From a8c06d21c5613b4d61ac9186ea8fb98b1b946944 Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Wed, 25 Feb 2026 19:04:53 +0000 Subject: [PATCH 1/2] Recharacterize EV4 for arbitrary vectors (not just zero vector). --- source/linear-algebra/source/02-EV/04.ptx | 280 +++++++++++++++------- 1 file changed, 188 insertions(+), 92 deletions(-) diff --git a/source/linear-algebra/source/02-EV/04.ptx b/source/linear-algebra/source/02-EV/04.ptx index b4c37ceda..0850aeb2f 100644 --- a/source/linear-algebra/source/02-EV/04.ptx +++ b/source/linear-algebra/source/02-EV/04.ptx @@ -162,137 +162,216 @@

- Let \vec w = 3\vec v_1 - \vec v_2 - 5 \vec v_3 = \left[\begin{array}{c}\unknown \\ \unknown \\ \unknown\end{array}\right]. - The set \{\vec v_1,\vec v_2,\vec v_3,\vec w\} is... + To solve the vector equation x_1\vec v_1 + x_2\vec v_2 = \vec v_3 , we could calculate +\RREF\left[\begin{array}{cc|c} +-2 & 1 & -2 \\ +0 & 3 & 0 \\ +-2 & 5 & 4 +\end{array}\right]= +\left[\begin{array}{cc|c} +1 & 0 & 0 \\ +0 & 1 & 0 \\ +0 & 0 & 1 +\end{array}\right] + . + What can we conclude?

    -

  1. linearly dependent: at least one vector is a linear combination of others

    2. -

  2. linearly independent: no vector is a linear combination of others

    3. +

  3. The equation is consistent, so \vec v_3 is a linear combination of \vec v_1,\vec v_2

    4. +

  4. The equation is consistent, so \vec v_3 is not a linear combination of \vec v_1,\vec v_2

    5. +

  5. The equation has no solutions, so \vec v_3 is a linear combination of \vec v_1,\vec v_2

    6. +

  6. The equation has no solutions, so \vec v_3 is not a linear combination of \vec v_1,\vec v_2

 -

A.

+

D.

-\vec w is a linear combination of the others, so the set is dependent. +The third row requires 0=1, a contradiction. Thus the equation has no +solutions, which means there is no linear combination of \vec v_1,\vec v_2 +which equals \vec v_3

- +

- Find \RREF \left[\begin{array}{ccc|c} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w \\ - \end{array}\right]= - \RREF \left[\begin{array}{ccc|c} - -2 & 1 &-2 & \unknown \\ - 0 & 3 & 5 & \unknown \\ - 0 &0 &4 & \unknown - \end{array}\right]= \unknown . + We can similarly show \vec v_2 is not a linear combination of + \vec v_1,\vec v_3, and \vec v_3 is not a linear combination + of \vec v_1,\vec v_2. + Therefore, the set \{\vec v_1,\vec v_2,\vec v_3\} is linearly... +

    +

  1. + independent, because one of the vectors cannot be written as a + linear combination of the others.

    2. +

  2. + independent, because none of the vectors can be written as a + linear combination of the others.

    3. +

  3. + dependent, because one of the vectors can be written as a + linear combination of the others.

    4. +

  4. + dependent, because every vector can be written as a + linear combination of the others.

    5. +

-

- What does this tell you about solution set for the vector equation - x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 =\vec w? + + +

D.

+

+The third row requires 0=1, a contradiction. Thus the equation has no +solutions, which means there is no linear combination of \vec v_1,\vec v_2 +which equals \vec v_3 +

+ + + + +

+ Let \vec w = 3\vec v_1 - \vec v_2 - 2 \vec v_3 = \left[\begin{array}{c}-3 \\ -13 \\ -8 \end{array}\right]. + The set \{\vec v_1,\vec v_2,\vec v_3,\vec w\} is linearly...

    -
  1. -

    - It is inconsistent. -

    -
    2. -
  2. -

    - It is consistent with one solution. -

    -
    3. -
  3. -

    - It is consistent with infinitely many solutions. -

    -
    4. +

  4. + independent, because one of the vectors cannot be written as a + linear combination of the others.

    5. +

  5. + independent, because none of the vectors can be written as a + linear combination of the others.

    6. +

  6. + dependent, because one of the vectors can be written as a + linear combination of the others.

    7. +

  7. + dependent, because every vector can be written as a + linear combination of the others.

 -

B.

-

Each variable is set equal to a specific value.

+

A.

+

+\vec w is a linear combination of the others, so the set is dependent. +

- Find \RREF \left[\begin{array}{cccc|c} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w & \vec 0\\ - \end{array}\right]= + To decide whether \{\vec v_1,\vec v_2,\vec v_3,\vec v_4\} spans + \mathbb R^4, we would consider how to solve the vector equation x_1\vec v_1 + x_2\vec v_2 +x_3 \vec v_3+x_4\vec w= \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right] for some arbitrary vector \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]: \RREF \left[\begin{array}{cccc|c} - -2 & 1 &-2 & \unknown & 0\\ - 0 & 3 & 5 & \unknown & 0 \\ - 0 &0 &4 & \unknown & 0 - \end{array}\right]= \unknown . + -2 & 1 &-2 & -3 & \unknown \\ + 0 & 3 & 5 & -13 & \unknown \\ + 0 &0 &4 & -8 & \unknown + \end{array}\right]= + \left[\begin{array}{cccc|c} + 1 & 0 &0 & 3 & \unknown \\ + 0 & 1 & 0 & -1 & \unknown \\ + 0 &0 &1 & 2 & \unknown + \end{array}\right] .

- What does this tell you about solution set for the vector equation - x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec w=\vec{0}? + What does this tell you about solution set for the vector equation?

  1. - It is inconsistent. + It is inconsistent, so the set does not span \mathbb R^4.

  2. - It is consistent with one solution. + It is consistent with one solution, so the set spans \mathbb R^4.

  3. - It is consistent with infinitely many solutions. + It is consistent with infinitely many solutions, so the + set spans \mathbb R^4.

- +

C.

-

\vec w's column is not a pivot, revealing a free variable and infinitely-many solutions.

+

The non-pivot column guarantees a free variable which could equal + any real number.

 - + + + +

+We can also use this same equation to explain why +\{\vec v_1,\vec v_2,\vec v_3,\vec w\} +is linearly dependent! The RREF matrix guarantees that \vec w is +a linear combination of the other vectors because it guarantees the matrix +will have... +

    +
  1. +

    +a pivot row +

    +
    2. +
  2. +

    +a non-pivot row +

    +
    3. +
  3. +

    +a pivot column +

    +
    4. +
  4. +

    +a non-pivot column +

    +
    5. +
+

+ + +

D.

+

+When we rewrite the matrix like so: + \RREF \left[\begin{array}{ccc|c} + -2 & 1 &-2 & -3 \\ + 0 & 3 & 5 & -13 \\ + 0 &0 &4 & -8 + \end{array}\right]= + \left[\begin{array}{ccc|c} + 1 & 0 &0 & 3 \\ + 0 & 1 & 0 & -1 \\ + 0 &0 &1 & 2 + \end{array}\right] +we see that a non-pivot column demonstrates how to construct +its vector as a linear combination of the others. +

+ +

-It follows that \{\vec v_1,\vec v_2,\vec v_3,\vec w\} -is linearly dependent because x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec w=\vec{0} -had the number of solutions you found in the previous task. -Which feature of which RREF matrix best reveals this? +Now, if all we knew was that the RREF matrix for the equation + x_1\vec v_1 + x_2\vec v_2 +x_3 \vec v_3+x_4\vec w= \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right] +had this one feature, then we would know this about its solutions:

  1. -The pivot column in the augmented matrix -\RREF \left[\begin{array}{ccc|c} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w -\end{array}\right]. +It must have a unique solution.

  2. -The pivot column in the coefficient matrix -\RREF \left[\begin{array}{cccc} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w -\end{array}\right]. +It must have infinitely-many solutions.

  3. -The non-pivot column in the augmented matrix -\RREF \left[\begin{array}{ccc|c} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w -\end{array}\right]. +It's possible for a solution to the equation to be unique.

  4. -The non-pivot column in the coefficient matrix -\RREF \left[\begin{array}{cccc} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w -\end{array}\right]. +It's impossible for a solution to the equation to be unique.

@@ -301,18 +380,37 @@ The non-pivot column in the coefficient matrix

D.

-\RREF \left[\begin{array}{cccc} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w -\end{array}\right] (or \RREF \left[\begin{array}{cccc|c} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w & \vec 0 -\end{array}\right]) can be used to solve -x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec w=\vec{0}, -and the non-pivot column reveals it has infinitely-many solutions. +The free variable guarantees that, as long as there are no contradictions, +solutions are not unique.

+ + + + +

+How would this change if \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right] was the vector with all zeros? +

    +
  1. +

    +The equation would have no solutions. +

    +
    2. +
  2. +

    +The equation must have a unique solution. +

    +
    3. +
  3. +

    +The equation must have infinitely-many solutions. +

    +
    4. +

    + + +

    C.

    -Note that (C) technically represents the equation -x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 =\vec w -which isn't what was asked about. +There cannot be contradictions, so its solutions are not unique.

    @@ -324,20 +422,18 @@ which isn't what was asked about.

    - For any vector space, - the set \{\vec v_1,\dots\vec v_n\} is linearly dependent if and only - if the vector equation x_1\vec v_1+ x_2 \vec v_2+\dots+x_n\vec v_n=\vec{0} is consistent with - infinitely many solutions. +The set \{\vec v_1,\dots,\vec v_n\} is linearly dependent +exactly when the vector equation + x_1 \vec{v}_1 + \cdots + x_n\vec{v}_n = \vec{w} +has infinitely-many solutions for some vector \vec{w}\in\IR^m +(in particular, \vec w=\vec 0).

    - Likewise, the set of vectors - \{\vec v_1,\dots\vec v_n\} is linearly independent - if and only the vector equation x_1\vec v_1+ x_2 \vec v_2 + \cdots + x_n\vec v_n = \vec{0} - has exactly one solution: \left[\begin{array}{c} - x_1 \\ \vdots \\ x_n - \end{array}\right]=\left[\begin{array}{c} - 0 \\ \vdots \\ 0 - \end{array}\right]. +On the other hand, +the set \{\vec v_1,\dots,\vec v_n\} is linearly independent +exactly when the vector equation + x_1 \vec{v}_1 + \cdots + x_n\vec{v}_n = \vec{w} +has at most one solution for every vector \vec{w}\in\IR^m.

    From d9689fec74d199539746501085fbb531f20862ac Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Wed, 25 Feb 2026 19:12:55 +0000 Subject: [PATCH 2/2] add zero vector observation --- source/linear-algebra/source/02-EV/04.ptx | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/source/linear-algebra/source/02-EV/04.ptx b/source/linear-algebra/source/02-EV/04.ptx index 0850aeb2f..bbf58ddbf 100644 --- a/source/linear-algebra/source/02-EV/04.ptx +++ b/source/linear-algebra/source/02-EV/04.ptx @@ -433,7 +433,8 @@ On the other hand, the set \{\vec v_1,\dots,\vec v_n\} is linearly independent exactly when the vector equation x_1 \vec{v}_1 + \cdots + x_n\vec{v}_n = \vec{w} -has at most one solution for every vector \vec{w}\in\IR^m. +has at most one solution for every vector \vec{w}\in\IR^m +(and exactly one for \vec w=\vec 0).