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src/posts/2025-02-15-linear-algebra.md
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---
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author: "Youwen Wu"
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authorTwitter: "@youwen"
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image: "https://wallpapercave.com/wp/wp12329537.png"
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keywords: "linear algebra, algebra, math"
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lang: "en"
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title: "An assortment of preliminaries on linear algebra"
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desc: "and also a test for pandoc"
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---
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This entire document was written entirely in [Typst](https://typst.app/) and
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directly translated to this file by Pandoc. It serves as a proof of concept of
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a way to do static site generation from Typst files instead of Markdown.
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---
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I figured I should write this stuff down before I forgot it.
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# Basic Notions
|
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## Vector spaces
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Before we can understand vectors, we need to first discuss *vector
|
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spaces*. Thus far, you have likely encountered vectors primarily in
|
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physics classes, generally in the two-dimensional plane. You may
|
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conceptualize them as arrows in space. For vectors of size $> 3$, a hand
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waving argument is made that they are essentially just arrows in higher
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dimensional spaces.
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It is helpful to take a step back from this primitive geometric
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understanding of the vector. Let us build up a rigorous idea of vectors
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from first principles.
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### Vector axioms
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The so-called *axioms* of a *vector space* (which we'll call the vector
|
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space $V$) are as follows:
|
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1. Commutativity: $u + v = v + u,\text{ }\forall u,v \in V$
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|
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2. Associativity:
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$(u + v) + w = u + (v + w),\text{ }\forall u,v,w \in V$
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3. Zero vector: $\exists$ a special vector, denoted $0$, such that
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$v + 0 = v,\text{ }\forall v \in V$
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4. Additive inverse:
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$\forall v \in V,\text{ }\exists w \in V\text{ such that }v + w = 0$.
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Such an additive inverse is generally denoted $- v$
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5. Multiplicative identity: $1v = v,\text{ }\forall v \in V$
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|
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6. Multiplicative associativity:
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$(\alpha\beta)v = \alpha(\beta v)\text{ }\forall v \in V,\text{ scalars }\alpha,\beta$
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7. Distributive property for vectors:
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$\alpha(u + v) = \alpha u + \alpha v\text{ }\forall u,v \in V,\text{ scalars }\alpha$
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8. Distributive property for scalars:
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$(\alpha + \beta)v = \alpha v + \beta v\text{ }\forall v \in V,\text{ scalars }\alpha,\beta$
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It is easy to show that the zero vector $0$ and the additive inverse
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$- v$ are *unique*. We leave the proof of this fact as an exercise.
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These may seem difficult to memorize, but they are essentially the same
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familiar algebraic properties of numbers you know from high school. The
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important thing to remember is which operations are valid for what
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objects. For example, you cannot add a vector and scalar, as it does not
|
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make sense.
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*Remark*. For those of you versed in computer science, you may recognize
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this as essentially saying that you must ensure your operations are
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*type-safe*. Adding a vector and scalar is not just "wrong" in the same
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sense that $1 + 1 = 3$ is wrong, it is an *invalid question* entirely
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because vectors and scalars and different types of mathematical objects.
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See [@chen2024digression] for more.
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### Vectors big and small
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In order to begin your descent into what mathematicians colloquially
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recognize as *abstract vapid nonsense*, let's discuss which fields
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constitute a vector space. We have the familiar field of $\mathbb{R}$
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where all scalars are real numbers, with corresponding vector spaces
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${\mathbb{R}}^{n}$, where $n$ is the length of the vector. We generally
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discuss 2D or 3D vectors, corresponding to vectors of length 2 or 3; in
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our case, ${\mathbb{R}}^{2}$ and ${\mathbb{R}}^{3}$.
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However, vectors in ${\mathbb{R}}^{n}$ can really be of any length.
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Vectors can be viewed as arbitrary length lists of numbers (for the
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computer science folk: think C++ `std::vector`).
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|
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*Example*. $$\begin{pmatrix}
|
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1 \\
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2 \\
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3 \\
|
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4 \\
|
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5 \\
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||||
6 \\
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7 \\
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8 \\
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9
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\end{pmatrix} \in {\mathbb{R}}^{9}$$
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Keep in mind that vectors need not be in ${\mathbb{R}}^{n}$ at all.
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Recall that a vector space need only satisfy the aforementioned *axioms
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of a vector space*.
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*Example*. The vector space ${\mathbb{C}}^{n}$ is similar to
|
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${\mathbb{R}}^{n}$, except it includes complex numbers. All complex
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vector spaces are real vector spaces (as you can simply restrict them to
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only use the real numbers), but not the other way around.
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From now on, let us refer to vector spaces ${\mathbb{R}}^{n}$ and
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${\mathbb{C}}^{n}$ as ${\mathbb{F}}^{n}$.
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In general, we can have a vector space where the scalars are in an
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arbitrary field, as long as the axioms are satisfied.
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*Example*. The vector space of all polynomials of at most degree 3, or
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${\mathbb{P}}^{3}$. It is not yet clear what this vector may look like.
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We shall return to this example once we discuss *basis*.
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## Vector addition. Multiplication
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Vector addition, represented by $+$ can be done entrywise.
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*Example.*
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|
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$$\begin{pmatrix}
|
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1 \\
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2 \\
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3
|
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\end{pmatrix} + \begin{pmatrix}
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4 \\
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5 \\
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6
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\end{pmatrix} = \begin{pmatrix}
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1 + 4 \\
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2 + 5 \\
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3 + 6
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\end{pmatrix} = \begin{pmatrix}
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5 \\
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7 \\
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9
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\end{pmatrix}$$ $$\begin{pmatrix}
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1 \\
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2 \\
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3
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\end{pmatrix} \cdot \begin{pmatrix}
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4 \\
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5 \\
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6
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\end{pmatrix} = \begin{pmatrix}
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1 \cdot 4 \\
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2 \cdot 5 \\
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3 \cdot 6
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\end{pmatrix} = \begin{pmatrix}
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4 \\
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10 \\
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18
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\end{pmatrix}$$
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This is simple enough to understand. Again, the difficulty is simply
|
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ensuring that you always perform operations with the correct *types*.
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For example, once we introduce matrices, it doesn't make sense to
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multiply or add vectors and matrices in this fashion.
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|
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## Vector-scalar multiplication
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Multiplying a vector by a scalar simply results in each entry of the
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vector being multiplied by the scalar.
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*Example*.
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|
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$$\beta\begin{pmatrix}
|
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a \\
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b \\
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c
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\end{pmatrix} = \begin{pmatrix}
|
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\beta \cdot a \\
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\beta \cdot b \\
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\beta \cdot c
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\end{pmatrix}$$
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## Linear combinations
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Given vector spaces $V$ and $W$ and vectors $v \in V$ and $w \in W$,
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$v + w$ is the *linear combination* of $v$ and $w$.
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|
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### Spanning systems
|
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We say that a set of vectors $v_{1},v_{2},\ldots,v_{n} \in V$ *span* $V$
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if the linear combination of the vectors can represent any arbitrary
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vector $v \in V$.
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Precisely, given scalars $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$,
|
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|
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$$\alpha_{1}v_{1} + \alpha_{2}v_{2} + \ldots + \alpha_{n}v_{n} = v,\forall v \in V$$
|
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|
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Note that any scalar $\alpha_{k}$ could be 0. Therefore, it is possible
|
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for a subset of a spanning system to also be a spanning system. The
|
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proof of this fact is left as an exercise.
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### Intuition for linear independence and dependence
|
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We say that $v$ and $w$ are linearly independent if $v$ cannot be
|
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represented by the scaling of $w$, and $w$ cannot be represented by the
|
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scaling of $v$. Otherwise, they are *linearly dependent*.
|
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|
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You may intuitively visualize linear dependence in the 2D plane as two
|
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vectors both pointing in the same direction. Clearly, scaling one vector
|
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will allow us to reach the other vector. Linear independence is
|
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therefore two vectors pointing in different directions.
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Of course, this definition applies to vectors in any ${\mathbb{F}}^{n}$.
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|
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### Formal definition of linear dependence and independence
|
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Let us formally define linear independence for arbitrary vectors in
|
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${\mathbb{F}}^{n}$. Given a set of vectors
|
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|
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$$v_{1},v_{2},\ldots,v_{n} \in V$$
|
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|
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we say they are linearly independent iff. the equation
|
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|
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$$\alpha_{1}v_{1} + \alpha_{2}v_{2} + \ldots + \alpha_{n}v_{n} = 0$$
|
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|
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has only a unique set of solutions
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$\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ such that all $\alpha_{n}$ are
|
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zero.
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|
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Equivalently,
|
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|
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$$\left| \alpha_{1} \right| + \left| \alpha_{2} \right| + \ldots + \left| \alpha_{n} \right| = 0$$
|
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|
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More precisely,
|
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|
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$$\sum_{i = 1}^{k}\left| \alpha_{i} \right| = 0$$
|
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|
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Therefore, a set of vectors $v_{1},v_{2},\ldots,v_{m}$ is linearly
|
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dependent if the opposite is true, that is there exists solution
|
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$\alpha_{1},\alpha_{2},\ldots,\alpha_{m}$ to the equation
|
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|
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$$\alpha_{1}v_{1} + \alpha_{2}v_{2} + \ldots + \alpha_{m}v_{m} = 0$$
|
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|
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such that
|
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|
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$$\sum_{i = 1}^{k}\left| \alpha_{i} \right| \neq 0$$
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|
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### Basis
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|
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We say a system of vectors $v_{1},v_{2},\ldots,v_{n} \in V$ is a *basis*
|
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in $V$ if the system is both linearly independent and spanning. That is,
|
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the system must be able to represent any vector in $V$ as well as
|
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satisfy our requirements for linear independence.
|
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|
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Equivalently, we may say that a system of vectors in $V$ is a basis in
|
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$V$ if any vector $v \in V$ admits a *unique representation* as a linear
|
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combination of vectors in the system. This is equivalent to our previous
|
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statement, that the system must be spanning and linearly independent.
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|
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### Standard basis
|
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|
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We may define a *standard basis* for a vector space. By convention, the
|
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standard basis in ${\mathbb{R}}^{2}$ is
|
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|
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$$\begin{pmatrix}
|
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1 \\
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0
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\end{pmatrix}\begin{pmatrix}
|
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0 \\
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1
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\end{pmatrix}$$
|
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|
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Verify that the above is in fact a basis (that is, linearly independent
|
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and generating).
|
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|
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Recalling the definition of the basis, we can represent any vector in
|
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${\mathbb{R}}^{2}$ as the linear combination of the standard basis.
|
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|
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Therefore, for any arbitrary vector $v \in {\mathbb{R}}^{2}$, we can
|
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represent it as
|
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|
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$$v = \alpha_{1}\begin{pmatrix}
|
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1 \\
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||||
0
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\end{pmatrix} + \alpha_{2}\begin{pmatrix}
|
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0 \\
|
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1
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\end{pmatrix}$$
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|
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Let us call $\alpha_{1}$ and $\alpha_{2}$ the *coordinates* of the
|
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vector. Then, we can write $v$ as
|
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|
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$$v = \begin{pmatrix}
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\alpha_{1} \\
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\alpha_{2}
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\end{pmatrix}$$
|
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|
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For example, the vector
|
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|
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$$\begin{pmatrix}
|
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1 \\
|
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2
|
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\end{pmatrix}$$
|
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|
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represents
|
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|
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$$1 \cdot \begin{pmatrix}
|
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1 \\
|
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0
|
||||
\end{pmatrix} + 2 \cdot \begin{pmatrix}
|
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0 \\
|
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1
|
||||
\end{pmatrix}$$
|
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|
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Verify that this aligns with your previous intuition of vectors.
|
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|
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You may recognize the standard basis in ${\mathbb{R}}^{2}$ as the
|
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familiar unit vectors
|
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|
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$$\hat{i},\hat{j}$$
|
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|
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This aligns with the fact that
|
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|
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$$\begin{pmatrix}
|
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\alpha \\
|
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\beta
|
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\end{pmatrix} = \alpha\hat{i} + \beta\hat{j}$$
|
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|
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However, we may define a standard basis in any arbitrary vector space.
|
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So, let
|
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|
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$$e_{1},e_{2},\ldots,e_{n}$$
|
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|
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be a standard basis in ${\mathbb{F}}^{n}$. Then, the coordinates
|
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$\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ of a vector
|
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$v \in {\mathbb{F}}^{n}$ represent the following
|
||||
|
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$$\begin{pmatrix}
|
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\alpha_{1} \\
|
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\alpha_{2} \\
|
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\vdots \\
|
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\alpha_{n}
|
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\end{pmatrix} = \alpha_{1}e_{1} + \alpha_{2} + e_{2} + \alpha_{n}e_{n}$$
|
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|
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Using our new notation, the standard basis in ${\mathbb{R}}^{2}$ is
|
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|
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$$e_{1} = \begin{pmatrix}
|
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1 \\
|
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0
|
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\end{pmatrix},e_{2} = \begin{pmatrix}
|
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0 \\
|
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1
|
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\end{pmatrix}$$
|
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|
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## Matrices
|
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|
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Before discussing any properties of matrices, let's simply reiterate
|
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what we learned in class about their notation. We say a matrix with rows
|
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of length $m$, and columns of size $n$ (in less precise terms, a matrix
|
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with length $m$ and height $n$) is a $m \times n$ matrix.
|
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|
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Given a matrix
|
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|
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$$A = \begin{pmatrix}
|
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1 & 2 & 3 \\
|
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4 & 5 & 6 \\
|
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7 & 8 & 9
|
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\end{pmatrix}$$
|
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|
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we refer to the entry in row $j$ and column $k$ as $A_{j,k}$ .
|
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|
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### Matrix transpose
|
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|
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A formalism that is useful later on is called the *transpose*, and we
|
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obtain it from a matrix $A$ by switching all the rows and columns. More
|
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precisely, each row becomes a column instead. We use the notation
|
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$A^{T}$ to represent the transpose of $A$.
|
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|
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$$\begin{pmatrix}
|
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1 & 2 & 3 \\
|
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4 & 5 & 6
|
||||
\end{pmatrix}^{T} = \begin{pmatrix}
|
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1 & 4 \\
|
||||
2 & 5 \\
|
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3 & 6
|
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\end{pmatrix}$$
|
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|
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Formally, we can say $\left( A^{T} \right)_{j,k} = A_{k,j}$
|
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|
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## Linear transformations
|
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|
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A linear transformation $T:V \rightarrow W$ is a mapping between two
|
||||
vector spaces $V \rightarrow W$, such that the following axioms are
|
||||
satisfied:
|
||||
|
||||
1. $T(v + w) = T(v) + T(w),\forall v \in V,\forall w \in W$
|
||||
|
||||
2. $T(\alpha v) + T(\beta w) = \alpha T(v) + \beta T(w),\forall v \in V,\forall w \in W$,
|
||||
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 $$T(v) = Tv,\forall v \in V$$
|
||||
|
||||
*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*.
|
||||
|
||||
1. Rotation for $V = W = {\mathbb{R}}^{2}$ (i.e. rotation in 2
|
||||
dimensions). Given $v,w \in {\mathbb{R}}^{2}$, 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.
|
||||
|
||||
2. Differentiation of polynomials. In this case $V = {\mathbb{P}}^{n}$
|
||||
and $W = {\mathbb{P}}^{n - 1}$, where ${\mathbb{P}}^{n}$ is the
|
||||
field of all polynomials of degree at most $n$.
|
||||
|
||||
$$\frac{d}{dx}(\alpha v + \beta w) = \alpha\frac{d}{dx}v + \beta\frac{d}{dx}w,\forall v \in V,w \in W,\forall\text{ scalars }\alpha,\beta$$
|
||||
|
||||
## Matrices represent linear transformations
|
||||
|
||||
Suppose we wanted to represent a linear transformation
|
||||
$T:{\mathbb{F}}^{n} \rightarrow {\mathbb{F}}^{m}$. I propose that we
|
||||
need encode how $T$ acts on the standard basis of ${\mathbb{F}}^{n}$.
|
||||
|
||||
Using our intuition from lower dimensional vector spaces, we know that
|
||||
the standard basis in ${\mathbb{R}}^{2}$ 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 $v \in {\mathbb{R}}^{2}$ 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 $v \in {\mathbb{R}}^{2}$ across the
|
||||
$y$-axis, we can simply change the standard basis to
|
||||
|
||||
$$\begin{pmatrix}
|
||||
- 1 \\
|
||||
0
|
||||
\end{pmatrix}\begin{pmatrix}
|
||||
0 \\
|
||||
1
|
||||
\end{pmatrix}$$
|
||||
|
||||
Then, any vector in ${\mathbb{R}}^{2}$ 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
|
||||
$T:{\mathbb{F}}^{m} \rightarrow {\mathbb{F}}^{n}$, 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
|
||||
${\mathbb{F}}^{m} \rightarrow {\mathbb{F}}^{n}$.
|
||||
|
||||
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
|
||||
|
||||
$$\begin{pmatrix}
|
||||
- 1 & 0 \\
|
||||
0 & 1
|
||||
\end{pmatrix}$$
|
||||
|
||||
### 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
|
||||
${\mathbb{R}}^{2}$ and then show how our matrix-vector multiplication
|
||||
naturally corresponds to the definition of the linear transformation.
|
||||
|
||||
$$\begin{pmatrix}
|
||||
1 \\
|
||||
2
|
||||
\end{pmatrix} \in {\mathbb{R}}^{2}$$
|
||||
|
||||
is the same as
|
||||
|
||||
$$1 \cdot \begin{pmatrix}
|
||||
1 \\
|
||||
0
|
||||
\end{pmatrix} + 2 \cdot \begin{pmatrix}
|
||||
0 \\
|
||||
1
|
||||
\end{pmatrix}$$
|
||||
|
||||
Then, to perform our reflection, we need only replace the basis vector
|
||||
$\begin{pmatrix}
|
||||
1 \\
|
||||
0
|
||||
\end{pmatrix}$ with $\begin{pmatrix}
|
||||
- 1 \\
|
||||
0
|
||||
\end{pmatrix}$.
|
||||
|
||||
Then, the reflected vector is given by
|
||||
|
||||
$$1 \cdot \begin{pmatrix}
|
||||
- 1 \\
|
||||
0
|
||||
\end{pmatrix} + 2 \cdot \begin{pmatrix}
|
||||
0 \\
|
||||
1
|
||||
\end{pmatrix} = \begin{pmatrix}
|
||||
- 1 \\
|
||||
2
|
||||
\end{pmatrix}$$
|
||||
|
||||
We can clearly see that this is exactly how the matrix multiplication
|
||||
|
||||
$$\begin{pmatrix}
|
||||
- 1 & 0 \\
|
||||
0 & 1
|
||||
\end{pmatrix} \cdot \begin{pmatrix}
|
||||
1 \\
|
||||
2
|
||||
\end{pmatrix}$$ is defined! The *column-by-coordinate* rule for
|
||||
matrix-vector multiplication says that we multiply the $n^{\text{th}}$
|
||||
entry of the vector by the corresponding $n^{\text{th}}$ 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 $A \cdot B$
|
||||
is essentially just taking each column of $B$, and applying the linear
|
||||
transformation defined by the matrix $A$!
|
||||
Loading…
Reference in a new issue