TBIL-LA Injective and Surjective Linear Maps (AT4)

Section 3.4 Injective and Surjective Linear Maps (AT4)

Learning Outcomes

Determine if a given linear map is injective and/or surjective.
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Subsection 3.4.1 Warm Up

Activity 3.4.1.

Consider the linear transformation \(T\colon\IR^4\to\IR^3\) that is represented by the standard matrix \(A=\left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right].\) Which of the following processes helps us compute a basis for \(\Im T\) and which helps us compute a basis for \(\ker T\text{?}\)

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Compute \(\RREF(A)\) and consider the set of columns of \(A\) that correspond to columns in \(\RREF(A)\) with pivots.
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Calculate a basis for the solution space to the homogenous system of equations for which \(A\) is the coefficient matrix.
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Subsection 3.4.2 Class Activities

Definition 3.4.2.

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 \(\vec{v} \neq \vec{w}\text{.}\)

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Figure 30. An injective transformation and a non-injective transformation
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Activity 3.4.3.

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?

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Yes, because \(T(\vec v)=T(\vec w)\) whenever \(\vec v=\vec w\text{.}\)
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Yes, because \(T(\vec v)\not=T(\vec w)\) whenever \(\vec v\not=\vec w\text{.}\)
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No, because \(T\left(\left[\begin{array}{c}0\\0\\1\end{array}\right]\right) \not= T\left(\left[\begin{array}{c}0\\0\\2\end{array}\right]\right) \text{.}\)
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No, because \(T\left(\left[\begin{array}{c}0\\0\\1\end{array}\right]\right) = T\left(\left[\begin{array}{c}0\\0\\2\end{array}\right]\right) \text{.}\)
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Activity 3.4.4.

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?

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Yes, because \(T(\vec v)=T(\vec w)\) whenever \(\vec v=\vec w\text{.}\)
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Yes, because \(T(\vec v)\not=T(\vec w)\) whenever \(\vec v\not=\vec w\text{.}\)
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No, because \(T\left(\left[\begin{array}{c}1\\2\end{array}\right]\right) \not= T\left(\left[\begin{array}{c}3\\4\end{array}\right]\right) \text{.}\)
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No, because \(T\left(\left[\begin{array}{c}1\\2\end{array}\right]\right) = T\left(\left[\begin{array}{c}3\\4\end{array}\right]\right) \text{.}\)
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Definition 3.4.5.

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

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Figure 31. A surjective transformation and a non-surjective transformation
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Activity 3.4.6.

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

Is \(T\) surjective?

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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 \(T(\vec v)=\vec w\text{.}\)
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No, because \(T\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)\) can never equal \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] \text{.}\)
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No, because \(T\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)\) can never equal \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right] \text{.}\)
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Activity 3.4.7.

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

Is \(T\) surjective?

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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 \(T(\vec v)=\vec w\text{.}\)
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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 \(T(\vec v)=\vec w\text{.}\)
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No, because \(T\left(\left[\begin{array}{c}x\\y\\z\end{array}\right]\right)\) can never equal \(\left[\begin{array}{c} 3\\-2 \end{array}\right] \text{.}\)
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Activity 3.4.8.

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

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\(T\) is injective
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\(T\) is not injective
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\(T\) is surjective
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\(T\) is not surjective
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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.

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Figure 32. A linear transformation with trivial kernel, which is therefore injective
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Activity 3.4.10.

Let \(T: V \rightarrow \IR^3\) be a linear transformation where \(\Im T\) may be spanned by only two vectors. What can you conclude?

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\(T\) is injective
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\(T\) is not injective
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\(T\) is surjective
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\(T\) is not surjective
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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.

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Figure 33. A linear transformation with identical codomain and image, which is therefore surjective; and a linear transformation with an image smaller than the codomain \(\IR^3\text{,}\) which is therefore not surjective.
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Definition 3.4.12.

A transformation that is both injective and surjective is said to be bijective.

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Activity 3.4.13.

Let \(T: \IR^n \rightarrow \IR^m\) be a linear map with standard matrix \(A\text{.}\) Determine whether each of the following statements means \(T\) is (A) injective, (B) surjective, or (C) bijective (both).

The kernel of \(T\) is trivial, i.e. \(\ker T=\{\vec 0\}\text{.}\)
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The image of \(T\) equals its codomain, i.e. \(\Im T=\IR^m\text{.}\)
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For every \(\vec w\in \IR^m\text{,}\) the set \(\{\vec v\in \IR^n|T(\vec v)=\vec w\}\) contains exactly one vector.
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Activity 3.4.14.

The columns of \(A\) span \(\IR^m\text{.}\)
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The columns of \(A\) form a basis for \(\IR^m\text{.}\)
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The columns of \(A\) are linearly independent.
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Activity 3.4.15.

\(\RREF(A)\) is the identity matrix.
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Every column of \(\RREF(A)\) has a pivot.
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Every row of \(\RREF(A)\) has a pivot.
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Activity 3.4.16.

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 \(\vec{b} \in \IR^m\text{.}\)
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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 all \(\vec{b} \in \IR^m\text{.}\)
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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.
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Observation 3.4.17.

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.

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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.

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Activity 3.4.18.

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

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Its standard matrix has more columns than rows, so \(T\) is not injective.
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Its standard matrix has more columns than rows, so \(T\) is injective.
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Its standard matrix has more rows than columns, so \(T\) is not surjective.
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Its standard matrix has more rows than columns, so \(T\) is surjective.
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Activity 3.4.19.

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

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Its standard matrix has more columns than rows, so \(T\) is not injective.
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Its standard matrix has more columns than rows, so \(T\) is injective.
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Its standard matrix has more rows than columns, so \(T\) is not surjective.
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Its standard matrix has more rows than columns, so \(T\) is surjective.
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Fact 3.4.20.

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

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

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Figure 34. A linear transformation whose domain has a larger dimension than its codomain, and is therefore not injective; and a linear transformation whose domain has a smaller dimension than its codomain, and is therefore not surjective.
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But dimension arguments cannot be used to prove a map is injective or surjective.

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Activity 3.4.21.

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 bijective.

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(a)

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

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

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(c)

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

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Activity 3.4.22.

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.

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\(\RREF(A)\) is the identity matrix.
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The columns of \(A\) form a basis for \(\IR^n\)
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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 \(\vec b \in \IR^n\text{.}\)
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Observation 3.4.23.

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.

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Activity 3.4.24.

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?

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\(T\) is neither injective nor surjective
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\(T\) is injective but not surjective
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\(T\) is surjective but not injective
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\(T\) is bijective.
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Activity 3.4.25.

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?

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\(T\) is neither injective nor surjective
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\(T\) is injective but not surjective
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\(T\) is surjective but not injective
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\(T\) is bijective.
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Activity 3.4.26.

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?

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\(T\) is neither injective nor surjective
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\(T\) is injective but not surjective
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\(T\) is surjective but not injective
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\(T\) is bijective.
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Activity 3.4.27.

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?

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\(T\) is neither injective nor surjective
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\(T\) is injective but not surjective
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\(T\) is surjective but not injective
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\(T\) is bijective.
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Subsection 3.4.3 Individual Practice

Activity 3.4.28.

Let \(T\colon\IR^n\to\IR^m\) be a linear transformation with standard matrix \(A\text{.}\) We reasoned during class that the following statements are logically equivalent:

The columns of \(A\) are linearly independent.
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\(\RREF(A)\) has a pivot in each column.
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The transformation \(T\) is injective.
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The system of equations given by \([A|\vec{0}]\) has a unique solution.
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While they are all logically equivalent, they are different statements that offer varied perspectives on our growing conceptual knowledge of linear algebra.

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(a)

If you are asked to decide if a transformation \(T\) is injective, which of the above statements do you think is the most useful?

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(b)

Can you think of some situations in which translating between these four statements might be useful to you?

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Activity 3.4.29.

Let \(T\colon\IR^n\to\IR^m\) be a linear transformation with standard matrix \(A\text{.}\) We reasoned during class that the following statements are logically equivalent:

The columns of \(A\) span all of \(\IR^m\text{.}\)
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\(\RREF(A)\) has a pivot in each row.
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The transformation \(T\) is surjective.
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The system of equations given by \([A|\vec{b}]\) is always consistent.
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While they are all logically equivalent, they are different statements that offer varied perspectives on our growing conceptual knowledge of linear algebra.

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(a)

If you are asked to decide if a transformation \(T\) is surjective, which of the above statements do you think is the most useful?

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(b)

Can you think of some situations in which translating between these four statements might be useful to you?

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Subsection 3.4.4 Videos

Figure 35. Video: The kernel and image of a linear transformation

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Figure 36. Video: Finding a basis of the image of a linear transformation

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Subsection 3.4.5 Exercises

Exercises available at https://library.tbil.org/2025/linear-algebra/exercises/#/bank/AT4/

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Subsection 3.4.6 Mathematical Writing Explorations

Exploration 3.4.30.

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.

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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 \(\vec{v}\text{.}\)

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Exploration 3.4.31.

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?

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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 \(j \neq i\text{.}\)

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

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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.

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

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Subsection 3.4.7 Sample Problem and Solution

Sample problem Example B.1.15.

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