Learning Outcomes
- Determine if a map between Euclidean vector spaces is linear or not.
Section 3.1 Linear Transformations (AT1)
Subsection 3.1.1 Warm Up
Activity 3.1.1.
(a)
What is our definition for a set \(S\) of vectors to be linearly independent?
(b)
What specific calculation would you perform to test is a set \(S\) of Euclidean vectors is linearly independent?
Activity 3.1.2.
(a)
What is our definition for a set \(S\) of vectors in \(\IR^n\) to span \(\IR^n\) ?
(b)
What specific calculation would you perform to test is a set \(S\) of Euclidean vectors spans all of \(\IR^n\) ?
Subsection 3.1.2 Class Activities
Definition 3.1.3.
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
- \(T(\vec{v}+\vec{w}) = T(\vec{v})+T(\vec{w})\) for any \(\vec{v},\vec{w} \in V\text{,}\) and
- \(T(c\vec{v}) = cT(\vec{v})\) 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 3.1.4.
Given a linear transformation \(T:V\to W\text{,}\) \(V\) is called the domain of \(T\) and \(W\) is called the co-domain of \(T\text{.}\)
Observation 3.1.5.
One example of a linear transformation \(\IR^3\to\IR^2\) is the projection of three-dimesional data onto a two-dimensional screen, as is necessary for computer animiation in film or video games.
Activity 3.1.6.
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-z \\ 3y \end{array}\right].
\end{equation*}
(a)
Compute the result of adding vectors before a \(T\) transformation:
\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)
\end{equation*}
- \(\displaystyle \left[\begin{array}{c} x-u+z-w \\ 3y-3v \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} x+u-z-w \\ 3y+3v \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} x+u \\ 3y+3v \\ z+w \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} x-u \\ 3y-3v \\ z-w \end{array}\right]\)
(b)
Compute the result of adding vectors after a \(T\) transformation:
\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]
\end{equation*}
- \(\displaystyle \left[\begin{array}{c} x-u+z-w \\ 3y-3v \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} x+u-z-w \\ 3y+3v \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} x+u \\ 3y+3v \\ z+w \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} x-u \\ 3y-3v \\ z-w \end{array}\right]\)
(c)
Is \(T\) a linear transformation?
- Yes.
- No.
- More work is necessary to know.
(d)
Compute the result of scalar multiplcation before a \(T\) transformation:
\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)
\end{equation*}
- \(\displaystyle \left[\begin{array}{c} cx-cz\\ 3cy \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} cx+cz \\ -3cy \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} x+c \\ 3y+c \\ z+c \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} x-c \\ 3y-c \\ z-c \end{array}\right]\)
(e)
Compute the result of scalar multiplcation after a \(T\) transformation:
\begin{equation*}
cT\left(\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)
=
c\left[\begin{array}{c} x-z \\ 3y \end{array}\right]
\end{equation*}
- \(\displaystyle \left[\begin{array}{c} cx-cz\\ 3cy \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} cx+cz \\ -3cy \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} x+c \\ 3y+c \\ z+c \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} x-c \\ 3y-c \\ z-c \end{array}\right]\)
(f)
Is \(T\) a linear transformation?
- Yes.
- No.
- More work is necessary to know.
Activity 3.1.7.
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*}
(a)
Compute
\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)
\end{equation*}
- \(\displaystyle \left[\begin{array}{c} 6 \\ 4 \\ 7 \\ 0 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} -3 \\ 0 \\ 1 \\ 5 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} -3 \\ -1 \\ 7 \\ 5 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} 6 \\ 4 \\ 10 \\ -1 \end{array}\right]\)
(b)
Compute
\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} 0+1 \\ 0^2 \\ 1+3 \\ 1-2^0 \end{array}\right] +
\left[\begin{array}{c} 2+3 \\ 2^2 \\ 3+3 \\ 3-2^2 \end{array}\right]
\end{equation*}
- \(\displaystyle \left[\begin{array}{c} 6 \\ 4 \\ 7 \\ 0 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} -3 \\ 0 \\ 1 \\ 5 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} -3 \\ -1 \\ 7 \\ 5 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} 6 \\ 4 \\ 10 \\ -1 \end{array}\right]\)
(c)
Is \(T\) a linear transformation?
- Yes.
- No.
- More work is necessary to know.
Activity 3.1.8.
Fill in the \(\unknown\)s, assuming \(T:\mathbb R^3\to\mathbb R^3\) is linear:
\begin{equation*}
T\left(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\right)
= T\left(\unknown \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\right)
= \unknown T\left(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\right)
= \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \end{array}\right]
\end{equation*}
Remark 3.1.9.
In summary, any one of the following is enough to prove that \(T:V\to W\) is not a linear transformation:
- Find specific values for \(\vec v,\vec w\in V\) such that \(T(\vec v+\vec w)\not=T(\vec v)+T(\vec w)\text{.}\)
- Find specific values for \(\vec v\in V\) and \(c\in \IR\) such that \(T(c\vec v)\not=cT(\vec v)\text{.}\)
- Show \(T(\vec 0)\not=\vec 0\text{.}\)
If you cannot do any of these, then \(T\) can be proven to be a linear transformation by doing both of the following:
- For all \(\vec v,\vec w\in V\) (not just specific values), \(T(\vec v+\vec w)=T(\vec v)+T(\vec w)\text{.}\)
- For all \(\vec v\in V\) and \(c\in \IR\) (not just specific values), \(T(c\vec v)=cT(\vec v)\text{.}\)
(Note the similarities between this process and showing that a subset of a vector space is or is not a subspace: Remark 2.3.11.)
Activity 3.1.10.
(a)
Consider the following maps of Euclidean vectors \(P:\mathbb R^3\rightarrow\mathbb R^3\) and \(Q:\mathbb R^3\rightarrow\mathbb R^3\) defined by
\begin{equation*}
P\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)=
\left[\begin{array}{c} -2 \, x - 3 \, y - 3 \, z \\ 3 \, x + 4 \, y + 4 \, z \\ 3 \, x + 4 \, y + 5 \, z \end{array}\right]
\hspace{1em} \text{and} \hspace{1em} Q\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)=
\left[\begin{array}{c} x - 4 \, y + 9 \, z \\ y - 2 \, z \\ 8 \, y^{2} - 3 \, x z \end{array}\right].
\end{equation*}
Which do you suspect?
- \(P\) is linear, but \(Q\) is not.
- \(Q\) is linear, but \(P\) is not.
- Both maps are linear.
- Neither map is linear.
(b)
Consider the following map of Euclidean vectors \(S:\mathbb R^2\rightarrow\mathbb R^2\)
\begin{equation*}
S\left( \left[\begin{array}{c} x \\ y \end{array}\right]\right)= \left[\begin{array}{c} x + 2 \, y \\ 9 \, x y \end{array}\right].
\end{equation*}
Prove that \(S\) is not a linear transformation.
(c)
Consider the following map of Euclidean vectors \(T:\mathbb R^2\rightarrow\mathbb R^2\)
\begin{equation*}
T\left( \left[\begin{array}{c} x \\ y \end{array}\right] \right)= \left[\begin{array}{c} 8 \, x - 6 \, y \\ 6 \, x - 4 \, y \end{array}\right].
\end{equation*}
Prove that \(T\) is a linear transformation.
Subsection 3.1.3 Cool Down
Activity 3.1.11.
Let \(f(x)=x^3-1\text{.}\) Then, \(f\colon\IR\to\IR\) is a function with domain and codomain equal to \(\IR\text{.}\) Is \(f(x)\) is a linear transformation?
Activity 3.1.12.
(a)
Is it the case that rotating \(\vec{u}+\vec{v}\) about the origin by \(\frac{\pi}{2}=90^\circ\) is the same as first rotating each of \(\vec{u},\vec{v}\) and then adding them together?
(b)
Is it the case that rotating \(5\vec{u}\) about the origin by \(\frac{\pi}{2}=90^\circ\) is the same as first rotating \(\vec{u}\) by \(\frac{\pi}{2}=90^\circ\) and then scaling by \(5\text{?}\)
(c)
Based on this, do you suspect that the transformation \(R\colon\IR^2\to\IR^2\) given by rotating vectors about the origin through an angle of \(\frac{\pi}{2}=90^\circ\) is linear? Do you think there is anything special about the angle \(\frac{\pi}{2}=90^\circ\text{?}\)
Activity 3.1.13.
In Activity 2.2.1, we made an analogy between vectors and linear combinations with ingredients and recipes. Let us think of cooking as a transformation of ingredients. In this analogy, would it be appropriate for us to consider "cooking" to be a linear transformation or not? Describe your reasoning.Subsection 3.1.4 Videos
Exercises 3.1.5 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2024/exercises/#/bank/AT1/.Subsection 3.1.6 Mathematical Writing Explorations
Exploration 3.1.14.
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 3.1.15.
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 3.1.7 Sample Problem and Solution
Sample problem Example B.1.12.
