Learning Outcomes
- Translate back and forth between a linear transformation of Euclidean spaces and its standard matrix, and perform related computations.
Section 3.2 Standard Matrices (AT2)
Subsection 3.2.1 Warm Up
Remark 3.2.1.
Recall that a linear map \(T:V\rightarrow W\) satisfies
- \(T(\vec{v}+\vec{w}) = T(\vec{v})+T(\vec{w})\) for any \(\vec{v},\vec{w} \in V\text{.}\)
- \(T(c\vec{v}) = cT(\vec{v})\) for any \(c \in \IR,\vec{v} \in V\text{.}\)
In other words, a map is linear when vector space operations can be applied before or after the transformation without affecting the result.
Activity 3.2.2.
Can you recall the following?
(a)
Given a transformation, what do the terms domain and codomain mean?
(b)
What does the notation \(T\colon V\to W\) mean?
Subsection 3.2.2 Class Activities
Activity 3.2.3.
Suppose \(T: \IR^3 \rightarrow \IR^2\) is a linear map, and you know \(T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right] \right)
=
\left[\begin{array}{c} 2 \\ 1 \end{array}\right]\) and \(T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right)
=
\left[\begin{array}{c} -3 \\ 2 \end{array}\right]
\text{.}\) What is \(T\left(\left[\begin{array}{c} 3 \\ 0 \\ 0 \end{array}\right]\right)\text{?}\)
- \(\displaystyle \left[\begin{array}{c} 6 \\ 3\end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} -9 \\ 6 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} -4 \\ -2 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} 6 \\ -4 \end{array}\right]\)
Activity 3.2.4.
Suppose \(T: \IR^3 \rightarrow \IR^2\) is a linear map, and you know \(T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right] \right)
=
\left[\begin{array}{c} 2 \\ 1 \end{array}\right]\) and \(T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right)
=
\left[\begin{array}{c} -3 \\ 2 \end{array}\right]
\text{.}\) What is \(T\left(\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right]\right)\text{?}\)
- \(\displaystyle \left[\begin{array}{c} 2 \\ 1\end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} 3 \\ -1 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} -1 \\ 3 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} 5 \\ -8 \end{array}\right]\)
Activity 3.2.5.
Suppose \(T: \IR^3 \rightarrow \IR^2\) is a linear map, and you know \(T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right] \right)
=
\left[\begin{array}{c} 2 \\ 1 \end{array}\right]\) and \(T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right)
=
\left[\begin{array}{c} -3 \\ 2 \end{array}\right]
\text{.}\) What is \(T\left(\left[\begin{array}{c} -2 \\ 0 \\ -3 \end{array}\right]\right)\text{?}\)
- \(\displaystyle \left[\begin{array}{c} 2 \\ 1\end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} 3 \\ -1 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} -1 \\ 3 \end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c} 5 \\ -8 \end{array}\right]\)
Activity 3.2.6.
Suppose \(T: \IR^3 \rightarrow \IR^2\) is a linear map, and you know \(T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right] \right)
=
\left[\begin{array}{c} 2 \\ 1 \end{array}\right]\) and \(T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right)
=
\left[\begin{array}{c} -3 \\ 2 \end{array}\right]
\text{.}\) What piece of information would help you compute \(T\left(\left[\begin{array}{c}0\\4\\-1\end{array}\right]\right)\text{?}\)
- The value of \(T\left(\left[\begin{array}{c} 0\\4\\0\end{array}\right]\right)= \left[\begin{array}{c} -4 \\ 16\end{array}\right]\text{.}\)
- The value of \(T\left(\left[\begin{array}{c} 0\\1\\0\end{array}\right]\right)= \left[\begin{array}{c} -1 \\ 4\end{array}\right]\text{.}\)
- The value of \(T\left(\left[\begin{array}{c} 1\\1\\1\end{array}\right]\right)= \left[\begin{array}{c} -2 \\ 7\end{array}\right]\text{.}\)
- Any of the above.
Observation 3.2.7.
Since all three choices in Activity 3.2.6 create a spanning and linearly independent set along with \(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right]\) and \(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right]\text{,}\) they each may be used to compute \(T\left(\left[\begin{array}{c}0\\4\\-1\end{array}\right]\right)\text{:}\)
\begin{equation*}
T\left(\left[\begin{array}{c}0\\4\\-1\end{array}\right]\right)
=
T\left(\left[\begin{array}{c}0\\4\\0\end{array}\right]\right)
-
T\left(\left[\begin{array}{c}0\\0\\1\end{array}\right]\right)
=
\left[\begin{array}{c} -4 \\ 16\end{array}\right]
-
\left[\begin{array}{c} -3 \\ 2 \end{array}\right]
=
\left[\begin{array}{c} -1 \\ 14 \end{array}\right]
\end{equation*}
\begin{equation*}
T\left(\left[\begin{array}{c}0\\4\\-1\end{array}\right]\right)
=
4
T\left(\left[\begin{array}{c}0\\1\\0\end{array}\right]\right)
-
T\left(\left[\begin{array}{c}0\\0\\1\end{array}\right]\right)
=
4
\left[\begin{array}{c} -1 \\ 4\end{array}\right]
-
\left[\begin{array}{c} -3 \\ 2 \end{array}\right]
=
\left[\begin{array}{c} -1 \\ 14 \end{array}\right]
\end{equation*}
\begin{equation*}
T\left(\left[\begin{array}{c}0\\4\\-1\end{array}\right]\right)
=
4
T\left(\left[\begin{array}{c}1\\1\\1\end{array}\right]\right)
-5
T\left(\left[\begin{array}{c}0\\0\\1\end{array}\right]\right)
-4
T\left(\left[\begin{array}{c}1\\0\\0\end{array}\right]\right)
\end{equation*}
\begin{equation*}
=
4
\left[\begin{array}{c} -2 \\ 7\end{array}\right]
-5
\left[\begin{array}{c} -3 \\ 2 \end{array}\right]
-4
\left[\begin{array}{c} 2 \\ 1 \end{array}\right]
=
\left[\begin{array}{c} -8+15-8 \\ 28-10-4 \end{array}\right]
=
\left[\begin{array}{c} -1 \\ 14 \end{array}\right]
\end{equation*}
Fact 3.2.8.
Consider any basis \(\{\vec b_1,\dots,\vec b_n\}\) for \(V\text{.}\) Since every vector \(\vec v\) can be written as a linear combination of basis vectors, \(\vec v = x_1\vec b_1+\dots+ x_n\vec b_n\text{,}\) we may compute \(T(\vec v)\) as follows:
\begin{equation*}
T(\vec v)=T(x_1\vec b_1+\dots+ x_n\vec b_n)=
x_1T(\vec b_1)+\dots+x_nT(\vec b_n)
.
\end{equation*}
Therefore any linear transformation \(T:V \rightarrow W\) can be defined by just describing the values of \(T(\vec b_i)\text{.}\)
Put another way, the images of the basis vectors completely determine the transformation \(T\text{.}\)
Definition 3.2.9.
Since a linear transformation \(T:\IR^n\to\IR^m\) is determined by its action on the standard basis \(\{\vec e_1,\dots,\vec e_n\}\text{,}\) it is convenient to store this information in an \(m\times n\) matrix, called the standard matrix of \(T\text{,}\) given by \([T(\vec e_1) \,\cdots\, T(\vec e_n)]\text{.}\)
For example, let \(T: \IR^3 \rightarrow \IR^2\) be the linear map determined by the following values for \(T\) applied to the standard basis of \(\IR^3\text{.}\)
\begin{equation*}
\scriptsize
T\left(\vec e_1 \right)
=
T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right] \right)
=
\left[\begin{array}{c} 2 \\ 1\end{array}\right]
\hspace{2em}
T\left(\vec e_2 \right)
=
T\left(\left[\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right] \right)
=
\left[\begin{array}{c} -1 \\ 4\end{array}\right]
\hspace{2em}
T\left(\vec e_3 \right)
=
T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right)
=
\left[\begin{array}{c} -3 \\ 2\end{array}\right]
\end{equation*}
Then the standard matrix corresponding to \(T\) is
\begin{equation*}
\left[\begin{array}{ccc}T(\vec e_1) & T(\vec e_2) & T(\vec e_3)\end{array}\right]
=
\left[\begin{array}{ccc}2 & -1 & -3 \\ 1 & 4 & 2 \end{array}\right]
.
\end{equation*}
Activity 3.2.10.
Let \(T: \IR^4 \rightarrow \IR^3\) be the linear transformation given by
\begin{equation*}
T\left(\vec e_1 \right)
=
\left[\begin{array}{c} 0 \\ 3 \\ -2\end{array}\right]
\hspace{2em}
T\left(\vec e_2 \right)
=
\left[\begin{array}{c} -3 \\ 0 \\ 1\end{array}\right]
\hspace{2em}
T\left(\vec e_3 \right)
=
\left[\begin{array}{c} 4 \\ -2 \\ 1\end{array}\right]
\hspace{2em}
T\left(\vec e_4 \right)
=
\left[\begin{array}{c} 2 \\ 0 \\ 0\end{array}\right]
\end{equation*}
Write the standard matrix \([T(\vec e_1) \,\cdots\, T(\vec e_n)]\) for \(T\text{.}\)
Activity 3.2.11.
Let \(T: \IR^3 \rightarrow \IR^2\) be the linear transformation given by
\begin{equation*}
T\left(\left[\begin{array}{c} x\\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x+3z \\ 2x-y-4z \end{array}\right]
\end{equation*}
(a)
Compute \(T(\vec e_1)\text{,}\) \(T(\vec e_2)\text{,}\) and \(T(\vec e_3)\text{.}\)
(b)
Find the standard matrix for \(T\text{.}\)
Fact 3.2.12.
Because every linear map \(T:\IR^n\to\IR^m\) has a linear combination of the variables in each component, and thus \(T(\vec e_i)\) yields exactly the coefficients of \(x_i\text{,}\) the standard matrix for \(T\) is simply an array of the coefficients of the \(x_i\text{:}\)
\begin{equation*}
T\left(\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]\right)
=
\left[\begin{array}{c}
ax+by+cz+dw \\
ex+fy+gz+hw
\end{array}\right]
\hspace{2em}
A
=
\left[\begin{array}{cccc}
a & b & c & d \\
e & f & g & h
\end{array}\right]
\end{equation*}
Since the formula for a linear transformation \(T\) and its standard matrix \(A\) may both be used to compute the transformation of a vector \(\vec x\text{,}\) we will often write \(T(\vec x)\) and \(A\vec x\) interchangably:
\begin{equation*}
T\left(\left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right]\right)
=
\left[\begin{array}{c}
ax_1+bx_2+cx_3+dx_4 \\
ex_1+fx_2+gx_3+hx_4
\end{array}\right]
=
A\left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right]
=
\left[\begin{array}{cccc}
a & b & c & d \\
e & f & g & h
\end{array}\right]
\left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right]
\end{equation*}
Activity 3.2.13.
Let \(T: \IR^4 \rightarrow \IR^3\) be the linear transformation given by
\begin{equation*}
T\left(\vec e_1 \right)
=
\left[\begin{array}{c} 0 \\ 3 \\ -2\end{array}\right]
\hspace{2em}
T\left(\vec e_2 \right)
=
\left[\begin{array}{c} -3 \\ 0 \\ 1\end{array}\right]
\hspace{2em}
T\left(\vec e_3 \right)
=
\left[\begin{array}{c} 4 \\ -2 \\ 1\end{array}\right]
\hspace{2em}
T\left(\vec e_4 \right)
=
\left[\begin{array}{c} 2 \\ 0 \\ 0\end{array}\right]
\end{equation*}
Write the standard matrix \([T(\vec e_1) \,\cdots\, T(\vec e_n)]\) for \(T\text{.}\)
Activity 3.2.14.
(a)
Explain and demonstrate how to compute the standard matrix for the linear transformation \(S:\mathbb{R}^2 \to \mathbb{R}^4\) given by
\begin{equation*}
S\left( \left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right] \right) = \left[\begin{array}{c} 9 \, x_{1} - 2 \, x_{2} \\ -3 \, x_{1} \\ 5 \, x_{1} - x_{2} \\ -6 \, x_{2} \end{array}\right]
\end{equation*}
by computing transformations of the standard basic vectors:
\begin{equation*}
S(\vec e_1)=\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right]\hspace{1em}S(\vec e_2)=\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right]\hspace{1em}\rightarrow\hspace{1em}\left[\begin{array}{cc} \unknown & \unknown \\ \unknown & \unknown \\ \unknown & \unknown \\\unknown & \unknown \end{array}\right]
\end{equation*}
Answer.
\begin{equation*}
\left[\begin{array}{cc} 9 & -2 \\ -3 & 0 \\ 5 & -1 \\ 0 & -6 \end{array}\right]
\end{equation*}
(b)
Let \(T:\mathbb{R}^4 \to \mathbb{R}^3\) be the linear transformation given by the standard matrix
\begin{equation*}
\left[\begin{array}{cccc} -2 & -4 & 2 & -2 \\ -4 & 3 & -3 & 2 \\ 5 & 0 & 2 & -6 \end{array}\right].
\end{equation*}
Explain and demonstrate how to compute \(T\left(\left[\begin{array}{c} -5 \\ 0 \\ -3 \\ -2 \end{array}\right]\right)\) by using the values of transformed standard basic vectors:
\begin{equation*}
T\left(\left[\begin{array}{c} -5 \\ 0 \\ -3 \\ -2 \end{array}\right]\right)=\unknown T(\vec e_1)+\unknown T(\vec e_2)+\unknown T(\vec e_3)+\unknown T(\vec e_4)
\end{equation*}
Answer.
\begin{equation*}
T\left(\left[\begin{array}{c} -5 \\ 0 \\ -3 \\ -2 \end{array}\right]\right)=\left[\begin{array}{c} 8 \\ 25 \\ -19 \end{array}\right]
\end{equation*}
Subsection 3.2.3 Individual Practice
Activity 3.2.15.
Consider the linear transformation \(R\colon\IR^2\to\IR^2\) given by rotating vectors about the origin through an angle of \(\frac{\pi}{4}=45^\circ\text{.}\)
(a)
If \(\vec{e}_1,\vec{e}_2\) are the standard basis vectors of \(\IR^2\text{,}\) calculate \(R(\vec{e}_1),R(\vec{e}_2)\text{.}\)
(b)
What is the standard matrix representing \(R\text{?}\)
Activity 3.2.16.
Consider the linear transformation \(S\colon\IR^2\to\IR^2\) given by reflecting vectors across the line \(x_1=x_2\text{.}\)
(a)
If \(\vec{e}_1,\vec{e}_2\) are the standard basis vectors of \(\IR^2\text{,}\) calculate \(S(\vec{e}_1),S(\vec{e}_2)\text{.}\)
(b)
What is the standard matrix representing \(S\text{?}\)
Subsection 3.2.4 Videos
Exercises 3.2.5 Exercises
Exercises available at
https://tbil.org/linear-algebra/preview/exercises/#/bank/AT2/.Subsection 3.2.6 Mathematical Writing Explorations
We can represent images in the plane \(\mathbb{R}^2\) using vectors, and manipulate those images with linear transformations. We introduce some notation in these explorations that is needed for their completion, but is not essential to the rest of the text. These have a geometric flair to them, and can be understood by thinking of geometric transformations in terms of standard matrices.
Given two vectors \(\vec{v} = \left[\begin{array}{c}v_1\\v_2\\ \vdots \\ v_n\end{array}\right]\) and \(\vec{w} = \left[\begin{array}{c}w_1 \\ w_2\\ \vdots \\ w_n\end{array}\right]\text{,}\) we define the dot product as
\begin{equation*}
\vec{v}\cdot \vec{w} = v_1w_1 + v_2w_2 + \cdots v_nw_n.
\end{equation*}
Exploration 3.2.17.
For each of the following properties, determine if it is held by the dot product. Either provide a proof it the property holds, or provide a counter-example if it does not.- Distributive over addition (e.g., (\(\vec{u} + \vec{v})\cdot \vec{w} = \vec{u}\cdot\vec{w} + \vec{v}\cdot\vec{w})?\)
- Associative?
- Commutative?
Exploration 3.2.18.
Given the properties you proved in the last exploration, could the dot product take the place of \(\oplus\) as a vector space operation on \(\mathbb{R}^n\text{?}\)Exploration 3.2.19.
Is the dot product a linear operator? That is, given vectors \(\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^n\text{,}\) and \(k,m \in \mathbb{R}\text{,}\) is it true that
\begin{equation*}
\vec{u} \cdot (k\vec{v} + m\vec{w}) = k(\vec{u} \cdot \vec{v}) + m(\vec{u}\cdot\vec{w}).
\end{equation*}
Prove or provide a counter-example.Exploration 3.2.20.
Assume \(\vec{v} = \left[\begin{array}{c}v_1\\v_2\\ \vdots \\ v_n\end{array}\right]\) and define the length of a vector by
\begin{equation*}
|\vec{v}| = \left(\sum_{i=1}^n v_i^2 \right)^{1/2}\text{.}
\end{equation*}
Prove that \(|\vec{u}| = |\vec{v}|\) if an only if \(\vec{u} + \vec{v}\) and \(\vec{u} - \vec{v}\) are perpendicular. You may use the fact (try and prove it!) that two vectors are perpendicular if and only if their dot product is zero.
Exploration 3.2.21.
- A dilation is given by by mapping a vector \(\vec{v} = \left[\begin{array}{c}x\\y\end{array}\right]\) to some scalar multiple of \(\vec{v}\text{.}\)
- A rotation is given by \(\vec{v} \mapsto \left[\begin{array}{c} \cos(\theta)x - \sin(\theta)y\\ \cos(\theta)y + \sin(\theta)x\end{array}\right].\)
- A reflection of \(\vec{v}\) over a line \(l\) can be found by first finding a vector \(\vec{l} = \left[\begin{array}{c} l_x\\l_y\end{array}\right]\) along \(l\text{,}\) then \(\vec{v} \mapsto 2\frac{\vec{l}\cdot\vec{v}}{\vec{l}\cdot\vec{l}}\vec{l} - \vec{v}.\)
- Rotation through an angle \(\theta\text{.}\)
- Reflection over a line \(l\) passing through the origin.
- Dilation by some scalar \(s\text{.}\)
Subsection 3.2.7 Sample Problem and Solution
Sample problem Example B.1.13.
