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updates to MX3 (#949)
* updates to MX3 * add motivating activity for B-coordinate matrix * Apply suggestion from @StevenClontz
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‎source/linear-algebra/source/04-MX/03.ptx‎

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@@ -49,96 +49,159 @@
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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>.
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We can explore alternative perspectives more easily if we expand our toolkit to analyze different bases.
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</p>
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<figure xml:id="MX3-fig-doenet-change-basis">
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<caption>Visualization of the change of basis in <m>\mathbb R^2</m></caption>
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<interactive label="MX3-interactive-doenet-change-basis" platform="doenetml" width="100%">
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<slate surface="doenetml">
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<xi:include parse="text" href="doenet/MX3-doenet-change-basis.xml"/>
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</slate>
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<description>
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<p>An interactive that visualizes the change of basis from the standard basis to a custom basis in the plane.</p>
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</description>
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</interactive>
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</figure>
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</remark>
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<activity>
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<introduction>
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<p>
56-
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>.
67+
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>.
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</p>
58-
</introduction>
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</introduction>
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<task>
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<statement>
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<p>
61-
Is <m>\cal{B}</m> a basis of <m>\IR^3</m>?
62-
</p>
63-
<p>
64-
<ol marker="A.">
65-
<li>
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<p>
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Yes.
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</p>
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</li>
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<li>
71-
<p>
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No.
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</p>
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</li>
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Since <m>\mathcal{B}</m> is a basis, how many ways can we write some
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arbitrary <m>\vec v\in \IR^3</m> in terms of <m>\mathcal B</m> vectors?
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<md>
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\vec v= x_1\vec{b}_1+x_2\vec{b}_2+x_3\vec{b}_3=
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x_1\begin{bmatrix}1\\0\\1\end{bmatrix}+
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x_2\begin{bmatrix}1\\-1\\1\end{bmatrix}+
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x_3\begin{bmatrix}0\\1\\1\end{bmatrix}
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</md>
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<ol label="A." cols="3">
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<li><p>Zero</p></li>
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<li><p>At most one</p></li>
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<li><p>Exactly one</p></li>
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<li><p>At least one</p></li>
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<li><p>Infinitely-many</p></li>
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</ol>
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</p>
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</statement>
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<answer>
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<p>
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C.
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</p>
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<p>
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Exactly one
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</p>
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</answer>
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</task>
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<task>
79-
<p>
80-
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:
81-
<me>x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3=\vec{v}</me>
82-
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.
83-
The map <m>C_{\mathcal{B}}</m> is a linear map.
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</p>
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<p>
86-
Compute <m>C_{\mathcal{B}}\left(\begin{bmatrix}1\\1\\1\end{bmatrix}\right)</m>, the unique solution to
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<me>
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x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3=\begin{bmatrix}1\\1\\1\end{bmatrix}.
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</me>
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</p>
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<statement>
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<p>
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If <m>\vec x=\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]</m>
103+
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>,
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which of these matrix equations can be used to find <m>x_1,x_2,x_3</m>?
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<ol label="A." cols="3">
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<li><p><m>\vec v=B\vec x</m></p></li>
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<li><p><m>B\vec v=\vec x</m></p></li>
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<li><p><m>\vec v=B^{-1}\vec x</m></p></li>
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<li><p><m>B^{-1}\vec v=\vec x</m></p></li>
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<li><p>A or D</p></li>
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<li><p>B or C</p></li>
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</ol>
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</p>
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</statement>
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<answer>
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<p>
117+
E.
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</p>
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<p>
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<m>B\vec v=\vec x</m> can be rewritten as <m>\vec v=B^{-1}\vec x</m>
121+
</p>
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</answer>
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</task>
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<task>
94-
<p>
95-
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>.
96-
</p>
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<sage language="octave">
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<input>
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100-
</input>
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<output>
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</output>
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</sage>
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<statement>
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<p>
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Let <m>\vec v=\begin{bmatrix}1\\2\\3\end{bmatrix}</m> and
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use this equation to find
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<md>x_1=\unknown,x_2=\unknown,x_3=\unknown</md>.
130+
</p>
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</statement>
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<answer>
133+
<p>
134+
<m>x_1=1,x_2=0,x_3=2</m> so
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<md>\begin{bmatrix}1\\2\\3\end{bmatrix}=
136+
1\begin{bmatrix}1\\0\\1\end{bmatrix}+
137+
0\begin{bmatrix}1\\-1\\1\end{bmatrix}+
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2\begin{bmatrix}0\\1\\1\end{bmatrix}</md>
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</p>
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</answer>
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</task>
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</activity>
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109144
<definition xml:id="def-change-of-basis">
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<statement>
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<p>
112-
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:
113-
<me>x_1\vec{v}_1+\dots+x_n\vec{v}_n=\vec{v}.</me>
114-
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>.
115-
It satisfies the following:
116-
<me>M_{\mathcal{B}}=[\vec{v}_1\ \cdots\ \vec{v}_n]^{-1}.</me>
147+
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
148+
<m>\vec v</m> in terms of the basis, that is,
149+
the unique solution to the vector equation:
150+
<md>\vec v=x_1\vec{b}_1+\dots+x_n\vec{b}_n</md>.
151+
</p>
152+
<p>
153+
Since the solution vector
154+
<m>C_{\mathcal B}(\vec v)=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}</m>
155+
describes the <q><m>\mathcal B</m>-coordinates</q> of <m>\vec v</m>,
156+
we will write
157+
<md>\vec v=x_1\vec{b}_1+\dots+x_n\vec{b}_n=
158+
B\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}=
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\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}_{\mathcal B}</md>
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</p>
118161
</statement>
119162
</definition>
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121164
<remark>
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<p>
123-
The vector <m>C_\mathcal{B}(\vec{v})</m> is the <m>\mathcal{B}</m>-coordinates of <m>\vec{v}</m>.
124-
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>.
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As was just shown, the standard matrix <m>M_{\mathcal B}</m> for this
167+
transformation is exactly the inverse matrix <m>B^{-1}</m>.
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</p>
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<p>
127-
In particular, notation as above, we would have:
128-
<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>
170+
The vector <m>C_\mathcal{B}(\vec{v})</m> describes the
171+
<q><m>\mathcal{B}</m>-coordinates</q> of <m>\vec{v}</m>.
172+
If you work with standard coordinates, and I work with
173+
<m>\mathcal{B}</m>-coordinates, then you might write
174+
<md>\vec{v}=a_1\vec{e}_1+\cdots+a_n\vec{e}_n=\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}</md>
175+
and I might instead write
176+
<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>
177+
</p>
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<p>
179+
To convert from your standard coordinates to my <m>\mathcal B</m>-coordinates,
180+
we need simply compute:
181+
<md>\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}=
182+
C_{\mathcal B}\left(\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}\right)=
183+
M_{\mathcal B}\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}=
184+
B^{-1}\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}</md>.
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</p>
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<p>
187+
And similarly we can convert backwards:
188+
<md>\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}=
189+
C_{\mathcal B}^{-1}\left(\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}\right)=
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M_{\mathcal B}^{-1}\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}=
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B\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}</md>.
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</p>
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</remark>
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<activity>
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<introduction>
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<p>
136-
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>
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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>
137200
</p>
138201
</introduction>
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<task>
140203
<p>
141-
Calculate <m>M_{\mathcal{B}}</m> using technology.
204+
Calculate <m>M_{\mathcal{B}}=B^{-1}</m> using technology.
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</p>
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<sage language="octave">
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<input>
@@ -151,28 +214,96 @@
151214
</task>
152215
<task>
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<p>
154-
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>.
217+
Use this matrix to write
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<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>
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as a linear combination of <m>\vec{b}_1,\vec{b}_2,\vec{b}_3</m>.
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</p>
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</task>
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</activity>
223+
224+
<activity>
225+
<introduction>
226+
<p>
227+
While defining linear transformations in terms of their standard
228+
matrix <m>A</m> is convenient when working with standard coordinates, it
229+
would be helpful to be able to apply transformations directly to
230+
non-standard bases/coordinates as well.
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</p>
232+
<p>
233+
Let <m>\mathcal B=\setList{\vec b_1,\cdots\vec b_n}</m> be a basis, and consider the matrix
234+
<m>B=\begin{bmatrix}\vec b_1&amp;\cdots&amp;\vec b_n\end{bmatrix}</m>
235+
<m>M_{\mathcal B}=B^{-1}</m>.
236+
</p>
237+
</introduction>
238+
<task>
239+
<statement>
240+
<p>
241+
Given <m>\vec v</m> representing <m>\mathcal B</m>-coordinates, which of these
242+
expressions would correctly compute the transformation of <m>\vec v</m> by a
243+
<em>standard</em> matrix <m>A</m> with output in <em>standard</em> coordinates?
244+
<ol marker="A.">
245+
<li><p><m>AB^{-1}\vec v=AM_{\mathcal B}\vec v</m></p></li>
246+
<li><p><m>AB\vec v=AM_{\mathcal B}^{-1}\vec v</m></p></li>
247+
<li><p><m>B^{-1}A\vec v=M_{\mathcal B}A\vec v</m></p></li>
248+
<li><p><m>BA\vec v=M_{\mathcal B}^{-1}A\vec v</m></p></li>
249+
</ol>
250+
</p>
251+
</statement>
252+
<answer>
253+
<p>
254+
B.
255+
</p>
256+
<p>
257+
Since <m>\vec v</m> represents <m>\mathcal B</m>-coordinates,
258+
the vector <m>B\vec v=M_{\mathcal B}^{-1}\vec v</m> represents
259+
its standard coordinates.
260+
</p>
261+
</answer>
262+
</task>
263+
<task>
264+
<statement>
265+
<p>
266+
Therefore, which matrix would directly calculate the transformation of
267+
vectors by a linear map with standard matrix <m>A</m>,
268+
but where inputs and outputs are all given in <m>\mathcal B</m>-coordinates?
269+
<ol marker="A.">
270+
<li><p><m>B^{-1}AB=M_{\mathcal B}AM_{\mathcal B}^{-1}</m></p></li>
271+
<li><p><m>BAB^{-1}=M_{\mathcal B}^{-1}AM_{\mathcal B}</m></p></li>
272+
<li><p><m>AB^{-1}B=AM_{\mathcal B}M_{\mathcal B}^{-1}</m></p></li>
273+
<li><p><m>ABB^{-1}=AM_{\mathcal B}^{-1}M_{\mathcal B}</m></p></li>
274+
</ol>
275+
</p>
276+
</statement>
277+
<answer>
278+
<p>
279+
A.
280+
</p>
281+
<p>
282+
We saw <m>AB</m> transforms <m>\mathcal B</m>-coordinates by the
283+
transformation, but outputs standard coordinates. Applying
284+
<m>B^{-1}=M_{\mathcal B}</m> on the left corrects the outputs to be
285+
in <m>\mathcal B</m>-coordinates.
286+
</p>
287+
</answer>
288+
</task>
289+
</activity>
290+
158291
<observation>
159292
<p>
160293
Let <m>T\colon\IR^n\to\IR^n</m> be a linear transformation and let <m>A</m> denote its standard matrix.
161-
If <m>\cal{B}=\setList{\vec{v}_1,\dots, \vec{v}_n}</m> is some other basis, then we have:
162-
<md>
163-
<mrow>M_\mathcal{B}AM_{\mathcal{B}}^{-1} \amp= M_\mathcal{B}A[\vec{v_1}\cdots\vec{v}_n] </mrow>
164-
<mrow> \amp= M_\mathcal{B}[T(\vec{v}_1)\cdots T(\vec{v}_n)]</mrow>
165-
<mrow> \amp= [C_\mathcal{B}(T(\vec{v}_1))\cdots C_\mathcal{B}(T(\vec{v}_n))]</mrow>
166-
</md>
167-
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>.
168-
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>.
294+
If <m>\mathcal{B}=\setList{\vec{b}_1,\dots, \vec{v}_n}</m> is some other basis
295+
and <m>B=\begin{bmatrix}\vec b_1&amp;\cdots&amp;\vec b_n\end{bmatrix}</m>,
296+
then <m>M_{\mathcal B}AM_{\mathcal B}^{-1}=B^{-1}AB</m> is the
297+
<term><m>\mathcal B</m>-coordinate matrix</term> for <m>T</m>,
298+
which applies the transformation <m>T</m> where inputs and outputs are
299+
all given in <m>\mathcal B</m>-coordinates.
169300
</p>
170301
</observation>
171302

172303
<activity>
173304
<introduction>
174305
<p>
175-
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.
306+
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.
176307
Let <m>T</m> denote the linear transformation whose standard matrix is given by:
177308
<me>A=\begin{bmatrix}9&amp;4&amp;4\\6&amp;9&amp;2\\-18&amp;-16&amp;-9\end{bmatrix}.</me>
178309
</p>

‎source/linear-algebra/source/04-MX/doenet/MX3-doenet-change-basis.xml‎

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<setup>
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<boolean name="standardBasis">$basisChoice.selectedIndices = 1</boolean>
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<point name="v">(-6,5)
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<constraints>
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<constrainToGrid/>
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</constraints>
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</point>
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<point name="b2">(-1,3)
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<constraints>
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<constrainToGrid/>
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</constraints>
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</point>
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<point name="b1">(4,1)
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<constraints>
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<constrainToGrid/>
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</constraints>
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</point>
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<math name="x" simplify>($v.x*$b2.y-$v.y*$b2.x)/($b1.x*$b2.y-$b2.x*$b1.y)</math>
19+
<math name="y" simplify>(-$v.x*$b1.y+$v.y*$b1.x)/($b1.x*$b2.y-$b2.x*$b1.y)</math>
20+
</setup>
21+
22+
<p>
23+
<choiceInput name="basisChoice" preselectChoice="1">
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<choice>Standard basis {e₁,e₂}</choice>
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<choice>Custom basis {b₁,b₂}</choice>
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</choiceInput>
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</p>
28+
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<graph xmin="-10" ymin="-10" xmax="10" ymax="10" displayXAxis="$standardBasis" displayYAxis="$standardBasis" displayXAxisTickLabels="false" displayYAxisTickLabels="false">
30+
<!-- both views -->
31+
<point coords="$v" styleNumber="1" hide="not $standardBasis">
32+
<label hide="not $standardBasis"><m>\vec v=\left[\begin{array}{c} $v.x \\ $v.y \end{array}\right]</m></label>
33+
</point>
34+
<point coords="$v" styleNumber="1" hide="$standardBasis">
35+
<label hide="$standardBasis"><m>\vec v=\left[\begin{array}{c} $x \\ $y \end{array}\right]_{\mathcal B}</m></label>
36+
</point>
37+
<lineSegment draggable="false" endpoints="(0,0) $v" styleNumber="1"/>
38+
<point coords="$b1" styleNumber="2">
39+
<label><m>\vec b_1</m></label>
40+
</point>
41+
<point coords="$b2" styleNumber="3">
42+
<label><m>\vec b_2</m></label>
43+
</point>
44+
<!-- standard basis -->
45+
<point draggable="false" hide="not $standardBasis" coords="($v.x,0)" styleNumber="1">
46+
<label><m>$v.x\vec e_1</m></label>
47+
</point>
48+
<lineSegment draggable="false" hide="not $standardBasis" endpoints="($v.x,0) $v" styleNumber="6"/>
49+
<point draggable="false" hide="not $standardBasis" coords="(0,$v.y)" styleNumber="1">
50+
<label><m>$v.y\vec e_2</m></label>
51+
</point>
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<lineSegment draggable="false" hide="not $standardBasis" endpoints="($0,$v.y) $v" styleNumber="6"/>
53+
<!-- nonstandard basis -->
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<line draggable="false" through="(0,0) $b1" styleNumber="2" hide="$standardBasis"/>
55+
<line draggable="false" through="(0,0) $b2" styleNumber="3" hide="$standardBasis"/>
56+
<point draggable="false" hide="$standardBasis" coords="$x*$b1" styleNumber="1">
57+
<label><m>$x\vec b_1</m></label>
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</point>
59+
<lineSegment draggable="false" hide="$standardBasis" endpoints="$x*$b1 $v" styleNumber="6"/>
60+
<point draggable="false" hide="$standardBasis" coords="$y*$b2" styleNumber="1">
61+
<label><m>$y\vec b_2</m></label>
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</point>
63+
<lineSegment draggable="false" hide="$standardBasis" endpoints="$y*$b2 $v" styleNumber="6"/>
64+
</graph>
65+

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