From f6ef825aaa67a66907321adbedc8d4606219c0cd Mon Sep 17 00:00:00 2001
From: Drew Lewis
Date: Tue, 12 May 2026 15:40:39 +0000
Subject: [PATCH 1/4] Typeset with dfrac
---
source/linear-algebra/source/03-AT/02.ptx | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/source/linear-algebra/source/03-AT/02.ptx b/source/linear-algebra/source/03-AT/02.ptx
index 2c1fec5ba..72fc78efc 100644
--- a/source/linear-algebra/source/03-AT/02.ptx
+++ b/source/linear-algebra/source/03-AT/02.ptx
@@ -752,7 +752,7 @@ T\left(\left[\begin{array}{c} -5 \\ 0 \\ -3 \\ -2 \end{array}\right]\right)
A rotation is given by \vec{v} \mapsto \left[\begin{array}{c} \cos(\theta)x - \sin(\theta)y\\ \cos(\theta)y + \sin(\theta)x\end{array}\right].
-A reflection of \vec{v} over a line l can be found by first finding a vector \vec{l} = \left[\begin{array}{c} l_x\\l_y\end{array}\right] along l, then \vec{v} \mapsto 2\frac{\vec{l}\cdot\vec{v}}{\vec{l}\cdot\vec{l}}\vec{l} - \vec{v}.
+A reflection of \vec{v} over a line l can be found by first finding a vector \vec{l} = \left[\begin{array}{c} l_x\\l_y\end{array}\right] along l, then \vec{v} \mapsto 2 \left(\dfrac{\vec{l}\cdot\vec{v}}{\vec{l}\cdot\vec{l}}\right) \vec{l} - \vec{v}.
Represent each of the following transformations with respect to the standard basis in \mathbb{R}^2.
From 858386683066373ab3ab3fd5779a8982dc730392 Mon Sep 17 00:00:00 2001
From: Drew Lewis
Date: Tue, 12 May 2026 15:42:06 +0000
Subject: [PATCH 2/4] Section 3
---
source/linear-algebra/source/03-AT/03.ptx | 4 ++--
1 file changed, 2 insertions(+), 2 deletions(-)
diff --git a/source/linear-algebra/source/03-AT/03.ptx b/source/linear-algebra/source/03-AT/03.ptx
index 79efb9eae..c9ff00589 100644
--- a/source/linear-algebra/source/03-AT/03.ptx
+++ b/source/linear-algebra/source/03-AT/03.ptx
@@ -84,7 +84,7 @@ 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 kernelkernelof T
+identity (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
@@ -107,7 +107,7 @@ is an important subspace of V defined by
\draw (-2,0) -- (2,0);
\draw (0,-2) -- (0,2);
\fill[blue] (0,0) circle (0.2)
- node[anchor=south east] {\(\vec{0}\)};
+ node[anchor=south east] {\(\vec{z}\)};
\end{scope}
\end{tikzpicture}
From 7fc0b477765b5ecedb333d9c5b34fa094b611f1a Mon Sep 17 00:00:00 2001
From: Drew Lewis
Date: Tue, 12 May 2026 15:47:34 +0000
Subject: [PATCH 3/4] Section 4
---
source/linear-algebra/source/03-AT/04.ptx | 6 +++---
1 file changed, 3 insertions(+), 3 deletions(-)
diff --git a/source/linear-algebra/source/03-AT/04.ptx b/source/linear-algebra/source/03-AT/04.ptx
index 50689ca60..61284e73c 100644
--- a/source/linear-algebra/source/03-AT/04.ptx
+++ b/source/linear-algebra/source/03-AT/04.ptx
@@ -1000,7 +1000,7 @@ Which of the following must be true?
-
- The system of equations given by [A|\vec{0}] has a unique solution.
+ The system of equations given by [A\,|\,\vec{0}] has a unique solution.
@@ -1046,7 +1046,7 @@ Which of the following must be true?
-
- The system of equations given by [A|\vec{b}] is always consistent.
+ The system of equations given by [A\,|\,\vec{b}] is always consistent.
@@ -1108,7 +1108,7 @@ Which of the following must be true?
Start with an n-dimensional vector space V. We can define the dual of V, denoted V^*, by
V^* = \{h:V \rightarrow \mathbb{R}: h \mbox{ is linear}\}.
-Prove that V is isomorphic toV^*. Here are some things to think about as you work through this.
+Prove that V is isomorphic to V^*. Here are some things to think about as you work through this.
- Start by assuming you have a basis for V. How many basis vectors should you have?
- For each basis vector in V, define a function that returns 1 if it's given that basis vector, and returns 0 if it's given any other basis vector. For example, if \vec{b_i} and \vec{b_j} are each members of the basis for V, and you'll need a function f_i:V \rightarrow \{0,1\}, where f_i(b_i) = 1 and f_i(b_j)= 0 for all j \neq i.
From a6622409526f545730543eaef230213ccab443bc Mon Sep 17 00:00:00 2001
From: Drew Lewis
Date: Tue, 12 May 2026 15:48:20 +0000
Subject: [PATCH 4/4] Section 6
---
source/linear-algebra/source/03-AT/06.ptx | 16 ++++++++--------
1 file changed, 8 insertions(+), 8 deletions(-)
diff --git a/source/linear-algebra/source/03-AT/06.ptx b/source/linear-algebra/source/03-AT/06.ptx
index 5d86e7e48..246c9a1b2 100644
--- a/source/linear-algebra/source/03-AT/06.ptx
+++ b/source/linear-algebra/source/03-AT/06.ptx
@@ -312,7 +312,7 @@ Consider the matrix space M_{2,2}=\left\{\left[\begin{array}{cc}
-Consider polynomial space \P^4=\left\{a+by+cy^2+dy^3+ey^4\middle| a,b,c,d,e\in\IR\right\} and the following set:
+Consider polynomial space \P_4=\left\{a+by+cy^2+dy^3+ey^4\middle| a,b,c,d,e\in\IR\right\} and the following set:
S=\setList{1,y,y^2,y^3,y^4}.
@@ -320,7 +320,7 @@ S=\setList{1,y,y^2,y^3,y^4}.
- Does the set S span \P^4?
+
Does the set S span \P_4?
-
@@ -339,7 +339,7 @@ S=\setList{1,y,y^2,y^3,y^4}.
-
- Yes; every polynomial in \P^4 is a linear combination of the polynomials in S.
+ Yes; every polynomial in \P_4 is a linear combination of the polynomials in S.
@@ -384,15 +384,15 @@ S=\setList{1,y,y^2,y^3,y^4}.
- S is linearly independent
- - S spans \P^4
- - S is a basis of \P^4
+ - S spans \P_4
+ - S is a basis of \P_4
- What is the dimension of \P^4?
+
What is the dimension of \P_4?
- 2
- 3
@@ -405,7 +405,7 @@ S=\setList{1,y,y^2,y^3,y^4}.
- Which Euclidean space is \P^4 isomorphic to?
+ Which Euclidean space is \P_4 isomorphic to?
- \IR^2
- \IR^3
@@ -418,7 +418,7 @@ S=\setList{1,y,y^2,y^3,y^4}.
- Describe an isomorphism T:\P^4\to\IR^{\unknown}:
+ Describe an isomorphism T:\P_4\to\IR^{\unknown}:
T\left(a+by+cy^2+dy^3+ey^4\right)=\left[\begin{array}{c}
\unknown\\\\\vdots\\\\\unknown