diff --git a/source/linear-algebra/source/05-GT/05.ptx b/source/linear-algebra/source/05-GT/05.ptx new file mode 100644 index 00000000..cf647fd0 --- /dev/null +++ b/source/linear-algebra/source/05-GT/05.ptx @@ -0,0 +1,233 @@ + +
+ Change of Basis (GT5) + + + + + + Warm Up + + + +Class Activities + +

+ So far, when working with the Euclidean vector space \IR^n, we have primarily worked with the standard basis \mathcal{E}=\setList{\vec{e}_1,\dots, \vec{e}_n}. + We can explore alternative perspectives more easily if we expand our toolkit to analyze different bases. +

+ + + +

+ Let \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}}. +

+ + +

+ Is \cal{B} a basis of \IR^3? +

+

+

    +
  1. +

    + Yes. +

    +
    2. +
  2. +

    + No. +

    +
    3. +
+

+ + +

+ Since \cal{B} is a basis, we know that if \vec{v}\in \IR^3, the following vector equation have will have a unique solution: + x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3=\vec{v} + Given this, we define a map C_{\mathcal{B}}\colon\IR^3\to\IR^3 via the rule that C_{\mathcal{B}}(\vec{v}) is equal to the unique solution to the above vector equation. + The map C_{\mathcal{B}} is a linear map. +

+

+ Compute C_{\mathcal{B}}\left(\begin{bmatrix}1\\1\\1\end{bmatrix}\right). +

+ + +

+ Compute C_\mathcal{B}(\vec{e}_1),C_\mathcal{B}(\vec{e}_2), C_\mathcal{B}(\vec{e}_3) and, in doing so, write down the standard matrix M_\mathcal{B} of C_\mathcal{B}. +

+ + + + + + + + + + + + + +

+ Note that one way to compute M_{\mathcal{B}} is calculate the RREF of the following matrix: + \left[\begin{array}{ccc|ccc}1&1&0&1&0&0\\ + 0&-1&1&0&1&0\\ + 1&1&1&0&0&1\end{array}\right]\sim\left[\begin{array}{ccc|ccc}1&0&0&2&1&-1\\ + 0&1&0&-1&-1&1\\ + 0&0&1&-1&0&1\end{array}\right] + Thus, the matrix M_{\mathcal{B}} is the inverse of the matrix [\vec{v}_1\ \vec{v}_2\ \vec{v}_3]. + That is: + M_{\mathcal{B}}^{-1}=[\vec{v}_1\ \vec{v}_2\ \vec{v}_3]. +

+ + + +

+ Given a basis \cal{B}=\setList{\vec{v}_1,\dots, \vec{v}_n} of \IR^n, the change of basis/coordinate transformation from the standard basis to \mathcal{B} is the transformation C_\mathcal{B}\colon\IR^n\to\IR^n defined by the property that, for any vector \vec{v}\in\IR^n, the vector C_\mathcal{B}(\vec{v}) is the unique solution to the vector equation: + x_1\vec{v}_1+\dots+x_n\vec{v}_n=\vec{v}. + Its standard matrix is called the change-of-basis matrix from the standard basis to \mathcal{B} and is denoted by M_{\mathcal{B}}. + It satisfies the following: + M_{\mathcal{B}}^{-1}=[\vec{v}_1\ \cdots\ \vec{v}_n]. +

+

+ The vector C_\mathcal{B}(\vec{v}) is the \mathcal{B}-coordinates of \vec{v}. + If you work with standard coordinates, and I work with \mathcal{B}-coordinates, then to build the vector that you call \vec{v}=\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}=a_1\vec{e}_1+\cdots+a_n\vec{e}_n, I would first compute C_\mathcal{B}(\vec{v})=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix} and then build \vec{v}=x_1\vec{v}_1+\cdots+x_n\vec{v}_n. +

+

+ In particular, notation as above, we would have: + a_1\vec{e}_1+\cdots a_n\vec{e}_n=\vec{v}=x_1\vec{v}_1+\cdots+x_n\vec{v}_n. +

+ + + + + + +

+ Let \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}, and \mathcal{B}=\setList{\vec{v}_1,\vec{v}_2,\vec{v}_3} +

+ + +

+ Calculate M_{\mathcal{B}} using technology. +

+ + + + + + + + + + +

+ Use your result to calculate C_\mathcal{B}\left(\begin{bmatrix}1\\1\\1\end{bmatrix}\right) and express the vector \begin{bmatrix}1\\1\\1\end{bmatrix} as a linear combination of \vec{v}_1,\vec{v}_2,\vec{v}_3. +

+ + + +

+ Let T\colon\IR^n\to\IR^n be a linear transformation and let A denote its standard matrix. + If \cal{B}=\setList{\vec{v}_1,\dots, \vec{v}_n} is some other basis, then we have: + + M_\mathcal{B}AM_{\mathcal{B}}^{-1} \amp= M_\mathcal{B}A[\vec{v_1}\cdots\vec{v}_n] + \amp= M_\mathcal{B}[T(\vec{v}_1)\cdots T(\vec{v}_n)] + \amp= [C_\mathcal{B}(T(\vec{v}_1))\cdots C_\mathcal{B}(T(\vec{v}_n))] + + In other words, the matrix M_{\mathcal{B}}AM_{\mathcal{B}}^{-1} is the matrix whose columns consist of \mathcal{B}-coordinate vectors of the image vectors T(\vec{v}_i). + The matrix M_{\mathcal{B}}AM_{\mathcal{B}}^{-1} is called the matrix of T with respect to \mathcal{B}-coordinates. +

+ + + + +

+ Let \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}} be basis from the previous Activity. + Let T denote the linear transformation whose standard matrix is given by: + A=\begin{bmatrix}9&4&4\\6&9&2\\-18&-16&-9\end{bmatrix}. +

+ + +

+ Calculate the matrix M_\mathcal{B}AM_{\mathcal{B}}^{-1}. +

+ + + + + + + + + + +

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

+ + + + + + + + Sample Problem and Solution + + +

+ Let \mathcal{B}=\setList{\begin{bmatrix}-2\\-2\\1\end{bmatrix},\begin{bmatrix}-1\\-2\\-1\end{bmatrix},\begin{bmatrix}1\\3\\2\end{bmatrix}}, and \vec{v}=\begin{bmatrix}1\\2\\3\end{bmatrix}. +

+ +

+ Explain and demonstrate how to verify that \mathcal{B} is a basis of \IR^3 and how to calculate M_\mathcal{B}, the change-of-basis matrix from the standard basis of \IR^3 to \mathcal{B}. +

+ + +

+ Explain and demonstrate how to use M_\mathcal{B} to express \vec{v} in terms of \mathcal{B}-basis vectors. +

+ + + + +

+ We can accomplish both tasks by calculating the RREF of the following matrix: + \RREF\left[\begin{array}{ccc|ccc}-2&-1&1&1&0&0\\ + -2&-2&3&0&1&0\\1&-1&2&0&0&1\end{array}\right] + = + \left[\begin{array}{ccc|ccc}1&0&0&1&-1&1\\ + 0&1&0&-7&5&-4\\ + 0&0&1&-4& 3&-2\end{array}\right]. + The fact that the matrix to the left of the vertical bar is the identity matrix tells that \mathcal{B} is a basis. + The matrix on the right hand side of the bar is equal to the change-of-basis matrix: + M_\mathcal{B}=\left[\begin{array}{ccc}1&-1&1\\ + -7&5&-4\\ + -4& 3&-2\end{array}\right]. +

+ + +

+ By definition of the change of basis matrix, if \vec{v}=\begin{bmatrix}1\\2\\3\end{bmatrix}, then the coordinates of \vec{v} with respect to \mathcal{B} are given by: + M_\mathcal{B}\vec{v}=M_\mathcal{B}=\left[\begin{array}{ccc}1&-1&1\\ + -7&5&-4\\ + -4& 3&-2\end{array}\right]\begin{bmatrix}1\\2\\3\end{bmatrix}=\begin{bmatrix}2\\-9\\-4\end{bmatrix}. + It follows that: + \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}. +

+ + + + + +
\ No newline at end of file diff --git a/source/linear-algebra/source/05-GT/main.ptx b/source/linear-algebra/source/05-GT/main.ptx index 4cf21668..299537a5 100644 --- a/source/linear-algebra/source/05-GT/main.ptx +++ b/source/linear-algebra/source/05-GT/main.ptx @@ -7,5 +7,5 @@ - +