documents: add selected solutions

2024/documents/by-course/math-4a/selected-solutions/main.typ

@@ -0,0 +1,98 @@
+#import "@preview/unequivocal-ams:0.1.1": ams-article, theorem, proof
+
+#show: ams-article.with(title: "Exam Solutions")
+
+Q6.
+
+The general clockwise rotation matrix in $RR^2$ is
+
+$
+  mat(cos theta, sin theta; -sin theta, cos theta)
+$
+
+We have $theta = (3pi)/2$, and
+
+$ cos((3pi) / 2) = 0, sin((3pi) / 2) = -1 $
+
+So our particular rotation matrix is
+
+$
+  T = mat(0, -1; 1, 0)
+$
+
+Clearly, the linear transformation that reflects a vector across the vertical
+axis changes the first standard basis vector, $vec(1, 0)$, to $vec(-1, 0)$.
+
+This corresponds to the linear transformation (matrix)
+
+$
+  S = mat(-1, 0; 0, 1)
+$
+
+Matrix composition, $compose$, is an equivalent notion to matrix multiplication. Therefore, we have the two compositions.
+
++ #[
+    $T compose S$, the linear transformation corresponding to a reflection followed by rotation:
+
+    $
+      T compose S &= mat(0, -1; 1, 0) mat(-1, 0; 0, 1) \
+      &#[using _column by coordinate_ rule] \
+      &= mat((-1 vec(0,1) + 0 vec(-1,0)), (0 vec(0,1) + 1 vec(-1,0))) \
+      &= mat(0, -1,;-1, 0)
+    $
+  ]
+
++ #[
+    $S compose T$, the linear transformation corresponding to rotation, followed by reflection. Since matrix composition is generally not commutative, we obtain a different matrix.
+
+    $
+      S compose T &= mat(-1,0; 0,1) mat(0,-1; 1,0) \
+      &#[using _column by coordinate_ rule] \
+      &= mat((0 vec(-1, 0) + 1 vec(0, 1)), (-1 vec(-1, 0) + 0 vec(0, -1))) \
+      &= mat(0, 1; 1, 0)
+    $
+  ]
+
+#pagebreak()
+
+Q7.
+
+#enum(
+  numbering: "7.1",
+  [
+    The matrix $A$ corresponds to a linear transformation
+    $T: RR^4 |-> RR^3$. $A$ has 3 rows and 4 columns, so its matrix-vector
+    multiplication is only defined when with vectors in $RR^4$. Accordingly, it
+    will output a vector in $RR^3$.
+
+    So, $p = 4$.
+  ],
+  [
+    See above explanation. $q = 3$.
+
+  ],
+  [
+    To find all vectors $arrow(x) in RR^4$ whose image under $T$ is $arrow(b)$, we seek all solutions $arrow(x) = (x_1, x_2, x_3, x_4, x_5)^T$ to the equation
+    $ T arrow(x) = arrow(b) $
+
+    We can do this using our usual row reduction methods.
+
+    $
+      mat(augment: #(-1), -2,3,7,-11,-3;1,0,-2,1,3; 1,-1,-3,4,2) \
+      mat(augment: #(-1), 1,-1,-3,4,2; 0, 1, 1, -3, 1; 0, 0, 0, 0, 0)
+    $
+
+    We now have the augmented matrix in echelon form. So, $x_4$ and $x_3$ are free. Then, let $s,t in RR$ be free variables
+    $
+      x_1 &= 3 - t + 2s \
+      x_2 &= 1 + 3t - s \
+      x_3 &= s \
+      x_4 &= t
+    $
+    So, all vectors $arrow(x)$ are of the form
+
+    $
+      arrow(x) = vec(3+2s - t, 1 - s + 3t, s, t)
+    $
+  ],
+)

2024/documents/by-course/math-4a/selected-solutions/package.nix

@@ -0,0 +1,34 @@
+{
+  typstPackagesCache,
+  typixLib,
+  cleanTypstSource,
+  ...
+}:
+let
+  src = cleanTypstSource ./.;
+  commonArgs = {
+    typstSource = "main.typ";
+
+    fontPaths = [
+      # Add paths to fonts here
+      # "${pkgs.roboto}/share/fonts/truetype"
+    ];
+
+    virtualPaths = [
+      # Add paths that must be locally accessible to typst here
+      # {
+      #   dest = "icons";
+      #   src = "${inputs.font-awesome}/svgs/regular";
+      # }
+    ];
+
+    XDG_CACHE_HOME = typstPackagesCache;
+  };
+
+in
+typixLib.buildTypstProject (
+  commonArgs
+  // {
+    inherit src;
+  }
+)

2024/documents/by-course/math-4a/selected-solutions/refs.bib

@@ -0,0 +1,9 @@
+@article{netwok2020,
+  title={At-scale impact of the {Net Wok}: A culinarically holistic investigation of distributed dumplings},
+  author={Astley, Rick and Morris, Linda},
+  journal={Armenian Journal of Proceedings},
+  volume={61},
+  pages={192--219},
+  year=2020,
+  publisher={Automatic Publishing Inc.}
+}

2024/documents/by-name/digression-linear-algebra/main.typ

@@ -15,21 +15,22 @@
 
 = Introduction
 
-Many introductory linear algebra classes focus on _application_. In general,
+Many introductory linear algebra classes focus on _application_. They teach you
-this is a red herring and is engineer-speak for "we will teach you how to
+how to perform trivial numerical operations such as the _matrix
-crunch numbers with no regard for conceptual understanding."
+multiplication_, _matrix-vector multiplication_, _row reduction_, and other
+trite tasks better suited for computers.
+
+This class is essentially useless. Linear algebra is really a much deeper
+subject, when viewed through the lens of _linear maps_ and _vector spaces_. In
+particular, taking an abstract point-free approach allows the freedom to prove
+theorems that generalize to linear algebra on arbitrary vector spaces, and
+indeed, even infinite vector spaces.
 
-If you are a math major (or math-adjacent, such as Computer Science), this
-class is essentially useless for you. You will learn how to perform trivial
-numerical operations such as the _matrix multiplication_, _matrix-vector
-multiplication_, _row reduction_, and other trite tasks better suited for
-computers.
 
 If you are taking this course, you might as well learn linear algebra properly.
 Otherwise, you will have to re-learn it later on, anyways. Completing a math
-course without gaining a theoretical appreciation for the topics at hand is an
+course without gaining a theoretical appreciation for the topics at hand is a
-unequivocal waste of time. I have prepared this brief crash course designed to
+complete and utter waste of time.
-fill in the theoretical gaps left by this class.
 
 = Basic Notions
 
@@ -78,37 +79,42 @@ example, you cannot add a vector and scalar, as it does not make sense.
 
 _Remark_. For those of you versed in computer science, you may recognize this
 as essentially saying that you must ensure your operations are _type-safe_.
-Adding a vector and scalar is not just false, it is an _invalid question_
+Adding a vector and scalar is not just "wrong" in the same sense that $1 + 1 =
-entirely because vectors and scalars and different types of mathematical
+3$ is wrong, it is an _invalid question_ entirely because vectors and scalars
-objects. See #cite(<chen2024digression>, form: "prose") for more.
+and different types of mathematical objects. See #cite(<chen2024digression>,
+form: "prose") for more.
 
 === Vectors big and small
 
 In order to begin your descent into what mathematicians colloquially recognize
-as _abstract vapid nonsense_, let's discuss which fields constitute a vector space. We
+as _abstract vapid nonsense_, let's discuss which fields constitute a vector
-have the familiar space where all scalars are real numbers, or $RR$. We
+space. We have the familiar field of $RR$ where all scalars are real numbers,
-generally discuss 2-D or 3-D vectors, corresponding to vectors of length 2 or
+with corresponding vector spaces $RR^n$, where $n$ is the length of the vector.
+We generally discuss 2D or 3D vectors, corresponding to vectors of length 2 or
 3; in our case, $RR^2$ and $RR^3$.
 
-However, vectors in $RR$ can really be of any length. Discard your primitive
+However, vectors in $RR^n$ can really be of any length. Vectors can be viewed
-conception of vectors as arrows in space. Vectors are simply arbitrary length
+as arbitrary length lists of numbers (for the computer science folk: think C++
-lists of numbers (for the computer science folk: think C++ `std::vector`).
+`std::vector`).
 
-_Example_. $ vec(1,2,3,4,5,6,7,8,9) $
+_Example_. $ vec(1,2,3,4,5,6,7,8,9) in RR^9 $
 
-Moreover, vectors need not be in $RR$ at all. Recall that a vector space need
+Keep in mind that vectors need not be in $RR^n$ at all. Recall that a vector
-only satisfy the aforementioned _axioms of a vector space_.
+space need only satisfy the aforementioned _axioms of a vector space_.
 
-_Example_. The vector space $CC$ is similar to $RR$, except it includes complex
+_Example_. The vector space $CC^n$ is similar to $RR^n$, except it includes
-numbers. All complex vector spaces are real vector spaces (as you can simply
+complex numbers. All complex vector spaces are real vector spaces (as you can
-restrict them to only use the real numbers), but not the other way around.
+simply restrict them to only use the real numbers), but not the other way
+around.
+
+From now on, let us refer to vector spaces $RR^n$ and $CC^n$ as $FF^n$.
 
 In general, we can have a vector space where the scalars are in an arbitrary
-field $FF$, as long as the axioms are satisfied.
+field, as long as the axioms are satisfied.
 
-_Example_. The vector space of all polynomials of degree 3, or $PP^3$. It is
+_Example_. The vector space of all polynomials of at most degree 3, or $PP^3$.
-not yet clear what this vector may look like. We shall return to this example
+It is not yet clear what this vector may look like. We shall return to this
-once we discuss _basis_.
+example once we discuss _basis_.
 
 == Vector addition. Multiplication
 
@@ -138,6 +144,129 @@ _Example_.
 
 $ beta vec(a, b, c) = vec(beta dot a, beta dot b, beta dot c) $
 
+== Linear combinations
+
+Given vector spaces $V$ and $W$ and vectors $v in V$ and $w in W$, $v + w$ is
+the _linear combination_ of $v$ and $w$.
+
+=== Spanning systems
+
+We say that a set of vectors $v_1, v_2, ..., v_n in V$ _span_ $V$ if the linear
+combination of the vectors can represent any arbitrary vector $v in V$.
+
+Precisely, given scalars $alpha_1, alpha_2, ..., alpha_n$,
+
+$ alpha_1 v_1 + alpha_2 v_2 + ... + alpha_n v_n = v, forall v in V $
+
+Note that any scalar $alpha_k$ could be 0. Therefore, it is possible for a
+subset of a spanning system to also be a spanning system. The proof of this
+fact is left as an exercise.
+
+=== Intuition for linear independence and dependence
+
+We say that $v$ and $w$ are linearly independent if $v$ cannot be represented by the scaling of $w$, and $w$ cannot be represented by the scaling of $v$. Otherwise, they are _linearly dependent_.
+
+You may intuitively visualize linear dependence in the 2D plane as two vectors
+both pointing in the same direction. Clearly, scaling one vector will allow us
+to reach the other vector. Linear independence is therefore two vectors
+pointing in different directions.
+
+Of course, this definition applies to vectors in any $FF^n$.
+
+=== Formal definition of linear dependence and independence
+
+Let us formally define linear independence for arbitrary vectors in $FF^n$. Given a set of vectors
+
+$ v_1, v_2, ..., v_n in V $
+
+we say they are linearly independent iff. the equation
+
+$ alpha_1 v_1 + alpha_2 v_2 + ... + alpha_n v_n = 0 $
+
+has only a unique set of solutions $alpha_1, alpha_2, ..., alpha_n$ such that
+all $alpha_n$ are zero.
+
+Equivalently,
+
+$ abs(alpha_1) + abs(alpha_2) + ... + abs(alpha_n) = 0 $
+
+More precisely,
+
+$ sum_(i=1)^k abs(alpha_i) = 0 $
+
+Therefore, a set of vectors $v_1, v_2, ..., v_m$ is linearly dependent if the opposite is true, that is there exists solution $alpha_1, alpha_2, ..., alpha_m$ to the equation
+
+$ alpha_1 v_1 + alpha_2 v_2 + ... + alpha_m v_m = 0 $
+
+such that
+
+$ sum_(i=1)^k abs(alpha_i) != 0 $
+
+=== Basis
+
+We say a system of vectors $v_1, v_2, ..., v_n in V$ is a _basis_ in $V$ if the
+system is both linearly independent and spanning. That is, the system must be
+able to represent any vector in $V$ as well as satisfy our requirements for
+linear independence.
+
+Equivalently, we may say that a system of vectors in $V$ is a basis in $V$ if
+any vector $v in V$ admits a _unique representation_ as a linear combination of
+vectors in the system. This is equivalent to our previous statement, that the
+system must be spanning and linearly independent.
+
+=== Standard basis
+
+We may define a _standard basis_ for a vector space. By convention, the
+standard basis in $RR^2$ is
+
+$ vec(1, 0) vec(0, 1) $
+
+Verify that the above is in fact a basis (that is, linearly independent and
+generating).
+
+Recalling the definition of the basis, we can represent any vector in $RR^2$ as
+the linear combination of the standard basis.
+
+Therefore, for any arbitrary vector $v in RR^2$, we can represent it as
+
+$ v = alpha_1 vec(1, 0) + alpha_2 vec(0,1) $
+
+Let us call $alpha_1$ and $alpha_2$ the _coordinates_ of the vector. Then, we can write $v$ as
+
+$ v = vec(alpha_1, alpha_2) $
+
+For example, the vector
+
+$ vec(1, 2) $
+
+represents
+
+$ 1 dot vec(1, 0) + 2 dot vec(0,1) $
+
+Verify that this aligns with your previous intuition of vectors.
+
+You may recognize the standard basis in $RR^2$ as the familiar unit vectors
+
+$ accent(i, hat), accent(j, hat) $
+
+This aligns with the fact that
+
+$ vec(alpha, beta) = alpha hat(i) + beta hat(j) $
+
+However, we may define a standard basis in any arbitrary vector space. So, let
+
+$ e_1, e_2, ..., e_n $
+
+be a standard basis in $FF^n$. Then, the coordinates $alpha_1, alpha_2, ..., alpha_n$ of a vector $v in FF^n$ represent the following
+
+$
+  vec(alpha_1, alpha_2, dots.v, alpha_n) = alpha_1 e_1 + alpha_2 + e_2 + alpha_n e_n
+$
+
+Using our new notation, the standard basis in $RR^2$ is
+
+$ e_1 = vec(1,0), e_2 = vec(0,1) $
+
 == Matrices
 
 Before discussing any properties of matrices, let's simply reiterate what we
@@ -162,4 +291,121 @@ $
   mat(1,2,3;4,5,6)^T = mat(1,4;2,5;3,6)
 $
 
-Formally, we can say $(A_(j,k))^T = A_(k,j)$.
+Formally, we can say $(A^T)_(j,k) = A_(k,j)$
+
+== Linear transformations
+
+A linear transformation $T : V -> W$ is a mapping between two vector spaces $V
+-> W$, such that the following axioms are satisfied:
+
++ $T(v + w) = T(v) + T(w), forall v in V, forall w in W$
++ $T(alpha v) + T(beta w) = alpha T(v) + beta T(w), forall v in V, forall w in W$, for all scalars $alpha, beta$
+
+_Definition_. $T$ is a linear transformation iff.
+
+$ T(alpha v + beta w) = alpha T(v) + beta T(w) $
+
+_Abuse of notation_. From now on, we may elide the parentheses and say that
+$ T(v) = T v, forall v in V $
+
+_Remark_. A phrase that you may commonly hear is that linear transformations
+preserve _linearity_. Essentially, straight lines remain straight, parallel
+lines remain parallel, and the origin remains fixed at 0. Take a moment to
+think about why this is true (at least, in lower dimensional spaces you can
+visualize).
+
+_Examples_.
+
++ #[Rotation for $V = W = RR^2$ (i.e. rotation in 2 dimensions). Given $v, w in
+RR^2$, and their linear combination $v + w$, a rotation of $gamma$ radians of
+    $v + w$ is equivalent to first rotating $v$ and $w$ individually by $gamma$ and
+    then taking their linear combination.]
+
++ #[Differentiation of polynomials. In this case $V = PP^n$ and $W = PP^(n -
+1)$, where $PP^n$ is the field of all polynomials of degree at most $n$.
+
+    $
+      dif / (dif x) (
+        alpha v + beta w
+      ) = alpha dif / (dif x) v + beta dif / (dif x) w, forall v in V, w in W, forall "scalars" alpha, beta
+    $
+  ]
+
+
+
+== Matrices represent linear transformations
+
+Suppose we wanted to represent a linear transformation $T: FF^n -> FF^m$. I
+propose that we need encode how $T$ acts on the standard basis of $FF^n$.
+
+Using our intuition from lower dimensional vector spaces, we know that the
+standard basis in $RR^2$ is the unit vectors $hat(i)$ and $hat(j)$. Because
+linear transformations preserve linearity (i.e. all straight lines remain
+straight and parallel lines remain parallel), we can encode any transformation
+as simply changing $hat(i)$ and $hat(j)$. And indeed, if any vector $v in RR^2$
+can be represented as the linear combination of $hat(i)$ and $hat(j)$ (this is
+the definition of a basis), it makes sense both symbolically and geometrically
+that we can represent all linear transformations as the transformations of the
+basis vectors.
+
+_Example_. To reflect all vectors $v in RR^2$ across the $y$-axis, we can simply change the standard basis to
+
+$ vec(-1, 0) vec(0,1) $
+
+Then, any vector in $RR^2$ using this new basis will be reflected across the
+$y$-axis. Take a moment to justify this geometrically.
+
+=== Writing a linear transformation as matrix
+
+For any linear transformation $T: FF^m -> FF^n$, we can write it as an $n times
+m$ matrix $A$. That is, there is a matrix $A$ with $n$ rows and $m$ columns
+that can represent any linear transformation from $FF^m -> FF^n$.
+
+How should we write this matrix? Naturally, from our previous discussion, we
+should write a matrix with each _column_ being one of our new transformed
+_basis_ vectors.
+
+_Example_. Our $y$-axis reflection transformation from earlier. We write the bases in a matrix
+
+$ mat(-1,0; 0,1) $
+
+=== Matrix-vector multiplication
+
+Perhaps you now see why the so-called matrix-vector multiplication is defined
+the way it is. Recalling our definition of a basis, given a basis in $V$, any
+vector $v in V$ can be written as the linear combination of the vectors in the
+basis. Then, given a linear transformation represented by the matrix containing
+the new basis, we simply write the linear combination with the new basis
+instead.
+
+_Example_. Let us first write a vector in the standard basis in $RR^2$ and then
+show how our matrix-vector multiplication naturally corresponds to the
+definition of the linear transformation.
+
+$ vec(1, 2) in RR^2 $
+
+is the same as
+
+$ 1 dot vec(1, 0) + 2 dot vec(0, 1) $
+
+Then, to perform our reflection, we need only replace the basis vector $vec(1,
+0)$ with $vec(-1, 0)$.
+
+Then, the reflected vector is given by
+
+$ 1 dot vec(-1, 0) + 2 dot vec(0,1) = vec(-1, 2) $
+
+We can clearly see that this is exactly how the matrix multiplication
+
+$ mat(-1, 0; 0, 1) dot vec(1, 2) $ is defined! The _column-by-coordinate_ rule
+for matrix-vector multiplication says that we multiply the $n^("th")$ entry of
+the vector by the corresponding $n^("th")$ column of the matrix and sum them
+all up (take their linear combination). This algorithm intuitively follows from
+our definition of matrices.
+
+=== Matrix-matrix multiplication
+
+As you may have noticed, a very similar natural definition arises for the
+_matrix-matrix_ multiplication. Multiplying two matrices $A dot B$ is
+essentially just taking each column of $B$, and applying the linear
+transformation defined by the matrix $A$!