diff --git a/source/linear-algebra/source/03-AT/03.ptx b/source/linear-algebra/source/03-AT/03.ptx index 26e2dd151..3b4fe43d2 100644 --- a/source/linear-algebra/source/03-AT/03.ptx +++ b/source/linear-algebra/source/03-AT/03.ptx @@ -50,8 +50,11 @@ Let T: \IR^2 \rightarrow \IR^3 be given by \text{with standard matrix } \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] -Which of these subspaces of \IR^2 describes -the set of all vectors that transform into \vec 0? +Which of these sets contain all vectors \left[\begin{array}{c}x \\ y \end{array}\right] +that transform into T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right) + = + \left[\begin{array}{c} x \\ y \\ 0 \end{array}\right]= + \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]?

  1. @@ -82,12 +85,12 @@ the set of all vectors that transform into \vec 0?

    -Let T: V \rightarrow W be a linear transformation, and let \vec{z} be the additive -identity (the zero vector) of W. The kernelkernel of T +Let T: V \rightarrow W be a linear transformation, and let \vec{0} be the zero vector of W. +The kernelkernel of T (also known as the null spacenull space of T) -is an important subspace of V defined by +is an important subset of V defined by -\ker T = \left\{ \vec{v} \in V\ \big|\ T(\vec{v})=\vec{z}\right\} +\ker T = \left\{ \vec{x} \in V\ \big|\ T(\vec{x})=\vec{0}\right\}

    @@ -128,8 +131,13 @@ Let T: \IR^3 \rightarrow \IR^2 be given by \text{with standard matrix } \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] -Which of these subspaces of \IR^3 describes \ker T, -the set of all vectors that transform into \vec 0? +Which of these subsets of \IR^3 is equal to +\ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\right\}, +the set of all vectors \left[\begin{array}{c}x \\ y \\ z\end{array}\right] +that transform into T\left(\left[\begin{array}{c}x \\ y \\ z \end{array}\right] \right) + = + \left[\begin{array}{c} x \\ y\end{array}\right]= + \left[\begin{array}{c} 0 \\ 0 \end{array}\right]?

    1. @@ -161,28 +169,101 @@ the set of all vectors that transform into \vec 0?

      Let T: \IR^3 \rightarrow \IR^2 be the linear transformation given by the standard matrix -T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{c} 3x+4y-z \\ x+2y+z \end{array}\right] + + A=\left[\begin{array}{ccc} 3 & 4 & -1 \\ 1 & 2 & 1 \end{array}\right] + =\left[\begin{array}{ccc} T(\vec e_1) & T(\vec e_2) & T(\vec e_3)\end{array}\right]. + +

      - + + +

      +Which of these is the most appropriate method to determine whether a \mathbb R^3 vector +\vec{x}=\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=x_1\vec{e}_1+x_2\vec{e}_2+x_3\vec{e}_3, +belongs to \ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\right\}? +

        +
      1. +

        + Determine if the set of vectors + \left\{T(\vec e_1), T(\vec e_2), T(\vec e_3), T(\vec x)\right\} + spans \IR^3. +

        +
        2. +
      2. +

        + Determine if the set of vectors + \left\{T(\vec e_1), T(\vec e_2), T(\vec e_3), T(\vec x)\right\} + is linearly dependent. +

        +
        3. +
      3. +

        + Determine if + \vec x=\left[\begin{array}{c} x_1 \\ x_2\\ x_3\end{array}\right] + belongs to the solution set of the vector equation + T(\vec x)=T\left(\left[\begin{array}{c} x_1 \\ x_2\\ x_3\end{array}\right]\right)= + x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)= + \left[\begin{array}{c} 0 \\ 0\end{array}\right]. +

        +
        4. +
      4. +

        + Determine if the equation + T(\vec 0)=T\left(\left[\begin{array}{c} 0 \\ 0\\ 0\end{array}\right]\right)= + 0T(\vec e_1)+0T(\vec e_2)+0T(\vec e_3)= + T(\vec x) is consistent. +

        +
        5. +
      +

      +       + +

      + B. This is exactly the same as finding the solution space for the + homoegeneous vector equation T(\vec x)=\vec 0. +

      +       +       + + +

      + Use this method to find the kernel of T. +

      + +

      + \ker T=\left\{\left[\begin{array}{c}3a\\-2a\\a\end{array}\right]\middle| + a\in\IR\right\} +

      +       +       +       + + + + + +

      -Set - - T\left(\left[\begin{array}{c}x\\y\\z\end{array}\right]\right) - = - \left[\begin{array}{c}0\\0\end{array}\right] - to find a linear system of equations whose solution set is the kernel. + The kernel of a transformation T + is exactly the solution space of + the homogeneous equation T(\vec{x})=\vec{0}. + If its standard matrix is A, then we may write + A\vec x=\vec 0 and use \RREF[A\,|\,\vec 0] to + find this kernel.

      -       -

      -Use \RREF(A) to solve this homogeneous system of equations and find a basis -for the kernel of T. + In particular, the kernel is a subspace of the transformation's + domain, and has a basis which may be found as in + : + + \ker T=\left\{\left[\begin{array}{c}3a\\-2a\\a\end{array}\right]\middle| + a\in\IR\right\} \hspace{2em} + \text{Basis for }\ker T=\left\{\left[\begin{array}{c}3\\-2\\1\end{array}\right]\right\}. +

      -       - - - + + @@ -245,9 +326,9 @@ Which of these subspaces of \IR^3 describes the set of all vectors that a

      Let T: V \rightarrow W be a linear transformation. The imageimage of T -is an important subspace of W defined by +is an important subset of W defined by -\Im T = \left\{ \vec{w} \in W\ \big|\ \text{there is some }\vec v\in V \text{ with } T(\vec{v})=\vec{w}\right\} +\Im T = \left\{ T(\vec v) \in W\ \big| \vec v\in V \right\}

      @@ -323,7 +404,7 @@ Let T: \IR^3 \rightarrow \IR^2 be given by \text{with standard matrix } \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] -Which of these subspaces of \IR^2 describes \Im T, +Which of these subsets of \IR^2 describes \Im T = \left\{ T(\vec v) \in \IR^2\ \big| \vec v\in \IR^3 \right\}, the set of all vectors that are the result of using T to transform \IR^3 vectors?

      @@ -366,9 +447,30 @@ the set of all vectors that are the result of using T to transform

      Consider the question: Which vectors \vec{w} in \IR^3 belong to - \Im T? + \Im T=\left\{ T(\vec v) \in \IR^3\ \big| \vec v\in \IR^4 \right\}?

      + + +

      + Recall that + + T\left(\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right] \right) = + x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)+x_4T(\vec e_4) + . + Complete the following vector equation which must be consistent + in order for \left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right] + to belong to \Im T: + + x_1 \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]+ + x_2 \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]+ + x_3\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]+ + x_4\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]= + \left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right] + +

      +       +

      @@ -377,6 +479,15 @@ the set of all vectors that are the result of using T to transform

      + + +

      + Write down the vector equation which must be consistent in + order for \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] + to belong to \Im T. +

      +       +

      @@ -390,12 +501,13 @@ the set of all vectors that are the result of using T to transform

      An arbitrary vector \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right] belongs to \Im T provided the equation -x_1 T(\vec{e}_1)+x_2 T(\vec{e}_2)+x_3T(\vec{e}_3)+x_4T(\vec{e}_4)=\vec{w} has... +x_1 T(\vec{e}_1)+x_2 T(\vec{e}_2)+x_3T(\vec{e}_3)+x_4T(\vec{e}_4)= +\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right] has...

      1. no solutions.
        2. +
      2. at most one solution.
      3. exactly one solution.
      4. at least one solution.
        5. -
      5. infinitely-many solutions.

       @@ -403,46 +515,68 @@ An arbitrary vector \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{ar

      - Based on this, how do \Im T and \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} relate to each other? + Based on this, what do we know?

        -
      1. The set \Im T contains \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} but is not equal to it.
        2. -
      2. The set \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} contains \Im T but is not equal to it.
        3. +
      3. + The set \left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} spans + \Im T whenever + \left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} + is linearly independent. +
        4. +
      4. + The set \Im T equals the codomain whenever + \left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} + is linearly independent. +
        5. +
      5. + The set \Im T is simply \{\vec 0\} whenever + \left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} + is linearly dependent. +
      6. The set \Im T and \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} are equal to each other.
        7. -
      7. There is no relation between these two sets.

             - +

      -Let T: \IR^4 \rightarrow \IR^3 be the linear transformation given by the +Let T: \IR^n \rightarrow \IR^m be the linear transformation given by the standard matrix A = - \left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right] + \left[\begin{array}{ccc} T(\vec e_1) & \cdots & T(\vec e_n) \end{array}\right] .

      -Since the set - - \setList{ - \left[\begin{array}{c}3\\-1\\2\end{array}\right], - \left[\begin{array}{c}4\\1\\1\end{array}\right], - \left[\begin{array}{c}7\\0\\3\end{array}\right], - \left[\begin{array}{c}1\\2\\-1\end{array}\right] +Then we have + + \Im T=\vspan\setList{ + T(\vec e_1), + \cdots, + T(\vec e_n) } - -spans \Im T, we can obtain a basis for \Im T by finding - - \RREF A + +and in particular, see that \Im T is a subspace of the codomain, and thus +has a kernel which may be found as in : +given + + \RREF \left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right] = - \left[\begin{array}{cccc} 1 & 0 & 1 & -1\\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] - -and only using the vectors corresponding to pivot columns: + \left[\begin{array}{cccc} \markedPivot{1} & 0 & 1 & -1\\ 0 & \markedPivot{1} & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] + +we see that we have +\Im T=\vspan\left\{ + \left[\begin{array}{c} 3 \\ -1 \\ 2 \end{array}\right], + \left[\begin{array}{c} 4 \\ 1 \\ 1 \end{array}\right], + \left[\begin{array}{c} 7\\ 0 \\ 3 \end{array}\right], + \left[\begin{array}{c} 1\\ 2 \\ -1 \end{array}\right], + \right\} + and +\text{Basis for }\Im T= \setList{ \left[\begin{array}{c}3\\-1\\2\end{array}\right], \left[\begin{array}{c}4\\1\\1\end{array}\right] @@ -450,9 +584,7 @@ and only using the vectors corresponding to pivot columns:

      - In general, the column spacecolumn space of a matrix M refers to the subspace obtained by considering the span of its column vectors. - Using this terminology, if the transformation T is represented by the matrix A, - then the image of T is the column spacecolumn space of A. +This justifies why the image of a transformation is also called the column spacecolumn space of its standard matrix.

      @@ -460,20 +592,20 @@ and only using the vectors corresponding to pivot columns:

      -Let T:\IR^n\to\IR^m be a linear transformation with standard matrix A. +To summarize, +let T:\IR^n\to\IR^m be a linear transformation with standard matrix A.

      • - The kernel of T is the solution set of the homogeneous system given -by the augmented matrix \left[\begin{array}{c|c}A&\vec 0\end{array}\right]. -Use the coefficients of its free variables to get a basis for the kernel (as in ). + The kernel of T is the solution set of the equation T(\vec{x})=\vec{0} or, equivalently, the linear system given by the augmented matrix \left[\begin{array}{c|c}A&\vec 0\end{array}\right]. + Use the coefficients of its free variables to get a basis for the kernel (as in ).

      • - The image of T is the span of the columns of A. Remove -the vectors creating non-pivot columns in \RREF A to get a basis -for the image (as in ). + The image of T is the span of the columns of its standard matrix A, + a.k.a. its column space. + Remove the vectors creating non-pivot columns in \RREF A to get a basis for the image (as in ).