From 668836f79bff8c78390d60d16f3c11d91cb1c068 Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Fri, 20 Feb 2026 10:32:23 -0600 Subject: [PATCH] Align EV7 activities with EV6 --- source/linear-algebra/source/02-EV/07.ptx | 129 +++++++++++++--------- 1 file changed, 74 insertions(+), 55 deletions(-) diff --git a/source/linear-algebra/source/02-EV/07.ptx b/source/linear-algebra/source/02-EV/07.ptx index b88e7d610..362ab312e 100644 --- a/source/linear-algebra/source/02-EV/07.ptx +++ b/source/linear-algebra/source/02-EV/07.ptx @@ -125,49 +125,75 @@ Find its solution set (a subspace of \IR^4).

-Rewrite this solution space in the form \setBuilder{ a \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right] + b \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right] }{a,b \in \IR}. +Rewrite this solution space in the following forms +\setBuilder{ a \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right] + b \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right] }{a,b \in \IR}= +\vspan \left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right]\right\}.

-Which of these choices best describes the set of two vectors -\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right]\right\} -used in this solution space? +So, how can we use this to find a basis for the solution space?

  1. - The set is linearly dependent. +Take RREF of an appropriate matrix and remove the vectors that caused +a non-pivot row.

    -
    2. +

  2. - The set is linearly independent. +Take RREF of an appropriate matrix and remove the vectors that caused +a non-pivot column.

    -
    3. -
  3. -

    - The set spans the solution space. -

    -
    4. -
  4. +
    5. +

  5. - The set is a basis of the solution space. +Take RREF of an appropriate matrix and remove the vectors that caused +a non-pivot row or column.

    -
    6. + +
  6. +

    +This set cannot be a basis for the solution space because it will always +have in a zero row in the RREF. +

    +

- D. + B. +

+

+This is exactly the technique used in to find +a basis from a set of spanning vectors.

- - + +

+To find a basis for the subspace + +\vspan \left\{\left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right], \left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right]\right\} + +we may compute + +\left[\begin{array}{cc} -2 & -1 \\ 1 & 0 \\ 0 & -4 \\ 0 & 1\end{array}\right] \sim +\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0\end{array}\right] +. +Because all columns are pivot columns, the set + +\left\{\left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right], \left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right]\right\} + +is already linearly independent and therefore already a basis for the subspace. +

+ + + @@ -195,44 +221,18 @@ Find its solution set (a subspace of \IR^7).

-Rewrite this solution space in the form \setBuilder{ a \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right] + b \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]+c \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]+d \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right] }{a,b,c,d \in \IR}. +Rewrite this solution space in the following forms: \setBuilder{ a \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right] + b \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]+c \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]+d \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right] }{a,b,c,d \in \IR}. + +=\vspan\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right],\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right],\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]\right\} +.

-Which of these choices best describes the set of vectors -\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right],\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right],\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]\right\} -used in this solution space? -

    -
  1. -

    - The set is linearly dependent. -

    -
    2. -
  2. -

    - The set is linearly independent. -

    -
    3. -
  3. -

    - The set spans the solution space. -

    -
    4. -
  4. -

    - The set is a basis for the solution space. -

    -
    5. -
+Find a basis for this solution space.

- -

- D. -

- @@ -318,6 +318,10 @@ solution space?

A.

+

+In we established that sets that contain the zero +vector cannot be independent. +

@@ -333,31 +337,46 @@ To create a computer-animated film, an animator first models a scene as a subset of \mathbb R^3. Then to transform this three-dimensional visual data for display on a two-dimensional movie screen or television set, the computer could apply a linear transformation that maps visual information -at the point (x,y,z)\in\mathbb R^3 onto the pixel located at -(x+y,y-z)\in\mathbb R^2. +at the vector \left[\begin{array}{c}x\\y\\z\end{array}\right]\in\mathbb R^3 onto the pixel located at +\left[\begin{array}{c}x+y\\y-z\end{array}\right]\in\mathbb R^2.

-What homogeneous linear system describes the positions (x,y,z) +What homogeneous linear system describes the positions +\left[\begin{array}{c}x\\y\\z\end{array}\right] within the original scene that would be aligned with the -pixel (0,0) on the screen? +pixel \left[\begin{array}{c}0\\0\end{array}\right] on the screen?

+ +

x+y=0y-z=0

+

-Solve this system to describe these locations. +Find a basis for the solution set of this system. What best describes the +shape of the data that gets projected onto this single point of the screen? +

    +

  1. A point

    2. +

  2. A line

    3. +

  3. A plane

    4. +

+ +

B.

+

The basis \left\{\left[\begin{array}{c}-1\\1\\1\end{array}\right]\right\} describes a one-dimensional line.

+ +