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…tcome and Activities for Change-of-Basis #457
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<?xml version='1.0' encoding='UTF-8'?>
<section xml:id="GT5" xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Change of Basis (GT5)</title>
<objectives>
<ul>
<li>
<p>
Calculate the change-of-basis matrix between two bases of <m>\IR^n</m>.
</p>
</li>
</ul>
</objectives>

<subsection>
<title>Warm Up</title>

</subsection>

<subsection><title>Class Activities</title>
<remark>
<p>
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>
</remark>
<activity>
<introduction>
<p>
Let <m>\cal{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>.
</p>
</introduction>
<task>
<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>
</ol>
</p>
</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 have 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>
</p>
</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>
</task>

</activity>

<observation>
<p>
Note that one way to compute <m>M_{\mathcal{B}}</m> is calculate the RREF of the following matrix:
<me>\left[\begin{array}{ccc|ccc}1&amp;1&amp;0&amp;1&amp;0&amp;0\\
0&amp;-1&amp;1&amp;0&amp;1&amp;0\\
1&amp;1&amp;1&amp;0&amp;0&amp;1\end{array}\right]\sim\left[\begin{array}{ccc|ccc}1&amp;0&amp;0&amp;2&amp;1&amp;-1\\
0&amp;1&amp;0&amp;-1&amp;-1&amp;1\\
0&amp;0&amp;1&amp;-1&amp;0&amp;1\end{array}\right]</me>
Thus, the matrix <m>M_{\mathcal{B}}</m> is the inverse of the matrix <m>[\vec{v}_1\ \vec{v}_2\ \vec{v}_3]</m>.
That is:
<me>M_{\mathcal{B}}^{-1}=[\vec{v}_1\ \vec{v}_2\ \vec{v}_3].</me>
</p>
</observation>
<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}}^{-1}=[\vec{v}_1\ \cdots\ \vec{v}_n].</me>
</p>
<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>.
</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>
</p>
</statement>
</definition>


<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>
</p>
</introduction>
<task>
<p>
Calculate <m>M_{\mathcal{B}}</m> using technology.
</p>
<sage language="octave">
<input>

</input>
<output>

</output>
</sage>
</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>.
</p>
</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>.
</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>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>
</introduction>
<task>
<p>
Calculate the matrix <m>M_\mathcal{B}AM_{\mathcal{B}}^{-1}</m>.
</p>
<sage language="octave">
<input>

</input>
<output>

</output>
</sage>
</task>
<task>
<p>
The matrix <m>A</m> describes how the standard basis of <m>\IR^3</m> is transformed by the linear transformation <m>T</m>.
The matrix <m>M_\mathcal{B}AM_{\mathcal{B}}^{-1}</m> describes how <m>\cal{B}</m> is transformed (in <m>\mathcal{B}</m>-coordinates).
Which of these two descriptions is easier for you to describe/understand/visualize and why?
</p>
</task>
</activity>


</subsection>

<subsection>
<title>Sample Problem and Solution</title>
<example>
<introduction>
<p>
Let <m>\mathcal{B}=\setList{\begin{bmatrix}-2\\-2\\1\end{bmatrix},\begin{bmatrix}-1\\-2\\-1\end{bmatrix},\begin{bmatrix}1\\3\\2\end{bmatrix}}</m>, and <m>\vec{v}=\begin{bmatrix}1\\2\\3\end{bmatrix}</m>.
</p>
<task>
<p>
Explain and demonstrate how to verify that <m>\mathcal{B}</m> is a basis of <m>\IR^3</m> and how to calculate <m>M_\mathcal{B}</m>, the change-of-basis matrix from the standard basis of <m>\IR^3</m> to <m>\mathcal{B}</m>.
</p>
</task>
<task>
<p>
Explain and demonstrate how to use <m>M_\mathcal{B}</m> to express <m>\vec{v}</m> in terms of <m>\mathcal{B}</m>-basis vectors.
</p>
</task>
</introduction>
<solution>
<task>
<p>
We can accomplish both tasks by calculating the RREF of the following matrix:
<me>\RREF\left[\begin{array}{ccc|ccc}-2&amp;-1&amp;1&amp;1&amp;0&amp;0\\
-2&amp;-2&amp;3&amp;0&amp;1&amp;0\\1&amp;-1&amp;2&amp;0&amp;0&amp;1\end{array}\right]
=
\left[\begin{array}{ccc|ccc}1&amp;0&amp;0&amp;1&amp;-1&amp;1\\
0&amp;1&amp;0&amp;-7&amp;5&amp;-4\\
0&amp;0&amp;1&amp;-4&amp; 3&amp;-2\end{array}\right].</me>
The fact that the matrix to the left of the vertical bar is the identity matrix tells that <m>\mathcal{B}</m> is a basis.
The matrix on the right hand side of the bar is equal to the change-of-basis matrix:
<me>M_\mathcal{B}=\left[\begin{array}{ccc}1&amp;-1&amp;1\\
-7&amp;5&amp;-4\\
-4&amp; 3&amp;-2\end{array}\right].</me>
</p>
</task>
<task>
<p>
By definition of the change of basis matrix, if <m>\vec{v}=\begin{bmatrix}1\\2\\3\end{bmatrix}</m>, then the coordinates of <m>\vec{v}</m> with respect to <m>\mathcal{B}</m> are given by:
<me>M_\mathcal{B}\vec{v}=M_\mathcal{B}=\left[\begin{array}{ccc}1&amp;-1&amp;1\\
-7&amp;5&amp;-4\\
-4&amp; 3&amp;-2\end{array}\right]\begin{bmatrix}1\\2\\3\end{bmatrix}=\begin{bmatrix}2\\-9\\-4\end{bmatrix}.</me>
It follows that:
<me>\begin{bmatrix}1\\2\\3\end{bmatrix}=2\begin{bmatrix}-2\\-2\\1\end{bmatrix} -9\begin{bmatrix}-1\\-2\\-1\end{bmatrix} -4\begin{bmatrix}1\\3\\2\end{bmatrix}. </me>
</p>
</task>
</solution>
</example>
</subsection>

</section>
2 changes: 1 addition & 1 deletion source/linear-algebra/source/05-GT/main.ptx
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<xi:include href="02.ptx"/>
<xi:include href="03.ptx"/>
<xi:include href="04.ptx"/>
<!-- <xi:include href="05.ptx"/> -->
<xi:include href="05.ptx"/>
</chapter>

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