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AT3 Incorporate Vector Equations #657

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AT3 Incorporate Vector Equations
jkostiuk Mar 12, 2025
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quick change to try out in 8 minutes
StevenClontz Mar 14, 2025
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on more tweak
StevenClontz Mar 14, 2025
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tweak kernel discovery
StevenClontz Mar 14, 2025
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more edits based on ad hoc adjustments made today
StevenClontz Mar 14, 2025
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254 changes: 193 additions & 61 deletions source/linear-algebra/source/03-AT/03.ptx
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Original file line number Diff line number Diff line change
Expand Up @@ -50,8 +50,11 @@ Let <m>T: \IR^2 \rightarrow \IR^3</m> be given by
\text{with standard matrix }
\left[\begin{array}{cc} 1 &amp; 0 \\ 0 &amp; 1 \\ 0 &amp; 0 \end{array}\right]
</me>
Which of these subspaces of <m>\IR^2</m> describes
the set of all vectors that transform into <m>\vec 0</m>?
Which of these sets contain all vectors <m>\left[\begin{array}{c}x \\ y \end{array}\right]</m>
that transform into <m>T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right)
=
\left[\begin{array}{c} x \\ y \\ 0 \end{array}\right]=
\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]</m>?
</p>
<ol marker="A." cols="2">
<li>
Expand Down Expand Up @@ -82,12 +85,12 @@ the set of all vectors that transform into <m>\vec 0</m>?
<definition xml:id="AT3-definition-kernel">
<statement>
<p>
Let <m>T: V \rightarrow W</m> be a linear transformation, and let <m>\vec{z}</m> be the additive
identity (the <q>zero vector</q>) of <m>W</m>. The <term>kernel</term><idx>kernel</idx> of <m>T</m>
Let <m>T: V \rightarrow W</m> be a linear transformation, and let <m>\vec{0}</m> be the zero vector of <m>W</m>.
The <term>kernel</term><idx>kernel</idx> of <m>T</m>
(also known as the <term>null space</term><idx>null space</idx> of <m>T</m>)
is an important subspace of <m>V</m> defined by
is an important subset of <m>V</m> defined by
<me>
\ker T = \left\{ \vec{v} \in V\ \big|\ T(\vec{v})=\vec{z}\right\}
\ker T = \left\{ \vec{x} \in V\ \big|\ T(\vec{x})=\vec{0}\right\}
</me>
</p>
<figure>
Expand Down Expand Up @@ -128,8 +131,13 @@ Let <m>T: \IR^3 \rightarrow \IR^2</m> be given by
\text{with standard matrix }
\left[\begin{array}{ccc} 1 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 \end{array}\right]
</me>
Which of these subspaces of <m>\IR^3</m> describes <m>\ker T</m>,
the set of all vectors that transform into <m>\vec 0</m>?
Which of these subsets of <m>\IR^3</m> is equal to
<m>\ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\right\}</m>,
the set of all vectors <m>\left[\begin{array}{c}x \\ y \\ z\end{array}\right]</m>
that transform into <m>T\left(\left[\begin{array}{c}x \\ y \\ z \end{array}\right] \right)
=
\left[\begin{array}{c} x \\ y\end{array}\right]=
\left[\begin{array}{c} 0 \\ 0 \end{array}\right]</m>?
</p>
<ol marker='A.' cols="2">
<li>
Expand Down Expand Up @@ -161,28 +169,101 @@ the set of all vectors that transform into <m>\vec 0</m>?
<p>
Let <m>T: \IR^3 \rightarrow \IR^2</m> be the linear transformation given by the
standard matrix
<me>T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{c} 3x+4y-z \\ x+2y+z \end{array}\right]</me>
<me>
A=\left[\begin{array}{ccc} 3 &amp; 4 &amp; -1 \\ 1 &amp; 2 &amp; 1 \end{array}\right]
=\left[\begin{array}{ccc} T(\vec e_1) &amp; T(\vec e_2) &amp; T(\vec e_3)\end{array}\right].
</me>

</p>
</introduction>
<task>
<task>
<statement>
<p>
Which of these is the most appropriate method to determine whether a <m>\mathbb R^3</m> vector
<m>\vec{x}=\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=x_1\vec{e}_1+x_2\vec{e}_2+x_3\vec{e}_3</m>,
belongs to <m>\ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\right\}</m>?
<ol marker="A.">
<li>
<p>
Determine if the set of vectors
<m>\left\{T(\vec e_1), T(\vec e_2), T(\vec e_3), T(\vec x)\right\}</m>
spans <m>\IR^3</m>.
</p>
</li>
<li>
<p>
Determine if the set of vectors
<m>\left\{T(\vec e_1), T(\vec e_2), T(\vec e_3), T(\vec x)\right\}</m>
is linearly dependent.
</p>
</li>
<li>
<p>
Determine if
<m>\vec x=\left[\begin{array}{c} x_1 \\ x_2\\ x_3\end{array}\right]</m>
belongs to the solution set of the vector equation
<me>T(\vec x)=T\left(\left[\begin{array}{c} x_1 \\ x_2\\ x_3\end{array}\right]\right)=
x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)=
\left[\begin{array}{c} 0 \\ 0\end{array}\right]</me>.
</p>
</li>
<li>
<p>
Determine if the equation
<me>T(\vec 0)=T\left(\left[\begin{array}{c} 0 \\ 0\\ 0\end{array}\right]\right)=
0T(\vec e_1)+0T(\vec e_2)+0T(\vec e_3)=
T(\vec x)</me> is consistent.
</p>
</li>
</ol>
</p>
</statement>
<answer>
<p>
B. This is exactly the same as finding the solution space for the
homoegeneous vector equation <m>T(\vec x)=\vec 0</m>.
</p>
</answer>
</task>
<task>
<statement>
<p>
Use this method to find the kernel of <m>T</m>.
</p>
<answer>
<p>
<m>\ker T=\left\{\left[\begin{array}{c}3a\\-2a\\a\end{array}\right]\middle|
a\in\IR\right\}</m>
</p>
</answer>
</statement>
</task>
</activity>
<sage language="octave">
</sage>

<observation xml:id="observation-kernel-homogeneous-solution">
<statement>
<p>
Set
<m>
T\left(\left[\begin{array}{c}x\\y\\z\end{array}\right]\right)
=
\left[\begin{array}{c}0\\0\end{array}\right]
</m> to find a linear system of equations whose solution set is the kernel.
The kernel of a transformation <m>T</m>
is exactly the solution space of
the homogeneous equation <m>T(\vec{x})=\vec{0}</m>.
If its standard matrix is <m>A</m>, then we may write
<m>A\vec x=\vec 0</m> and use <m>\RREF[A\,|\,\vec 0]</m> to
find this kernel.
</p>
</task>
<task>
<p>
Use <m>\RREF(A)</m> to solve this homogeneous system of equations and find a basis
for the kernel of <m>T</m>.
In particular, the kernel is a subspace of the transformation's
domain, and has a basis which may be found as in
<xref ref="fact-solution-space-basis"/>:
<me>
\ker T=\left\{\left[\begin{array}{c}3a\\-2a\\a\end{array}\right]\middle|
a\in\IR\right\} \hspace{2em}
\text{Basis for }\ker T=\left\{\left[\begin{array}{c}3\\-2\\1\end{array}\right]\right\}.
</me>
</p>
</task>
</activity>
<sage language="octave">
</sage>
</statement>
</observation>

<activity estimated-time='10'>
<statement>
Expand Down Expand Up @@ -245,9 +326,9 @@ Which of these subspaces of <m>\IR^3</m> describes the set of all vectors that a
<p>
Let <m>T: V \rightarrow W</m> be a linear transformation.
The <term>image</term><idx>image</idx> of <m>T</m>
is an important subspace of <m>W</m> defined by
is an important subset of <m>W</m> defined by
<me>
\Im T = \left\{ \vec{w} \in W\ \big|\ \text{there is some }\vec v\in V \text{ with } T(\vec{v})=\vec{w}\right\}
\Im T = \left\{ T(\vec v) \in W\ \big| \vec v\in V \right\}
</me>
</p>
<p>
Expand Down Expand Up @@ -323,7 +404,7 @@ Let <m>T: \IR^3 \rightarrow \IR^2</m> be given by
\text{with standard matrix }
\left[\begin{array}{ccc} 1 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 \end{array}\right]
</me>
Which of these subspaces of <m>\IR^2</m> describes <m>\Im T</m>,
Which of these subsets of <m>\IR^2</m> describes <m>\Im T = \left\{ T(\vec v) \in \IR^2\ \big| \vec v\in \IR^3 \right\}</m>,
the set of all vectors that are the result of using <m>T</m> to transform
<m>\IR^3</m> vectors?
</p>
Expand Down Expand Up @@ -366,9 +447,30 @@ the set of all vectors that are the result of using <m>T</m> to transform
</p>
<p>
Consider the question: Which vectors <m>\vec{w}</m> in <m>\IR^3</m> belong to
<m>\Im T</m>?
<m>\Im T=\left\{ T(\vec v) \in \IR^3\ \big| \vec v\in \IR^4 \right\}</m>?
</p>
</introduction>
<task>
<statement>
<p>
Recall that
<me>
T\left(\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right] \right) =
x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)+x_4T(\vec e_4)
</me>.
Complete the following vector equation which must be consistent
in order for <m>\left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right]</m>
to belong to <m>\Im T</m>:
<me>
x_1 \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]+
x_2 \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]+
x_3\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]+
x_4\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]=
\left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right]
</me>
</p>
</statement>
</task>
<task>
<statement>
<p>
Expand All @@ -377,6 +479,15 @@ the set of all vectors that are the result of using <m>T</m> to transform
</p>
</statement>
</task>
<task>
<statement>
<p>
Write down the vector equation which must be consistent in
order for <m>\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]</m>
to belong to <m>\Im T</m>.
</p>
</statement>
</task>
<task>
<statement>
<p>
Expand All @@ -390,90 +501,111 @@ the set of all vectors that are the result of using <m>T</m> to transform
<p>
An arbitrary vector <m>\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]</m> belongs to
<m>\Im T</m> provided the equation
<me>x_1 T(\vec{e}_1)+x_2 T(\vec{e}_2)+x_3T(\vec{e}_3)+x_4T(\vec{e}_4)=\vec{w}</me> has...
<me>x_1 T(\vec{e}_1)+x_2 T(\vec{e}_2)+x_3T(\vec{e}_3)+x_4T(\vec{e}_4)=
\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]</me> has...
<ol marker="A.">
<li>no solutions.</li>
<li>at most one solution.</li>
<li>exactly one solution.</li>
<li>at least one solution.</li>
<li>infinitely-many solutions.</li>
</ol>
</p>
</statement>
</task>
<task>
<statement>
<p>
Based on this, how do <m>\Im T</m> and <m>\vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}</m> relate to each other?
Based on this, what do we know?
<ol marker="A.">
<li>The set <m>\Im T</m> contains <m>\vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}</m> but is not equal to it.</li>
<li>The set <m>\vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}</m> contains <m>\Im T</m> but is not equal to it.</li>
<li>
The set <m>\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}</m> spans
<m>\Im T</m> whenever
<m>\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}</m>
is linearly independent.
</li>
<li>
The set <m>\Im T</m> equals the codomain whenever
<m>\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}</m>
is linearly independent.
</li>
<li>
The set <m>\Im T</m> is simply <m>\{\vec 0\}</m> whenever
<m>\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}</m>
is linearly dependent.
</li>
<li>The set <m>\Im T</m> and <m>\vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}</m> are equal to each other.</li>
<li>There is no relation between these two sets.</li>
</ol>
</p>
</statement>
</task>
</activity>

<observation>
<observation xml:id="AT3-columnspace-observation">
<p>
Let <m>T: \IR^4 \rightarrow \IR^3</m> be the linear transformation given by the
Let <m>T: \IR^n \rightarrow \IR^m</m> be the linear transformation given by the
standard matrix
<me>
A
=
\left[\begin{array}{cccc} 3 &amp; 4 &amp; 7 &amp; 1\\ -1 &amp; 1 &amp; 0 &amp; 2 \\ 2 &amp; 1 &amp; 3 &amp; -1 \end{array}\right]
\left[\begin{array}{ccc} T(\vec e_1) &amp; \cdots &amp; T(\vec e_n) \end{array}\right]
.</me>
</p>
<p>
Since the set
<m>
\setList{
\left[\begin{array}{c}3\\-1\\2\end{array}\right],
\left[\begin{array}{c}4\\1\\1\end{array}\right],
\left[\begin{array}{c}7\\0\\3\end{array}\right],
\left[\begin{array}{c}1\\2\\-1\end{array}\right]
Then we have
<me>
\Im T=\vspan\setList{
T(\vec e_1),
\cdots,
T(\vec e_n)
}
</m>
spans <m>\Im T</m>, we can obtain a basis for <m>\Im T</m> by finding
<m>
\RREF A
</me>
and in particular, see that <m>\Im T</m> is a subspace of the codomain, and thus
has a kernel which may be found as in <xref ref="observation-subspace-basis"/>:
given
<me>
\RREF \left[\begin{array}{cccc} 3 &amp; 4 &amp; 7 &amp; 1\\ -1 &amp; 1 &amp; 0 &amp; 2 \\ 2 &amp; 1 &amp; 3 &amp; -1 \end{array}\right]
=
\left[\begin{array}{cccc} 1 &amp; 0 &amp; 1 &amp; -1\\ 0 &amp; 1 &amp; 1 &amp; 1 \\ 0 &amp; 0 &amp; 0 &amp; 0 \end{array}\right]
</m>
and only using the vectors corresponding to pivot columns:
\left[\begin{array}{cccc} \markedPivot{1} &amp; 0 &amp; 1 &amp; -1\\ 0 &amp; \markedPivot{1} &amp; 1 &amp; 1 \\ 0 &amp; 0 &amp; 0 &amp; 0 \end{array}\right]
</me>
we see that we have
<me>
\Im T=\vspan\left\{
\left[\begin{array}{c} 3 \\ -1 \\ 2 \end{array}\right],
\left[\begin{array}{c} 4 \\ 1 \\ 1 \end{array}\right],
\left[\begin{array}{c} 7\\ 0 \\ 3 \end{array}\right],
\left[\begin{array}{c} 1\\ 2 \\ -1 \end{array}\right],
\right\}
</me> and <me>
\text{Basis for }\Im T=
\setList{
\left[\begin{array}{c}3\\-1\\2\end{array}\right],
\left[\begin{array}{c}4\\1\\1\end{array}\right]
}.
</me>
</p>
<p>
In general, the <term>column space</term><idx>column space</idx> of a matrix <m>M</m> refers to the subspace obtained by considering the span of its column vectors.
Using this terminology, if the transformation <m>T</m> is represented by the matrix <m>A</m>,
then the image of <m>T</m> is the <term>column space</term><idx>column space</idx> of <m>A</m>.
This justifies why the image of a transformation is also called the <term>column space</term><idx>column space</idx> of its standard matrix.
</p>

</observation>

<fact>
<statement>
<p>
Let <m>T:\IR^n\to\IR^m</m> be a linear transformation with standard matrix <m>A</m>.
To summarize,
let <m>T:\IR^n\to\IR^m</m> be a linear transformation with standard matrix <m>A</m>.
<ul>
<li>
<p>
The kernel of <m>T</m> is the solution set of the homogeneous system given
by the augmented matrix <m>\left[\begin{array}{c|c}A&amp;\vec 0\end{array}\right]</m>.
Use the coefficients of its free variables to get a basis for the kernel (as in <xref ref="fact-solution-space-basis"/>).
The kernel of <m>T</m> is the solution set of the equation <m>T(\vec{x})=\vec{0}</m> or, equivalently, the linear system given by the augmented matrix <m>\left[\begin{array}{c|c}A&amp;\vec 0\end{array}\right]</m>.
Use the coefficients of its free variables to get a basis for the kernel (as in <xref ref="fact-solution-space-basis"/>).
</p>
</li>
<li>
<p>
The image of <m>T</m> is the span of the columns of <m>A</m>. Remove
the vectors creating non-pivot columns in <m>\RREF A</m> to get a basis
for the image (as in <xref ref="observation-subspace-basis"/>).
The image of <m>T</m> is the span of the columns of its standard matrix <m>A</m>,
a.k.a. its column space.
Remove the vectors creating non-pivot columns in <m>\RREF A</m> to get a basis for the image (as in <xref ref="observation-subspace-basis"/>).
</p>
</li>
</ul>
Expand Down