COORDINATES & BASES

Intuition: Coordinates

Given a set of vectors $P_1,\dots,P_n \in \mathbb{R}^m $ organized as the columns of a matrix $P \in \mathbb{R}^{ m \times n}$ $$ P = \begin{bmatrix} | & & | \\ P_1 & \cdots & P_n \\ | & & | \end{bmatrix} $$ we can represent other vectors as linear combinations of those vectors. This is illustrated in the basic figure shown. The vector $y$ is a linear combination of the columns of $P$ with coefficients $x$, ie. $y = Px$. Here, we say that $x$ is the "coordinates of $y$ with respect to the columns of $P$" (or just "with respect to $P$" for short.)

Intuitively, we can see that if there are less columns than the dimension of $y$, ie. $m < n$ than we will not be able to represent every possible $y$ as a linear combination of the columns. In this case we say that the columns of $P$ do not span all of $\mathbb{R}^m$. We can also see that if there are more columns than the dimension of $y$, there are likely multiple sets of coordinates $x$ that will reach any given $y$. We will see that this is because the columns of $P$ are not linearly indepedent. These two cases are illustrated below.

Bases

When a set of vectors $P$ has just enough vectors to span the space $\mathbb{R}^m$ without having any redundant vectors, we say that (the columns of) $P$ is a basis for $\mathbb{R}^m$. Mathematically, "just enough vectors to span the space without being redundant" means that the columns of $P$ span $\mathbb{R}^m$ and they are linearly independent from each other. When this is the case, every vector $y \in \mathbb{R}^m$ can be uniquely represented by a set of coordinates $x \in \mathbb{R}^n$ and for practical purposes we can think of $y$ and $x$ as representing the same vector (albeit in two different ways). Perhaps unsurprisingly, one basic requirement for this to be possible is that $m = n$

Speaking a bit more abstractly, it's not actually possible to represent a general vector without first defining a basis. Our intuitive notation of a vector as a string of numbers $\begin{bmatrix}x_1 & \cdots & x_n \end{bmatrix}^T $ is actually just the coordinates of that vector with respect to the standard basis $$ I_1 = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, I_2 = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \cdots I_n = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} $$ Since the standard basis is just the columns of the identity matrix, the equation $y = Px$ just becomes $x = Ix$. If we define a vector space more abstractly than $\mathbb{R}^m$, before we can write down general vectors in that vector space we have to first define a basis and then write other vectors as linear combinations (ie. coordinates) with respect to those basis vectors.

Span and Linear Independence

For a set of vectors $P$ to be a basis for $\mathbb{R}^m$ it must satisfy two requirements. It is must span all of $\mathbb{R}^m$ and it must also be linearly independent. The span requirement guarantees that any vector $y \in \mathbb{R}^m$ can be represented as coordinates with respect to $P$ and the linear independence requirement guarantees that there is a unique set of coordintes $x$ that represent $y$, ie. there is only one $x$ such that $y = Px$. If we satisfy these two requirements than we can switch back and forth freely between the "$y$-representation" of $y$ and the "$x$-representation" of $y$ without issue. The process of switching back and forth between $y$ and $x$ involves the idea of a matrix inverse which we will cover next. If we have the equation $y = Px$, the inverse of $P$, denoted $P^{-1}$ is the matrix we use to figure out what $x$ should be for any given $y$, ie. the matrix that satisfies $$ x = P^{-1}y $$

More Math: Constructing an Abstract Basis (Difficulty: 6/10)

When we talk about bases in an abstract sense we have to have a way of constructing them from elements in a vectors space, ie. do a bunch of abstact nonsense to prove obvious things

Steinitz Exchange Lemma

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