diff --git a/source/linear-algebra/source/04-MX/03.ptx b/source/linear-algebra/source/04-MX/03.ptx index e1547f6f1..92d6321ad 100644 --- a/source/linear-algebra/source/04-MX/03.ptx +++ b/source/linear-algebra/source/04-MX/03.ptx @@ -49,83 +49,146 @@ 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.

+
+ Visualization of the change of basis in \mathbb R^2 + + + + + +

An interactive that visualizes the change of basis from the standard basis to a custom basis in the plane.

+ + +

- Let \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}}. + Consider the non-standard basis \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}} for \mathbb R^3.

- + +

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

-

-

    -
  1. -

    - Yes. -

    -
    2. -
  2. -

    - No. -

    -
    3. + Since \mathcal{B} is a basis, how many ways can we write some + arbitrary \vec v\in \IR^3 in terms of \mathcal B vectors? + + \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} + +
      +

    1. Zero

      2. +

    2. At most one

      3. +

    3. Exactly one

      4. +

    4. At least one

      5. +

    5. Infinitely-many

    + + +

    + C. +

    +

    + Exactly one +

    +     -

    - Since \cal{B} is a basis, we know that if \vec{v}\in \IR^3, the following vector equation 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), the unique solution to - - x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3=\begin{bmatrix}1\\1\\1\end{bmatrix}. - - -

    + +

    + If \vec x=\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] + and B = \left[\begin{array}{ccc}\vec b_1& \vec b_2&\vec b_3\end{array}\right]=\left[\begin{array}{ccc}1&1&0\\0&-1&1\\1&1&1\end{array}\right], + which of these matrix equations can be used to find x_1,x_2,x_3? +

      +

    1. \vec v=B\vec x

      2. +

    2. B\vec v=\vec x

      3. +

    3. \vec v=B^{-1}\vec x

      4. +

    4. B^{-1}\vec v=\vec x

      5. +

    5. A or D

      6. +

    6. B or C

      7. +
    +

    +     + +

    + E. +

    +

    + B\vec v=\vec x can be rewritten as \vec v=B^{-1}\vec x +

    +     -

    - 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}. -

    - - - - - - - - + +

    + Let \vec v=\begin{bmatrix}1\\2\\3\end{bmatrix} and + use this equation to find + x_1=\unknown,x_2=\unknown,x_3=\unknown. +

    +     + +

    + x_1=1,x_2=0,x_3=2 so + \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} +

    +     -

    - 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}}=[\vec{v}_1\ \cdots\ \vec{v}_n]^{-1}. + Given a basis \mathcal{B}=\setList{\vec{b}_1,\dots, \vec{b}_n} of \IR^n and corresponding matrix B=\begin{bmatrix}\vec b_1&\cdots&\vec b_n\end{bmatrix}, 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}) describes the unique way to write + \vec v in terms of the basis, that is, + the unique solution to the vector equation: + \vec v=x_1\vec{b}_1+\dots+x_n\vec{b}_n. +

    +

    + Since the solution vector + C_{\mathcal B}(\vec v)=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix} + describes the \mathcal B-coordinates of \vec v, + we will write + \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}

    - 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. + As was just shown, the standard matrix M_{\mathcal B} for this + transformation is exactly the inverse matrix B^{-1}.

    - 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. + The vector C_\mathcal{B}(\vec{v}) describes the + \mathcal{B}-coordinates of \vec{v}. + If you work with standard coordinates, and I work with + \mathcal{B}-coordinates, then you might write + \vec{v}=a_1\vec{e}_1+\cdots+a_n\vec{e}_n=\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix} + and I might instead write + \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} +

    +

    + To convert from your standard coordinates to my \mathcal B-coordinates, + we need simply compute: + \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}. +

    +

    + And similarly we can convert backwards: + \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}.

    @@ -133,12 +196,12 @@

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

    - Calculate M_{\mathcal{B}} using technology. + Calculate M_{\mathcal{B}}=B^{-1} using technology.

    @@ -151,28 +214,96 @@

    - 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. + Use this matrix to write + \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 + as a linear combination of \vec{b}_1,\vec{b}_2,\vec{b}_3.

    + + + +

    + While defining linear transformations in terms of their standard + matrix A 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. +

    +

    +Let \mathcal B=\setList{\vec b_1,\cdots\vec b_n} be a basis, and consider the matrix +B=\begin{bmatrix}\vec b_1&\cdots&\vec b_n\end{bmatrix} +M_{\mathcal B}=B^{-1}. +

    +     + + +

    +Given \vec v representing \mathcal B-coordinates, which of these +expressions would correctly compute the transformation of \vec v by a +standard matrix A with output in standard coordinates? +

      +

    1. AB^{-1}\vec v=AM_{\mathcal B}\vec v

      2. +

    2. AB\vec v=AM_{\mathcal B}^{-1}\vec v

      3. +

    3. B^{-1}A\vec v=M_{\mathcal B}A\vec v

      4. +

    4. BA\vec v=M_{\mathcal B}^{-1}A\vec v

      5. +
    +

    +     + +

    + B. +

    +

    + Since \vec v represents \mathcal B-coordinates, + the vector B\vec v=M_{\mathcal B}^{-1}\vec v represents + its standard coordinates. +

    +     +     + + +

    +Therefore, which matrix would directly calculate the transformation of +vectors by a linear map with standard matrix A, +but where inputs and outputs are all given in \mathcal B-coordinates? +

      +

    1. B^{-1}AB=M_{\mathcal B}AM_{\mathcal B}^{-1}

      2. +

    2. BAB^{-1}=M_{\mathcal B}^{-1}AM_{\mathcal B}

      3. +

    3. AB^{-1}B=AM_{\mathcal B}M_{\mathcal B}^{-1}

      4. +

    4. ABB^{-1}=AM_{\mathcal B}^{-1}M_{\mathcal B}

      5. +
    +

    +     + +

    + A. +

    +

    + We saw AB transforms \mathcal B-coordinates by the + transformation, but outputs standard coordinates. Applying + B^{-1}=M_{\mathcal B} on the left corrects the outputs to be + in \mathcal B-coordinates. +

    +     +     +     +

    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. + If \mathcal{B}=\setList{\vec{b}_1,\dots, \vec{v}_n} is some other basis + and B=\begin{bmatrix}\vec b_1&\cdots&\vec b_n\end{bmatrix}, + then M_{\mathcal B}AM_{\mathcal B}^{-1}=B^{-1}AB is the + \mathcal B-coordinate matrix for T, + which applies the transformation T where inputs and outputs are + all given in \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 \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}} 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}.

    diff --git a/source/linear-algebra/source/04-MX/doenet/MX3-doenet-change-basis.xml b/source/linear-algebra/source/04-MX/doenet/MX3-doenet-change-basis.xml new file mode 100644 index 000000000..0efea36eb --- /dev/null +++ b/source/linear-algebra/source/04-MX/doenet/MX3-doenet-change-basis.xml @@ -0,0 +1,65 @@ + + $basisChoice.selectedIndices = 1 + (-6,5) + + + + + (-1,3) + + + + + (4,1) + + + + + ($v.x*$b2.y-$v.y*$b2.x)/($b1.x*$b2.y-$b2.x*$b1.y) + (-$v.x*$b1.y+$v.y*$b1.x)/($b1.x*$b2.y-$b2.x*$b1.y) + + +

    + + Standard basis {e₁,e₂} + Custom basis {b₁,b₂} + +

    + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + \ No newline at end of file