{array}\right] \right) = \left[\begin{array}{c} x \\ y \end{array}\right] \hspace{3em} \text{with standard matrix } \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] \end{equation*}

Is $$T$$ surjective?

1. 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 $$T(\vec v)=\vec w\text{.}$$

2. 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 $$T(\vec v)=\vec w\text{.}$$

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

#### Observation3.4.7.

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

\begin{equation*} \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] \end{equation*}

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

\begin{equation*} \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] \end{equation*}

has a pivot in each row.

#### Activity3.4.8.

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

1. $$T$$ is injective

2. $$T$$ is not injective

3. $$T$$ is surjective

4. $$T$$ is not surjective

#### Activity3.4.10.

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

1. $$T$$ is injective

2. $$T$$ is not injective

3. $$T$$ is surjective

4. $$T$$ is not surjective

#### Activity3.4.12.

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.

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

2. The columns of $$A$$ span $$\IR^m\text{.}$$

3. The columns of $$A$$ are linearly independent.

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

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

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

7. 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 $$\vec{b} \in \IR^m\text{.}$$

8. 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.

#### Observation3.4.13.

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.

#### Activity3.4.14.

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

1. Its standard matrix has more columns than rows, so $$T$$ is not injective.

2. Its standard matrix has more columns than rows, so $$T$$ is injective.

3. Its standard matrix has more rows than columns, so $$T$$ is not surjective.

4. Its standard matrix has more rows than columns, so $$T$$ is surjective.

#### Activity3.4.15.

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

1. Its standard matrix has more columns than rows, so $$T$$ is not injective.

2. Its standard matrix has more columns than rows, so $$T$$ is injective.

3. Its standard matrix has more rows than columns, so $$T$$ is not surjective.

4. Its standard matrix has more rows than columns, so $$T$$ is surjective.

#### Activity3.4.17.

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 $$\RREF A\text{?}$$

#### Activity3.4.18.

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.

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

2. The columns of $$A$$ form a basis for $$\IR^n$$

3. 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 $$\vec b \in \IR^n\text{.}$$

#### Observation3.4.19.

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.

#### Activity3.4.20.

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

\begin{equation*} A=\left[\begin{array}{ccc} 2&1&-1 \\ 4&1&1 \\ 6&2&1\end{array}\right]. \end{equation*}

Which of the following must be true?

1. $$T$$ is neither injective nor surjective

2. $$T$$ is injective but not surjective

3. $$T$$ is surjective but not injective

4. $$T$$ is bijective.

rref([2,1,-1; 4,1,1; 6,2,1])


#### Activity3.4.21.

Let $$T: \IR^3 \rightarrow \IR^3$$ be given by

\begin{equation*} T\left(\left[\begin{array}{ccc} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} 2x+y-z \\ 4x+y+z \\ 6x+2y\end{array}\right]. \end{equation*}

Which of the following must be true?

1. $$T$$ is neither injective nor surjective

2. $$T$$ is injective but not surjective

3. $$T$$ is surjective but not injective

4. $$T$$ is bijective.

rref([2,1,-1; 4,1,1; 6,2,0])


#### Activity3.4.22.

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

\begin{equation*} T\left(\left[\begin{array}{c} x \\ y \end{array}\right] \right) = \left[\begin{array}{c} 2x+3y \\ x-y \\ x+3y\end{array}\right]. \end{equation*}

Which of the following must be true?

1. $$T$$ is neither injective nor surjective

2. $$T$$ is injective but not surjective

3. $$T$$ is surjective but not injective

4. $$T$$ is bijective.

rref([2,3;1,-1;1,3])


#### Activity3.4.23.

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

\begin{equation*} T\left(\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} 2x+y-z \\ 4x+y+z\end{array}\right]. \end{equation*}

Which of the following must be true?

1. $$T$$ is neither injective nor surjective

2. $$T$$ is injective but not surjective

3. $$T$$ is surjective but not injective

4. $$T$$ is bijective.

rref([2,1,-1;4,1,1])


### Subsection3.4.3Slideshow

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

### Subsection3.4.5Mathematical Writing Explorations

#### Exploration3.4.24.

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 $$\vec{v}\text{.}$$

#### Exploration3.4.25.

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

\begin{equation*} V^* = \{h:V \rightarrow \mathbb{R}: h \mbox{ is linear}\}. \end{equation*}

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?

### Subsection3.4.6Sample Problem and Solution

Sample problem Example B.1.17.

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