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updates to MX3
StevenClontz Apr 8, 2026
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add motivating activity for B-coordinate matrix
StevenClontz Apr 10, 2026
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255 changes: 193 additions & 62 deletions source/linear-algebra/source/04-MX/03.ptx
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Expand Up @@ -49,96 +49,159 @@
So far, when working with the Euclidean vector space <m>\IR^n</m>, we have primarily worked with the standard basis <m>\mathcal{E}=\setList{\vec{e}_1,\dots, \vec{e}_n}</m>.
We can explore alternative perspectives more easily if we expand our toolkit to analyze different bases.
</p>
<figure xml:id="MX3-fig-doenet-change-basis">
<caption>Visualization of the change of basis in <m>\mathbb R^2</m></caption>
<interactive label="MX3-interactive-doenet-change-basis" platform="doenetml" width="100%">
<slate surface="doenetml">
<xi:include parse="text" href="doenet/MX3-doenet-change-basis.xml"/>
</slate>
<description>
<p>An interactive that visualizes the change of basis from the standard basis to a custom basis in the plane.</p>
</description>
</interactive>
</figure>
</remark>
<activity>
<introduction>
<p>
Let <m>\mathcal{B}=\setList{\vec{v}_1,\vec{v}_2,\vec{v}_3}=\setList{\begin{bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}1\\-1\\1\end{bmatrix},\begin{bmatrix}0\\1\\1\end{bmatrix}}</m>.
Consider the non-standard basis <md>\mathcal{B}=\setList{\vec{b}_1,\vec{b}_2,\vec{b}_3}=\setList{\begin{bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}1\\-1\\1\end{bmatrix},\begin{bmatrix}0\\1\\1\end{bmatrix}}</md> for <m>\mathbb R^3</m>.
</p>
</introduction>
</introduction>
<task>
<statement>
<p>
Is <m>\cal{B}</m> a basis of <m>\IR^3</m>?
</p>
<p>
<ol marker="A.">
<li>
<p>
Yes.
</p>
</li>
<li>
<p>
No.
</p>
</li>
Since <m>\mathcal{B}</m> is a basis, how many ways can we write some
arbitrary <m>\vec v\in \IR^3</m> in terms of <m>\mathcal B</m> vectors?
<md>
\vec v= x_1\vec{b}_1+x_2\vec{b}_2+x_3\vec{b}_3=
x_1\begin{bmatrix}1\\0\\1\end{bmatrix}+
x_2\begin{bmatrix}1\\-1\\1\end{bmatrix}+
x_3\begin{bmatrix}0\\1\\1\end{bmatrix}
</md>
<ol label="A." cols="3">
<li><p>Zero</p></li>
<li><p>At most one</p></li>
<li><p>Exactly one</p></li>
<li><p>At least one</p></li>
<li><p>Infinitely-many</p></li>
</ol>
</p>
</statement>
<answer>
<p>
C.
</p>
<p>
Exactly one
</p>
</answer>
</task>
<task>
<p>
Since <m>\cal{B}</m> is a basis, we know that if <m>\vec{v}\in \IR^3</m>, the following vector equation will have a unique solution:
<me>x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3=\vec{v}</me>
Given this, we define a map <m>C_{\mathcal{B}}\colon\IR^3\to\IR^3</m> via the rule that <m>C_{\mathcal{B}}(\vec{v})</m> is equal to the unique solution to the above vector equation.
The map <m>C_{\mathcal{B}}</m> is a linear map.
</p>
<p>
Compute <m>C_{\mathcal{B}}\left(\begin{bmatrix}1\\1\\1\end{bmatrix}\right)</m>, the unique solution to
<me>
x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3=\begin{bmatrix}1\\1\\1\end{bmatrix}.
</me>

</p>
<statement>
<p>
If <m>\vec x=\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]</m>
and <m>B = \left[\begin{array}{ccc}\vec b_1&amp; \vec b_2&amp;\vec b_3\end{array}\right]=\left[\begin{array}{ccc}1&amp;1&amp;0\\0&amp;-1&amp;1\\1&amp;1&amp;1\end{array}\right]</m>,
which of these matrix equations can be used to find <m>x_1,x_2,x_3</m>?
<ol label="A." cols="3">
<li><p><m>\vec v=B\vec x</m></p></li>
<li><p><m>B\vec v=\vec x</m></p></li>
<li><p><m>\vec v=B^{-1}\vec x</m></p></li>
<li><p><m>B^{-1}\vec v=\vec x</m></p></li>
<li><p>A or D</p></li>
<li><p>B or C</p></li>
</ol>
</p>
</statement>
<answer>
<p>
E.
</p>
<p>
<m>B\vec v=\vec x</m> can be rewritten as <m>\vec v=B^{-1}\vec x</m>
</p>
</answer>
</task>
<task>
<p>
Compute <m>C_\mathcal{B}(\vec{e}_1),C_\mathcal{B}(\vec{e}_2), C_\mathcal{B}(\vec{e}_3)</m> and, in doing so, write down the standard matrix <m>M_\mathcal{B}</m> of <m>C_\mathcal{B}</m>.
</p>
<sage language="octave">
<input>

</input>
<output>

</output>
</sage>
<statement>
<p>
Let <m>\vec v=\begin{bmatrix}1\\2\\3\end{bmatrix}</m> and
use this equation to find
<md>x_1=\unknown,x_2=\unknown,x_3=\unknown</md>.
</p>
</statement>
<answer>
<p>
<m>x_1=1,x_2=0,x_3=2</m> so
<md>\begin{bmatrix}1\\2\\3\end{bmatrix}=
1\begin{bmatrix}1\\0\\1\end{bmatrix}+
0\begin{bmatrix}1\\-1\\1\end{bmatrix}+
2\begin{bmatrix}0\\1\\1\end{bmatrix}</md>
</p>
</answer>
</task>

</activity>

<definition xml:id="def-change-of-basis">
<statement>
<p>
Given a basis <m>\cal{B}=\setList{\vec{v}_1,\dots, \vec{v}_n}</m> of <m>\IR^n</m>, the <term>change of basis/coordinate</term> transformation <em>from</em> the standard basis <em>to</em> <m>\mathcal{B}</m> is the transformation <m>C_\mathcal{B}\colon\IR^n\to\IR^n</m> defined by the property that, for any vector <m>\vec{v}\in\IR^n</m>, the vector <m>C_\mathcal{B}(\vec{v})</m> is the unique solution to the vector equation:
<me>x_1\vec{v}_1+\dots+x_n\vec{v}_n=\vec{v}.</me>
Its standard matrix is called the change-of-basis matrix from the standard basis to <m>\mathcal{B}</m> and is denoted by <m>M_{\mathcal{B}}</m>.
It satisfies the following:
<me>M_{\mathcal{B}}=[\vec{v}_1\ \cdots\ \vec{v}_n]^{-1}.</me>
Given a basis <m>\mathcal{B}=\setList{\vec{b}_1,\dots, \vec{b}_n}</m> of <m>\IR^n</m> and corresponding matrix <m>B=\begin{bmatrix}\vec b_1&amp;\cdots&amp;\vec b_n\end{bmatrix}</m>, the <term>change of basis/coordinate</term> transformation <em>from</em> the standard basis <em>to</em> <m>\mathcal{B}</m> is the transformation <m>C_\mathcal{B}\colon\IR^n\to\IR^n</m> defined by the property that, for any vector <m>\vec{v}\in\IR^n</m>, the vector <m>C_\mathcal{B}(\vec{v})</m> describes the unique way to write
<m>\vec v</m> in terms of the basis, that is,
the unique solution to the vector equation:
<md>\vec v=x_1\vec{b}_1+\dots+x_n\vec{b}_n</md>.
</p>
<p>
Since the solution vector
<m>C_{\mathcal B}(\vec v)=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}</m>
describes the <q><m>\mathcal B</m>-coordinates</q> of <m>\vec v</m>,
we will write
<md>\vec v=x_1\vec{b}_1+\dots+x_n\vec{b}_n=
B\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}=
\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}_{\mathcal B}</md>
</p>
</statement>
</definition>

<remark>
<p>
The vector <m>C_\mathcal{B}(\vec{v})</m> is the <m>\mathcal{B}</m>-coordinates of <m>\vec{v}</m>.
If you work with standard coordinates, and I work with <m>\mathcal{B}</m>-coordinates, then to build the vector that you call <m>\vec{v}=\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}=a_1\vec{e}_1+\cdots+a_n\vec{e}_n</m>, I would first compute <m>C_\mathcal{B}(\vec{v})=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}</m> and then build <m>\vec{v}=x_1\vec{v}_1+\cdots+x_n\vec{v}_n</m>.
As was just shown, the standard matrix <m>M_{\mathcal B}</m> for this
transformation is exactly the inverse matrix <m>B^{-1}</m>.
</p>
<p>
In particular, notation as above, we would have:
<me>a_1\vec{e}_1+\cdots a_n\vec{e}_n=\vec{v}=x_1\vec{v}_1+\cdots+x_n\vec{v}_n.</me>
The vector <m>C_\mathcal{B}(\vec{v})</m> describes the
<q><m>\mathcal{B}</m>-coordinates</q> of <m>\vec{v}</m>.
If you work with standard coordinates, and I work with
<m>\mathcal{B}</m>-coordinates, then you might write
<md>\vec{v}=a_1\vec{e}_1+\cdots+a_n\vec{e}_n=\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}</md>
and I might instead write
<md>\vec{v}=x_1\vec{b}_1+\cdots+x_n\vec{b}_n=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}_{\mathcal B}=B\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}</md>
</p>
<p>
To convert from your standard coordinates to my <m>\mathcal B</m>-coordinates,
we need simply compute:
<md>\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}=
C_{\mathcal B}\left(\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}\right)=
M_{\mathcal B}\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}=
B^{-1}\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}</md>.
</p>
<p>
And similarly we can convert backwards:
<md>\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}=
C_{\mathcal B}^{-1}\left(\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}\right)=
M_{\mathcal B}^{-1}\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}=
B\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}</md>.
</p>
</remark>


<activity>
<introduction>
<p>
Let <m>\vec{v}_1=\begin{bmatrix}1\\-2\\1\end{bmatrix},\ \vec{v}_2=\begin{bmatrix}-1\\0\\3\end{bmatrix},\ \vec{v}_3=\begin{bmatrix}0\\1\\-1\end{bmatrix}</m>, and <m>\mathcal{B}=\setList{\vec{v}_1,\vec{v}_2,\vec{v}_3}</m>
Let <m>\vec{b}_1=\begin{bmatrix}-1\\1\\2\end{bmatrix},\ \vec{b}_2=\begin{bmatrix}0\\-1\\-5\end{bmatrix},\ \vec{b}_3=\begin{bmatrix}-4\\2\\-1\end{bmatrix}</m>, and <m>\mathcal{B}=\setList{\vec{b}_1,\vec{b}_2,\vec{b}_3}</m>
</p>
</introduction>
<task>
<p>
Calculate <m>M_{\mathcal{B}}</m> using technology.
Calculate <m>M_{\mathcal{B}}=B^{-1}</m> using technology.
</p>
<sage language="octave">
<input>
Expand All @@ -151,28 +214,96 @@
</task>
<task>
<p>
Use your result to calculate <m>C_\mathcal{B}\left(\begin{bmatrix}1\\1\\1\end{bmatrix}\right)</m> and express the vector <m>\begin{bmatrix}1\\1\\1\end{bmatrix}</m> as a linear combination of <m>\vec{v}_1,\vec{v}_2,\vec{v}_3</m>.
Use this matrix to write
<md>\begin{bmatrix}-1\\0\\3\end{bmatrix}=\begin{bmatrix}\unknown\\\unknown\\\unknown\end{bmatrix}_{\mathcal B}=\unknown\vec b_1+\unknown\vec b_2+\unknown\vec b_3</md>
as a linear combination of <m>\vec{b}_1,\vec{b}_2,\vec{b}_3</m>.
</p>
</task>
</activity>

<activity>
<introduction>
<p>
While defining linear transformations in terms of their standard
matrix <m>A</m> is convenient when working with standard coordinates, it
would be helpful to be able to apply transformations directly to
non-standard bases/coordinates as well.
</p>
<p>
Let <m>\mathcal B=\setList{\vec b_1,\cdots\vec b_n}</m> be a basis, and consider the matrix
<m>B=\begin{bmatrix}\vec b_1&amp;\cdots&amp;\vec b_n\end{bmatrix}</m>
<m>M_{\mathcal B}=B^{-1}</m>.
</p>
</introduction>
<task>
<statement>
<p>
Given <m>\vec v</m> representing <m>\mathcal B</m>-coordinates, which of these
expressions would correctly compute the transformation of <m>\vec v</m> by a
<em>standard</em> matrix <m>A</m> with output in <em>standard</em> coordinates?
<ol marker="A.">
<li><p><m>AB^{-1}\vec v=AM_{\mathcal B}\vec v</m></p></li>
<li><p><m>AB\vec v=AM_{\mathcal B}^{-1}\vec v</m></p></li>
<li><p><m>B^{-1}A\vec v=M_{\mathcal B}A\vec v</m></p></li>
<li><p><m>BA\vec v=M_{\mathcal B}^{-1}A\vec v</m></p></li>
</ol>
</p>
</statement>
<answer>
<p>
B.
</p>
<p>
Since <m>\vec v</m> represents <m>\mathcal B</m>-coordinates,
the vector <m>B\vec v=M_{\mathcal B}^{-1}\vec v</m> represents
its standard coordinates.
</p>
</answer>
</task>
<task>
<statement>
<p>
Therefore, which matrix would directly calculate the transformation of
vectors by a linear map with standard matrix <m>A</m>,
but where inputs and outputs are all given in <m>\mathcal B</m>-coordinates?
<ol marker="A.">
<li><p><m>B^{-1}AB=M_{\mathcal B}AM_{\mathcal B}^{-1}</m></p></li>
<li><p><m>BAB^{-1}=M_{\mathcal B}^{-1}AM_{\mathcal B}</m></p></li>
<li><p><m>AB^{-1}B=AM_{\mathcal B}M_{\mathcal B}^{-1}</m></p></li>
<li><p><m>ABB^{-1}=AM_{\mathcal B}^{-1}M_{\mathcal B}</m></p></li>
</ol>
</p>
</statement>
<answer>
<p>
A.
</p>
<p>
We saw <m>AB</m> transforms <m>\mathcal B</m>-coordinates by the
transformation, but outputs standard coordinates. Applying
<m>B^{-1}=M_{\mathcal B}</m> on the left corrects the outputs to be
in <m>\mathcal B</m>-coordinates.
</p>
</answer>
</task>
</activity>

<observation>
<p>
Let <m>T\colon\IR^n\to\IR^n</m> be a linear transformation and let <m>A</m> denote its standard matrix.
If <m>\cal{B}=\setList{\vec{v}_1,\dots, \vec{v}_n}</m> is some other basis, then we have:
<md>
<mrow>M_\mathcal{B}AM_{\mathcal{B}}^{-1} \amp= M_\mathcal{B}A[\vec{v_1}\cdots\vec{v}_n] </mrow>
<mrow> \amp= M_\mathcal{B}[T(\vec{v}_1)\cdots T(\vec{v}_n)]</mrow>
<mrow> \amp= [C_\mathcal{B}(T(\vec{v}_1))\cdots C_\mathcal{B}(T(\vec{v}_n))]</mrow>
</md>
In other words, the matrix <m>M_{\mathcal{B}}AM_{\mathcal{B}}^{-1}</m> is the matrix whose columns consist of <em><m>\mathcal{B}</m>-coordinate</em> vectors of the image vectors <m>T(\vec{v}_i)</m>.
The matrix <m>M_{\mathcal{B}}AM_{\mathcal{B}}^{-1}</m> is called the <alert>matrix of <m>T</m> with respect to <m>\mathcal{B}</m>-coordinates</alert>.
If <m>\mathcal{B}=\setList{\vec{b}_1,\dots, \vec{v}_n}</m> is some other basis
and <m>B=\begin{bmatrix}\vec b_1&amp;\cdots&amp;\vec b_n\end{bmatrix}</m>,
then <m>M_{\mathcal B}AM_{\mathcal B}^{-1}=B^{-1}AB</m> is the
<term><m>\mathcal B</m>-coordinate matrix</term> for <m>T</m>,
which applies the transformation <m>T</m> where inputs and outputs are
all given in <m>\mathcal B</m>-coordinates.
</p>
</observation>

<activity>
<introduction>
<p>
Let <m>\mathcal{B}=\setList{\vec{v}_1,\vec{v}_2,\vec{v}_3}=\setList{\begin{bmatrix}1\\-2\\1\end{bmatrix},\begin{bmatrix}-1\\0\\3\end{bmatrix},\begin{bmatrix}0\\1\\-1\end{bmatrix}}</m> be basis from the previous Activity.
Let <m>\mathcal{B}=\setList{\vec{b}_1,\vec{b}_2,\vec{b}_3}=\setList{\begin{bmatrix}1\\-2\\1\end{bmatrix},\begin{bmatrix}-1\\0\\3\end{bmatrix},\begin{bmatrix}0\\1\\-1\end{bmatrix}}</m> be basis from the previous Activity.
Let <m>T</m> denote the linear transformation whose standard matrix is given by:
<me>A=\begin{bmatrix}9&amp;4&amp;4\\6&amp;9&amp;2\\-18&amp;-16&amp;-9\end{bmatrix}.</me>
</p>
Expand Down
65 changes: 65 additions & 0 deletions source/linear-algebra/source/04-MX/doenet/MX3-doenet-change-basis.xml
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@@ -0,0 +1,65 @@
<setup>
<boolean name="standardBasis">$basisChoice.selectedIndices = 1</boolean>
<point name="v">(-6,5)
<constraints>
<constrainToGrid/>
</constraints>
</point>
<point name="b2">(-1,3)
<constraints>
<constrainToGrid/>
</constraints>
</point>
<point name="b1">(4,1)
<constraints>
<constrainToGrid/>
</constraints>
</point>
<math name="x" simplify>($v.x*$b2.y-$v.y*$b2.x)/($b1.x*$b2.y-$b2.x*$b1.y)</math>
<math name="y" simplify>(-$v.x*$b1.y+$v.y*$b1.x)/($b1.x*$b2.y-$b2.x*$b1.y)</math>
</setup>

<p>
<choiceInput name="basisChoice" preselectChoice="1">
<choice>Standard basis {e₁,e₂}</choice>
<choice>Custom basis {b₁,b₂}</choice>
</choiceInput>
</p>

<graph xmin="-10" ymin="-10" xmax="10" ymax="10" displayXAxis="$standardBasis" displayYAxis="$standardBasis" displayXAxisTickLabels="false" displayYAxisTickLabels="false">
<!-- both views -->
<point coords="$v" styleNumber="1" hide="not $standardBasis">
<label hide="not $standardBasis"><m>\vec v=\left[\begin{array}{c} $v.x \\ $v.y \end{array}\right]</m></label>
</point>
<point coords="$v" styleNumber="1" hide="$standardBasis">
<label hide="$standardBasis"><m>\vec v=\left[\begin{array}{c} $x \\ $y \end{array}\right]_{\mathcal B}</m></label>
</point>
<lineSegment draggable="false" endpoints="(0,0) $v" styleNumber="1"/>
<point coords="$b1" styleNumber="2">
<label><m>\vec b_1</m></label>
</point>
<point coords="$b2" styleNumber="3">
<label><m>\vec b_2</m></label>
</point>
<!-- standard basis -->
<point draggable="false" hide="not $standardBasis" coords="($v.x,0)" styleNumber="1">
<label><m>$v.x\vec e_1</m></label>
</point>
<lineSegment draggable="false" hide="not $standardBasis" endpoints="($v.x,0) $v" styleNumber="6"/>
<point draggable="false" hide="not $standardBasis" coords="(0,$v.y)" styleNumber="1">
<label><m>$v.y\vec e_2</m></label>
</point>
<lineSegment draggable="false" hide="not $standardBasis" endpoints="($0,$v.y) $v" styleNumber="6"/>
<!-- nonstandard basis -->
<line draggable="false" through="(0,0) $b1" styleNumber="2" hide="$standardBasis"/>
<line draggable="false" through="(0,0) $b2" styleNumber="3" hide="$standardBasis"/>
<point draggable="false" hide="$standardBasis" coords="$x*$b1" styleNumber="1">
<label><m>$x\vec b_1</m></label>
</point>
<lineSegment draggable="false" hide="$standardBasis" endpoints="$x*$b1 $v" styleNumber="6"/>
<point draggable="false" hide="$standardBasis" coords="$y*$b2" styleNumber="1">
<label><m>$y\vec b_2</m></label>
</point>
<lineSegment draggable="false" hide="$standardBasis" endpoints="$y*$b2 $v" styleNumber="6"/>
</graph>

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