documents: add selected solutions
This commit is contained in:
parent
b28b8455a9
commit
866b86c69a
GPG key ID:
865658ED1FE61EC3
5 changed files with 418 additions and 31 deletions
BIN
2024/documents/by-course/math-4a/selected-solutions/main.pdf
Normal file
BIN
2024/documents/by-course/math-4a/selected-solutions/main.pdf
Normal file
Binary file not shown.
98
2024/documents/by-course/math-4a/selected-solutions/main.typ
Normal file
98
2024/documents/by-course/math-4a/selected-solutions/main.typ
Normal file
|
|
@ -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)
|
||||
$
|
||||
],
|
||||
)
|
||||
|
|
@ -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;
|
||||
}
|
||||
)
|
||||
|
|
@ -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.}
|
||||
}
|
||||
|
|
@ -15,21 +15,22 @@
|
|||
|
||||
= Introduction
|
||||
|
||||
Many introductory linear algebra classes focus on _application_. In general,
|
||||
this is a red herring and is engineer-speak for "we will teach you how to
|
||||
crunch numbers with no regard for conceptual understanding."
|
||||
Many introductory linear algebra classes focus on _application_. They teach you
|
||||
how to perform trivial numerical operations such as the _matrix
|
||||
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
|
||||
unequivocal waste of time. I have prepared this brief crash course designed to
|
||||
fill in the theoretical gaps left by this class.
|
||||
course without gaining a theoretical appreciation for the topics at hand is a
|
||||
complete and utter waste of time.
|
||||
|
||||
= 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_
|
||||
entirely because vectors and scalars and different types of mathematical
|
||||
objects. See #cite(<chen2024digression>, form: "prose") for more.
|
||||
Adding a vector and scalar is not just "wrong" in the same sense that $1 + 1 =
|
||||
3$ is wrong, it is an _invalid question_ entirely because vectors and scalars
|
||||
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
|
||||
have the familiar space where all scalars are real numbers, or $RR$. We
|
||||
generally discuss 2-D or 3-D vectors, corresponding to vectors of length 2 or
|
||||
as _abstract vapid nonsense_, let's discuss which fields constitute a vector
|
||||
space. We have the familiar field of $RR$ where all scalars are real numbers,
|
||||
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
|
||||
conception of vectors as arrows in space. Vectors are simply arbitrary length
|
||||
lists of numbers (for the computer science folk: think C++ `std::vector`).
|
||||
However, vectors in $RR^n$ can really be of any length. Vectors can be viewed
|
||||
as arbitrary length lists of numbers (for the computer science folk: think C++
|
||||
`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
|
||||
only satisfy the aforementioned _axioms of a vector space_.
|
||||
Keep in mind that vectors need not be in $RR^n$ at all. Recall that a vector
|
||||
space need only satisfy the aforementioned _axioms of a vector space_.
|
||||
|
||||
_Example_. The vector space $CC$ is similar to $RR$, except it includes complex
|
||||
numbers. All complex vector spaces are real vector spaces (as you can simply
|
||||
restrict them to only use the real numbers), but not the other way around.
|
||||
_Example_. The vector space $CC^n$ is similar to $RR^n$, except it includes
|
||||
complex numbers. All complex vector spaces are real vector spaces (as you can
|
||||
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
|
||||
not yet clear what this vector may look like. We shall return to this example
|
||||
once we discuss _basis_.
|
||||
_Example_. The vector space of all polynomials of at most degree 3, or $PP^3$.
|
||||
It is not yet clear what this vector may look like. We shall return to this
|
||||
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 $A$!
|
||||
|
|
|
|||
Loading…
Reference in a new issue