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Section 3.1 Linear Transformations (AT1)

Subsection 3.1.1 Class Activities

Definition 3.1.1.

A linear transformation (also called a linear map) is a map between vector spaces that preserves the vector space operations. More precisely, if \(V\) and \(W\) are vector spaces, a map \(T:V\rightarrow W\) is called a linear transformation if

  1. \(T(\vec{v}+\vec{w}) = T(\vec{v})+T(\vec{w})\) for any \(\vec{v},\vec{w} \in V\text{,}\) and

  2. \(T(c\vec{v}) = cT(\vec{v})\) for any \(c \in \IR,\) and \(\vec{v} \in V\text{.}\)

In other words, a map is linear when vector space operations can be applied before or after the transformation without affecting the result.

Definition 3.1.2.

Given a linear transformation \(T:V\to W\text{,}\) \(V\) is called the domain of \(T\) and \(W\) is called the co-domain of \(T\text{.}\)

Figure 23. A linear transformation with a domain of \(\IR^3\) and a co-domain of \(\IR^2\)

Example 3.1.3.

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-z \\ 3y \end{array}\right]. \end{equation*}

To show that \(T\) is a linear transformation, we must verify that \(T(\vec{v}+\vec{w}) = T(\vec{v})+T(\vec{w})\) by computing

\begin{equation*} T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] + \left[\begin{array}{c} u \\ v \\ w \end{array}\right] \right) = T\left( \left[\begin{array}{c} x+u \\ y+v \\ z+w \end{array}\right] \right) = \left[\begin{array}{c} (x+u)-(z+w) \\ 3(y+v) \end{array}\right] \end{equation*}


\begin{equation*} T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) + T\left( \left[\begin{array}{c} u \\ v \\ w \end{array}\right] \right) = \left[\begin{array}{c} x-z \\ 3y \end{array}\right] + \left[\begin{array}{c} u-w \\ 3v \end{array}\right]= \left[\begin{array}{c} (x+u)-(z+w) \\ 3(y+v) \end{array}\right]\text{,} \end{equation*}

and we must verify that \(T(c\vec{v}) = cT(\vec{v})\) by computing

\begin{equation*} T\left(c\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = T\left(\left[\begin{array}{c} cx \\ cy \\ cz \end{array}\right] \right) = \left[\begin{array}{c} cx-cz \\ 3cy \end{array}\right] \text{ and } cT\left(\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = c\left[\begin{array}{c} x-z \\ 3y \end{array}\right] = \left[\begin{array}{c} cx-cz \\ 3cy \end{array}\right]\text{.} \end{equation*}

Therefore \(T\) is a linear transformation.

Example 3.1.4.

Let \(S : \IR^2 \rightarrow \IR^4\) be given by

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

To show that \(S\) is not linear, we only need to find one counterexample.

\begin{equation*} S\left( \left[\begin{array}{c} 0 \\ 1 \end{array}\right] + \left[\begin{array}{c} 2 \\ 3 \end{array}\right] \right) = S\left( \left[\begin{array}{c} 2 \\ 4 \end{array}\right] \right) = \left[\begin{array}{c} 6 \\ 4 \\ 7 \\ 0 \end{array}\right] \end{equation*}
\begin{equation*} S\left( \left[\begin{array}{c} 0 \\ 1 \end{array}\right] \right) + S\left( \left[\begin{array}{c} 2 \\ 3\end{array}\right] \right) = \left[\begin{array}{c} 1 \\ 0 \\ 4 \\ 0 \end{array}\right] + \left[\begin{array}{c} 5 \\ 4 \\ 6 \\ -1 \end{array}\right] = \left[\begin{array}{c} 6 \\ 4 \\ 10 \\ -1 \end{array}\right] \end{equation*}

Since the resulting vectors are different, \(S\) is not a linear transformation.

Activity 3.1.6.

Let \(D:\P\to\P\) be the derivative map defined by \(D(f(x))=f'(x)\) for each polynomial \(f \in \P\text{.}\) We recall from calculus that

\begin{equation*} D(f(x)+g(x))=f'(x)+g'(x)\text{,} \end{equation*}
\begin{equation*} D(cf(x))=cf'(x)\text{.} \end{equation*}

Which of the following can we conclude from these calculus rules?

  1. \(\P\) is not a vector space

  2. \(D\) is a linear map

  3. \(D\) is not a linear map

Activity 3.1.7.

Let the polynomial maps \(S: \P_4 \rightarrow \P_3\) and \(T: \P_4 \rightarrow \P_3\) be defined by

\begin{equation*} S(f(x)) = 2f'(x)-f''(x) \hspace{3em} T(f(x)) = f'(x)+x^3\text{.} \end{equation*}

Compute \(S(x^4+x)\text{,}\) \(S(x^4)+S(x)\text{,}\) \(T(x^4+x)\text{,}\) and \(T(x^4)+T(x)\text{.}\) Based on these computations, can you conclude that either \(S\) or \(T\) is definitely not a linear transformation?

Observation 3.1.9.

Showing \(T:V\to W\) is not a linear transformation can be done by finding an example for any one of the following.

  • Show \(T(\vec z)\not=\vec z\) (where \(\vec z\) is the additive identity of \(V\) and \(W\)).

  • Find \(\vec v,\vec w\in V\) such that \(T(\vec v+\vec w)\not=T(\vec v)+T(\vec w)\text{.}\)

  • Find \(\vec v\in V\) and \(c\in \IR\) such that \(T(c\vec v)\not=cT(\vec v)\text{.}\)

Otherwise, \(T\) can be shown to be linear by proving the following in general.

  • For all \(\vec v,\vec w\in V\text{,}\) \(T(\vec v+\vec w)=T(\vec v)+T(\vec w)\text{.}\)

  • For all \(\vec v\in V\) and \(c\in \IR\text{,}\) \(T(c\vec v)=cT(\vec v)\text{.}\)

Note the similarities between this process and showing that a subset of a vector space is or is not a subspace.

Activity 3.1.10.

Continue to consider \(S: \P_4 \rightarrow \P_3\) defined by

\begin{equation*} S(f(x)) = 2f'(x)-f''(x)\text{.} \end{equation*}

Verify that

\begin{equation*} S(f(x)+g(x))=2f'(x)+2g'(x)-f''(x)-g''(x) \end{equation*}

is equal to \(S(f(x))+S(g(x))\) for all polynomials \(f,g \in \P_4\text{.}\)


Verify that \(S(cf(x))\) is equal to \(cS(f(x))\) for all real numbers \(c\) and polynomials \(f \in \P_4\text{.}\)


Is \(S\) linear?

Activity 3.1.11.

Let polynomial maps \(S: \P \rightarrow \P\) and \(T: \P \rightarrow \P\) be defined by

\begin{equation*} S(f(x)) = (f(x))^2 \hspace{3em} T(f(x)) = 3xf(x^2) \end{equation*}

Note that \(S(0)=0\) and \(T(0)=0\text{.}\) So instead, show that \(S(x+1)\not= S(x)+S(1)\) to verify that \(S\) is not linear.


Prove that \(T\) is linear by verifying that \(T(f(x)+g(x))=T(f(x))+T(g(x))\) and \(T(cf(x))=cT(f(x))\text{.}\)

Subsection 3.1.2 Videos

Figure 24. Video: Showing a transformation is linear
Figure 25. Video: Showing a transformation is not linear

Subsection 3.1.3 Slideshow

Slideshow of activities available at

Exercises 3.1.4 Exercises

Exercises available at

Subsection 3.1.5 Mathematical Writing Explorations

Exploration 3.1.12.

If \(V,W\) are vectors spaces, with associated zero vectors \(\vec{0}_V\) and \(\vec{0}_W\text{,}\) and \(T:V \rightarrow W\) is a linear transformation, does \(T(\vec{0}_V) = \vec{0}_W\text{?}\) Prove this is true, or find a counterexample.

Exploration 3.1.13.

Assume \(f: V \rightarrow W\) is a linear transformation between vector spaces. Let \(\vec{v} \in V\) with additive inverse \(\vec{v}^{-1}\text{.}\) Prove that \(f(\vec{v}^{-1}) = [f(\vec{v})]^{-1}\text{.}\)

Subsection 3.1.6 Sample Problem and Solution

Sample problem Example B.1.14.

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