diff --git a/source/linear-algebra/source/03-AT/03.ptx b/source/linear-algebra/source/03-AT/03.ptx index c6e5ed791..aef8e9c31 100644 --- a/source/linear-algebra/source/03-AT/03.ptx +++ b/source/linear-algebra/source/03-AT/03.ptx @@ -159,35 +159,77 @@ 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] +Let T: \IR^3 \rightarrow \IR^2 be the linear transformation with the following standard matrix: +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 the following vectors is an element of \ker T? +

+
    +
  1. +

    + \left[\begin{array}{c}1\\1\\1\end{array}\right] +

    +
    2. +
  2. +

    + \left[\begin{array}{c}3\\-2\\1\end{array}\right] +

    +
    3. +
  3. +

    + \left[\begin{array}{c}4\\-3\\1\\1\end{array}\right] +

    +
    4. +
+ + + + +

+ In general, \ker T is the set of solutions to the equation: + + T\left(\vec{x}\right)=T\left(\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]\right)=\left[\begin{array}{c}0\\0\end{array}\right]. + + Write down an equivalent vector equation, solve it, and describe \ker T. +

+ + + + + + + +

-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\}. +

- - - - + +

-Let T: \IR^4 \rightarrow \IR^3 be the linear transformation given by +Let T: \IR^4 \rightarrow \IR^3 be the linear transformation whose standard matrix is: T\left(\left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right] \right) = \left[\begin{array}{c} 2x+4y+2z-4w \\ -2x-4y+z+w \\ 3x+6y-z-4w\end{array}\right].

@@ -372,23 +414,50 @@ the set of all vectors that are the result of using T to transform

- Determine if \left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right] belongs to - \Im T. + Which of the following statements is most helpful in deciding if \left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right] is an element of \Im T?

+
    +
  1. +

    + The equation T\left(\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right]\right)=\left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right] has infinitely many solutions. +

    +
    2. +
  2. +

    + The equation T\left(\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right]\right)=\left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right] has at least one solution. +

    +
    3. +
  3. +

    + The equation T\left(\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right]\right)=\left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right] has no solutions. +

    +
    4. +
  4. +

    + The equation T\left(\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right]\right)=\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right] has at least one solution. +

    +
    5. +
- -

- Determine if \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] belongs to - \Im T. -

- + +

+ Translate your choice into a statement about a specific vector equation and use it to determine whether or not the vector \left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right] is an element of \Im T. +

+ + + + +

+ Determine whether or not the vector \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] is an element of \Im T by analyzing an appropriate vector equation. +

+

-An arbitrary vector \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right] belongs to +In general, 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...