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Recharacterize EV4 for arbitrary vectors (not just zero vector). #941

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a8c06d2
Recharacterize EV4 for arbitrary vectors (not just zero vector).
StevenClontz Feb 25, 2026
d9689fe
add zero vector observation
StevenClontz Feb 25, 2026
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281 changes: 189 additions & 92 deletions source/linear-algebra/source/02-EV/04.ptx
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Original file line number Diff line number Diff line change
Expand Up @@ -162,137 +162,216 @@
<task>
<statement>
<p>
Let <m> \vec w = 3\vec v_1 - \vec v_2 - 5 \vec v_3 = \left[\begin{array}{c}\unknown \\ \unknown \\ \unknown\end{array}\right]</m>.
The set <m>\{\vec v_1,\vec v_2,\vec v_3,\vec w\}</m> is...
To solve the vector equation <m> x_1\vec v_1 + x_2\vec v_2 = \vec v_3 </m>, we could calculate <md>
\RREF\left[\begin{array}{cc|c}
-2 &amp; 1 &amp; -2 \\
0 &amp; 3 &amp; 0 \\
-2 &amp; 5 &amp; 4
\end{array}\right]=
\left[\begin{array}{cc|c}
1 &amp; 0 &amp; 0 \\
0 &amp; 1 &amp; 0 \\
0 &amp; 0 &amp; 1
\end{array}\right]
</md>.
What can we conclude?
<ol marker="A.">
<li><p>linearly dependent: at least one vector is a linear combination of others</p></li>
<li><p>linearly independent: no vector is a linear combination of others</p></li>
<li><p>The equation is consistent, so <m>\vec v_3</m> is a linear combination of <m>\vec v_1,\vec v_2</m></p></li>
<li><p>The equation is consistent, so <m>\vec v_3</m> is <em>not</em> a linear combination of <m>\vec v_1,\vec v_2</m></p></li>
<li><p>The equation has no solutions, so <m>\vec v_3</m> is a linear combination of <m>\vec v_1,\vec v_2</m></p></li>
<li><p>The equation has no solutions, so <m>\vec v_3</m> is <em>not</em> a linear combination of <m>\vec v_1,\vec v_2</m></p></li>
</ol>
</p>
</statement>
<answer>
<p>A.</p>
<p>D.</p>
<p>
<m>\vec w</m> is a linear combination of the others, so the set is dependent.
The third row requires <m>0=1</m>, a contradiction. Thus the equation has no
solutions, which means there is no linear combination of <m>\vec v_1,\vec v_2</m>
which equals <m>\vec v_3</m>
</p>
</answer>
</task>
<task>
<task>
<statement>
<p>
Find <me>\RREF \left[\begin{array}{ccc|c}
\vec v_1 &amp; \vec v_2 &amp; \vec v_3 &amp; \vec w \\
\end{array}\right]=
\RREF \left[\begin{array}{ccc|c}
-2 &amp; 1 &amp;-2 &amp; \unknown \\
0 &amp; 3 &amp; 5 &amp; \unknown \\
0 &amp;0 &amp;4 &amp; \unknown
\end{array}\right]= \unknown .</me>
We can similarly show <m>\vec v_2</m> is not a linear combination of
<m>\vec v_1,\vec v_3</m>, and <m>\vec v_3</m> is not a linear combination
of <m>\vec v_1,\vec v_2</m>.
Therefore, the set <m>\{\vec v_1,\vec v_2,\vec v_3\}</m> is linearly...
<ol marker="A.">
<li><p>
<em>independent</em>, because one of the vectors cannot be written as a
linear combination of the others.</p></li>
<li><p>
<em>independent</em>, because none of the vectors can be written as a
linear combination of the others.</p></li>
<li><p>
<em>dependent</em>, because one of the vectors can be written as a
linear combination of the others.</p></li>
<li><p>
<em>dependent</em>, because every vector can be written as a
linear combination of the others.</p></li>
</ol>
</p>
<p>
What does this tell you about solution set for the vector equation
<m>x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 =\vec w</m>?
</statement>
<answer>
<p>D.</p>
<p>
The third row requires <m>0=1</m>, a contradiction. Thus the equation has no
solutions, which means there is no linear combination of <m>\vec v_1,\vec v_2</m>
which equals <m>\vec v_3</m>
</p>
</answer>
</task>
<task>
<statement>
<p>
Let <m> \vec w = 3\vec v_1 - \vec v_2 - 2 \vec v_3 = \left[\begin{array}{c}-3 \\ -13 \\ -8 \end{array}\right]</m>.
The set <m>\{\vec v_1,\vec v_2,\vec v_3,\vec w\}</m> is linearly...
<ol marker="A.">
<li>
<p>
It is inconsistent.
</p>
</li>
<li>
<p>
It is consistent with one solution.
</p>
</li>
<li>
<p>
It is consistent with infinitely many solutions.
</p>
</li>
<li><p>
<em>independent</em>, because one of the vectors cannot be written as a
linear combination of the others.</p></li>
<li><p>
<em>independent</em>, because none of the vectors can be written as a
linear combination of the others.</p></li>
<li><p>
<em>dependent</em>, because one of the vectors can be written as a
linear combination of the others.</p></li>
<li><p>
<em>dependent</em>, because every vector can be written as a
linear combination of the others.</p></li>
</ol>
</p>
</statement>
<answer>
<p>B.</p>
<p>Each variable is set equal to a specific value.</p>
<p>A.</p>
<p>
<m>\vec w</m> is a linear combination of the others, so the set is dependent.
</p>
</answer>
</task>
<task>
<statement>
<p>
Find <me>\RREF \left[\begin{array}{cccc|c}
\vec v_1 &amp; \vec v_2 &amp; \vec v_3 &amp; \vec w &amp; \vec 0\\
\end{array}\right]=
To decide whether <m>\{\vec v_1,\vec v_2,\vec v_3,\vec v_4\}</m> spans
<m>\mathbb R^4</m>, we would consider how to solve the vector equation <m> 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]</m> for some arbitrary vector <m>\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]</m>: <md>
\RREF \left[\begin{array}{cccc|c}
-2 &amp; 1 &amp;-2 &amp; \unknown &amp; 0\\
0 &amp; 3 &amp; 5 &amp; \unknown &amp; 0 \\
0 &amp;0 &amp;4 &amp; \unknown &amp; 0
\end{array}\right]= \unknown .</me>
-2 &amp; 1 &amp;-2 &amp; -3 &amp; \unknown \\
0 &amp; 3 &amp; 5 &amp; -13 &amp; \unknown \\
0 &amp;0 &amp;4 &amp; -8 &amp; \unknown
\end{array}\right]=
\left[\begin{array}{cccc|c}
1 &amp; 0 &amp;0 &amp; 3 &amp; \unknown \\
0 &amp; 1 &amp; 0 &amp; -1 &amp; \unknown \\
0 &amp;0 &amp;1 &amp; 2 &amp; \unknown
\end{array}\right] </md>.
</p>
<p>
What does this tell you about solution set for the vector equation
<m>x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec w=\vec{0}</m>?
What does this tell you about solution set for the vector equation?
<ol marker="A.">
<li>
<p>
It is inconsistent.
It is inconsistent, so the set does not span <m>\mathbb R^4</m>.
</p>
</li>
<li>
<p>
It is consistent with one solution.
It is consistent with one solution, so the set spans <m>\mathbb R^4</m>.
</p>
</li>
<li>
<p>
It is consistent with infinitely many solutions.
It is consistent with infinitely many solutions, so the
set spans <m>\mathbb R^4</m>.
</p>
</li>
</ol>
</p>
</statement>
</statement>
<answer>
<p>C.</p>
<p><m>\vec w</m>'s column is not a pivot, revealing a free variable and infinitely-many solutions.</p>
<p>The non-pivot column guarantees a free variable which could equal
any real number.</p>
</answer>
</task>
</task>
<task>
<statement>
<p>
We can <em>also</em> use this same equation to explain why
<m>\{\vec v_1,\vec v_2,\vec v_3,\vec w\}</m>
is linearly dependent! The RREF matrix guarantees that <m>\vec w</m> is
a linear combination of the other vectors because it guarantees the matrix
will have...
<ol marker="A.">
<li>
<p>
a pivot row
</p>
</li>
<li>
<p>
a non-pivot row
</p>
</li>
<li>
<p>
a pivot column
</p>
</li>
<li>
<p>
a non-pivot column
</p>
</li>
</ol>
</p>
</statement>
<answer>
<p>D.</p>
<p>
When we rewrite the matrix like so:<md>
\RREF \left[\begin{array}{ccc|c}
-2 &amp; 1 &amp;-2 &amp; -3 \\
0 &amp; 3 &amp; 5 &amp; -13 \\
0 &amp;0 &amp;4 &amp; -8
\end{array}\right]=
\left[\begin{array}{ccc|c}
1 &amp; 0 &amp;0 &amp; 3 \\
0 &amp; 1 &amp; 0 &amp; -1 \\
0 &amp;0 &amp;1 &amp; 2
\end{array}\right] </md>
we see that a non-pivot column demonstrates how to construct
its vector as a linear combination of the others.
</p>
</answer>
</task>
<task>
<statement>
<p>
It follows that <m>\{\vec v_1,\vec v_2,\vec v_3,\vec w\}</m>
is linearly dependent because <m>x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec w=\vec{0}</m>
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
<m> 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]</m>
had this one feature, then we would know this about its solutions:
<ol marker="A.">
<li>
<p>
The <em>pivot</em> column in the <em>augmented</em> matrix
<m>\RREF \left[\begin{array}{ccc|c}
\vec v_1 &amp; \vec v_2 &amp; \vec v_3 &amp; \vec w
\end{array}\right]</m>.
It must have a unique solution.
</p>
</li>
<li>
<p>
The <em>pivot</em> column in the <em>coefficient</em> matrix
<m>\RREF \left[\begin{array}{cccc}
\vec v_1 &amp; \vec v_2 &amp; \vec v_3 &amp; \vec w
\end{array}\right]</m>.
It must have infinitely-many solutions.
</p>
</li>
<li>
<p>
The <em>non-pivot</em> column in the <em>augmented</em> matrix
<m>\RREF \left[\begin{array}{ccc|c}
\vec v_1 &amp; \vec v_2 &amp; \vec v_3 &amp; \vec w
\end{array}\right]</m>.
It's possible for a solution to the equation to be unique.
</p>
</li>
<li>
<p>
The <em>non-pivot</em> column in the <em>coefficient</em> matrix
<m>\RREF \left[\begin{array}{cccc}
\vec v_1 &amp; \vec v_2 &amp; \vec v_3 &amp; \vec w
\end{array}\right]</m>.
It's impossible for a solution to the equation to be unique.
</p>
</li>
</ol>
Expand All @@ -301,18 +380,37 @@ The <em>non-pivot</em> column in the <em>coefficient</em> matrix
<answer>
<p>D.</p>
<p>
<m>\RREF \left[\begin{array}{cccc}
\vec v_1 &amp; \vec v_2 &amp; \vec v_3 &amp; \vec w
\end{array}\right]</m> (or <m>\RREF \left[\begin{array}{cccc|c}
\vec v_1 &amp; \vec v_2 &amp; \vec v_3 &amp; \vec w &amp; \vec 0
\end{array}\right]</m>) can be used to solve
<m>x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec w=\vec{0}</m>,
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.
</p>
</answer>
</task>
<task>
<statement>
<p>
How would this change if <m>\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]</m> was the vector with all zeros?
<ol marker="A.">
<li>
<p>
The equation would have no solutions.
</p>
</li>
<li>
<p>
The equation must have a unique solution.
</p>
</li>
<li>
<p>
The equation must have infinitely-many solutions.
</p>
</li>
</p>
</statement>
<answer>
<p>C.</p>
<p>
Note that (C) technically represents the equation
<m>x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 =\vec w</m>
which isn't what was asked about.
There cannot be contradictions, so its solutions are not unique.
</p>
</answer>
</task>
Expand All @@ -324,20 +422,19 @@ which isn't what was asked about.
<fact>
<statement>
<p>
For any vector space,
the set <m>\{\vec v_1,\dots\vec v_n\}</m> is linearly dependent if and only
if the vector equation <m>x_1\vec v_1+ x_2 \vec v_2+\dots+x_n\vec v_n=\vec{0}</m> is consistent with
infinitely many solutions.
The set <m>\{\vec v_1,\dots,\vec v_n\}</m> is <em>linearly dependent</em>
exactly when the vector equation
<me> x_1 \vec{v}_1 + \cdots + x_n\vec{v}_n = \vec{w} </me>
has infinitely-many solutions for <em>some</em> vector <m>\vec{w}\in\IR^m</m>
(in particular, <m>\vec w=\vec 0</m>).
</p>
<p>
Likewise, the set of vectors
<m>\{\vec v_1,\dots\vec v_n\}</m> is linearly independent
if and only the vector equation <me>x_1\vec v_1+ x_2 \vec v_2 + \cdots + x_n\vec v_n = \vec{0}</me>
has exactly one solution: <m>\left[\begin{array}{c}
x_1 \\ \vdots \\ x_n
\end{array}\right]=\left[\begin{array}{c}
0 \\ \vdots \\ 0
\end{array}\right]</m>.
On the other hand,
the set <m>\{\vec v_1,\dots,\vec v_n\}</m> is <em>linearly independent</em>
exactly when the vector equation
<me> x_1 \vec{v}_1 + \cdots + x_n\vec{v}_n = \vec{w} </me>
has at most one solution for <em>every</em> vector <m>\vec{w}\in\IR^m</m>
(and exactly one for <m>\vec w=\vec 0</m>).
</p>
</statement>
</fact>
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