---
title: Coordinate Systems
module: Vector Spaces
moduleNumber: 4
lessonNumber: 4
order: 404
summary: >
  Fixing a basis assigns every vector a unique list of coordinates, turning an
  abstract space into $\mathbb{R}^n$. The coordinate mapping is a one-to-one linear
  transformation onto $\mathbb{R}^n$ — an isomorphism — so any $n$-dimensional
  space is indistinguishable from $\mathbb{R}^n$ as far as vector-space
  computations go. In $\mathbb{R}^n$ the change-of-coordinates matrix $P_B$ and its
  inverse convert between basis coordinates and standard coordinates.
topics: [Vector Spaces]
draft: false
sources:
  - book: Lay
    ref: "§4.4 Coordinate Systems"
---

Choosing a basis for a vector space gives it a coordinate system. Each vector
acquires a unique list of weights, and
those weights behave exactly like a column in $\mathbb{R}^n$. This is why every
finite-dimensional space, however abstract, can be computed with as if it were
$\mathbb{R}^n$: the correspondence preserves all vector-space structure.[^lay-coord]

## Unique representation

The uniqueness of coordinates is what makes them a coordinate system rather than
just one of many ways to write a vector.

> **Theorem (Unique representation).** Let $\mathcal{B} = \{b_1, \dots, b_n\}$ be
> a basis for a vector space $V$. Then for each $x$ in $V$ there is a unique set of
> scalars $c_1, \dots, c_n$ with
> $$
> x = c_1 b_1 + \cdots + c_n b_n.
> $$

Existence comes from spanning; uniqueness from independence. If $x$ also equalled
$d_1 b_1 + \cdots + d_n b_n$, subtracting gives $0 = (c_1 - d_1)b_1 + \cdots + (c_n -
d_n)b_n$, and independence forces every $c_j - d_j = 0$. So the weights are
determined by $x$ and $\mathcal{B}$ alone.

> **Definition (Coordinates).** With $\mathcal{B} = \{b_1, \dots, b_n\}$ a basis
> for $V$ and $x = c_1 b_1 + \cdots + c_n b_n$, the scalars $c_1, \dots, c_n$ are
> the coordinates of $x$ relative to $\mathcal{B}$, and
> $$
> [x]_{\mathcal{B}} = \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix}
> $$
> is the $\mathcal{B}$-coordinate vector of $x$. The map $x \mapsto
> [x]_{\mathcal{B}}$ is the coordinate mapping determined by $\mathcal{B}$.

The basis must be an _ordered_ set, so that the entries of $[x]_{\mathcal{B}}$ have
a fixed meaning. For the standard basis $\mathcal{E}$ of $\mathbb{R}^n$, the
coordinate vector is the vector itself: $[x]_{\mathcal{E}} = x$.

## Coordinates as a change of grid

Coordinates are the readout of a grid drawn from the basis vectors. Standard graph
paper is the grid of the standard basis; a different basis rules a different grid
over the same plane, and the same point gets different coordinates on each.

> **Worked example.** For $\mathcal{B} = \{b_1, b_2\}$ with $b_1 = (1, 0)$ and
> $b_2 = (1, 2)$, a vector $x$ has $[x]_{\mathcal{B}} = (-2, 3)$. Find $x$.
>
> The $\mathcal{B}$-coordinates say how to build $x$ from the basis vectors:
> $$
> x = -2b_1 + 3b_2 = -2\begin{bmatrix} 1 \\ 0 \end{bmatrix} + 3\begin{bmatrix} 1 \\ 2 \end{bmatrix}
> = \begin{bmatrix} 1 \\ 6 \end{bmatrix}.
> $$
>
> On the standard grid $x$ reads $(1, 6)$; on the skewed $\mathcal{B}$-grid the same
> point reads $(-2, 3)$, two steps along $b_1$ and three along $b_2$.

$$
% caption: One point, two grids. On the standard grid the vector reads $(1,6)$; on
% the skewed grid ruled by $b_1, b_2$ the same point reads $(-2,3)$.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=0.62]
\definecolor{acc}{HTML}{2F6DB5}
\definecolor{red}{HTML}{B23A48}
% standard grid
\draw[black] (-1,-0.5) grid (3,7);
\draw[->, black] (-1,0) -- (3,0);
\draw[->, black] (0,-0.5) -- (0,7.2);
\coordinate (x) at (1,6);
\draw[->, acc, very thick] (0,0) -- (1,0) node[below] {$b_1$};
\draw[->, acc, very thick] (0,0) -- (1,2) node[left] {$b_2$};
\fill[red] (x) circle (2.6pt);
\node[red, anchor=west] at (x) {$x =$ (1, 6)};
\node[black, anchor=south] at (1.5,7.2) {standard grid};
% B grid: skew a standard grid by the linear map e1->b1, e2->b2
\begin{scope}[xshift=9cm]
  \begin{scope}[cm={1,0,1,2,(0,0)}]
    \draw[black] (-2,-1) grid (2,4);
    \draw[->, acc, very thick] (0,0) -- (1,0);
    \draw[->, acc, very thick] (0,0) -- (0,1);
  \end{scope}
  \node[acc, anchor=north] at (1,0) {$b_1$};
  \node[acc, anchor=east] at (1,2) {$b_2$};
  \fill[red] (1,6) circle (2.6pt);
  \node[red, anchor=west] at (1,6) {$[x]_{B} =$ (-2, 3)};
  \node[black, anchor=south] at (1.5,7.2) {$B$-grid};
\end{scope}
\end{tikzpicture}
$$

## The change-of-coordinates matrix in $\mathbb{R}^n$

When $\mathcal{B} = \{b_1, \dots, b_n\}$ is a basis for $\mathbb{R}^n$, collecting
the basis vectors as columns gives a matrix that converts $\mathcal{B}$-coordinates
back to standard coordinates. Let

$$
P_{\mathcal{B}} = [\,b_1 \;\; b_2 \;\; \cdots \;\; b_n\,].
$$

Then $x = c_1 b_1 + \cdots + c_n b_n$ rewrites as the matrix statement

$$
x = P_{\mathcal{B}}\,[x]_{\mathcal{B}}.
$$

$P_{\mathcal{B}}$ is the **change-of-coordinates matrix from $\mathcal{B}$ to the
standard basis**. Its columns are a basis, so it is invertible, and

$$
[x]_{\mathcal{B}} = P_{\mathcal{B}}^{-1}\, x.
$$

Left-multiplication by $P_{\mathcal{B}}$ goes from basis coordinates to standard
coordinates; by $P_{\mathcal{B}}^{-1}$ goes the other way.

> **Worked example.** For $b_1 = (2, 1)$, $b_2 = (-1, 1)$, and $x = (4, 5)$, find
> $[x]_{\mathcal{B}}$.
>
> The coordinates $c_1, c_2$ solve $c_1 b_1 + c_2 b_2 = x$, which is
> $P_{\mathcal{B}}\,[x]_{\mathcal{B}} = x$. Row reduce $[\,P_{\mathcal{B}} \;\; x\,]$:
> $$
> \left[\begin{array}{cc|c} 2 & -1 & 4 \\ 1 & 1 & 5 \end{array}\right]
> \sim
> \left[\begin{array}{cc|c} 1 & 1 & 5 \\ 0 & -3 & -6 \end{array}\right]
> \sim
> \left[\begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & 2 \end{array}\right].
> $$
>
> So $[x]_{\mathcal{B}} = (3, 2)$, and $x = 3b_1 + 2b_2$.

Converting between two _nonstandard_ bases is the subject of
[change of basis](/linear-algebra/vector-spaces/change-of-basis).

## The coordinate mapping is an isomorphism

The coordinate mapping respects every vector-space operation.

> **Theorem (Coordinate mapping).** Let $\mathcal{B} = \{b_1, \dots, b_n\}$ be a
> basis for a vector space $V$. Then the coordinate mapping $x \mapsto
> [x]_{\mathcal{B}}$ is a one-to-one linear transformation from $V$ onto
> $\mathbb{R}^n$.

Linearity is a direct computation. If $u = c_1 b_1 + \cdots + c_n b_n$ and $w = d_1
b_1 + \cdots + d_n b_n$, then $u + w = (c_1 + d_1)b_1 + \cdots + (c_n + d_n)b_n$, so

$$
[u + w]_{\mathcal{B}} = [u]_{\mathcal{B}} + [w]_{\mathcal{B}},
\qquad
[ru]_{\mathcal{B}} = r\,[u]_{\mathcal{B}}.
$$

The mapping preserves addition and scalar multiplication, and by uniqueness it is
one-to-one and onto. More generally the coordinate vector of a linear combination
is the same combination of the coordinate vectors:

$$
[\,c_1 u_1 + \cdots + c_p u_p\,]_{\mathcal{B}} = c_1 [u_1]_{\mathcal{B}} + \cdots + c_p [u_p]_{\mathcal{B}}.
$$

> **Definition (Isomorphism).** A one-to-one linear transformation from a vector
> space $V$ onto a vector space $W$ is an isomorphism from $V$ onto $W$. When one
> exists, $V$ and $W$ are indistinguishable as vector spaces: every calculation in
> one is reproduced in the other.

The coordinate-mapping theorem says the coordinate mapping is an isomorphism from $V$ onto
$\mathbb{R}^n$. Any real vector space with a basis of $n$ vectors is isomorphic to
$\mathbb{R}^n$, so the abstract space and the concrete one carry identical
vector-space structure under different names.

$$
% caption: The coordinate mapping is an isomorphism: it carries an abstract space
% $V$ one-to-one onto $\mathbb{R}^n$, and every vector-space operation in $V$ is
% mirrored exactly in $\mathbb{R}^n$.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{2F6DB5}
\draw[black, fill=acc!6] (-0.3,-1.4) rectangle (2.9,1.4);
\node[font=\small] at (1.3,0.9) {$V$};
\node[black, font=\scriptsize] at (1.3,-0.1) {a combination of $b_i$};
\begin{scope}[xshift=6.5cm]
  \draw[black, fill=acc!6] (-0.3,-1.4) rectangle (2.9,1.4);
  \node[font=\small] at (1.3,0.9) {$\mathbb{R}^n$};
  \node[black, font=\scriptsize] at (1.3,-0.1) {($c_1$, ..., $c_n$)};
\end{scope}
\draw[->, acc, very thick] (3.1,0.3) -- (6.0,0.3) node[midway, above, font=\scriptsize] {$x$ to $[x]_{B}$};
\draw[<-, black, thick] (3.1,-0.4) -- (6.0,-0.4) node[midway, below, font=\scriptsize] {one-to-one, onto};
\end{tikzpicture}
$$

### Polynomials become columns

Let $\mathcal{B} = \{1, t, t^2, t^3\}$ be the standard basis of $\mathbb{P}_3$. A
polynomial $p(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3$ is already written in the basis,
so $[p]_{\mathcal{B}} = (a_0, a_1, a_2, a_3)$, and $p \mapsto [p]_{\mathcal{B}}$ is
an isomorphism $\mathbb{P}_3 \to \mathbb{R}^4$. Every polynomial operation
corresponds to a column operation.

$$
% caption: The coordinate map relative to $\{1, t, t^2, t^3\}$ sends a polynomial to
% the column of its coefficients, an isomorphism $\mathbb{P}_3 \to \mathbb{R}^4$
% under which every polynomial operation becomes a column operation.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{2F6DB5}
\draw[black, fill=acc!6] (-0.3,-1.4) rectangle (3.6,1.4);
\node[font=\small] at (1.65,0.95) {$\mathbb{P}_3$};
\node[black, font=\scriptsize] at (1.65,-0.1) {$a_0 + a_1 t + a_2 t^2 + a_3 t^3$};
\begin{scope}[xshift=7.2cm]
  \draw[black, fill=acc!6] (-0.3,-1.4) rectangle (2.5,1.4);
  \node[font=\small] at (1.1,0.95) {$\mathbb{R}^4$};
  \node[black, font=\scriptsize] at (1.1,-0.1) {($a_0$, $a_1$, $a_2$, $a_3$)};
\end{scope}
\draw[->, acc, very thick] (3.8,0) -- (6.7,0) node[midway, above, font=\scriptsize] {coordinate map};
\end{tikzpicture}
$$

Questions about polynomials become questions about columns.

> **Worked example.** Are $1 + 2t^2$, $4 + t + 5t^2$, and $3 + 2t$ linearly dependent
> in $\mathbb{P}_2$?
>
> Take coordinate vectors relative to $\{1, t, t^2\}$: $(1, 0, 2)$, $(4, 1, 5)$,
> $(3, 2, 0)$. Row reduce the matrix with these columns:
> $$
> \begin{bmatrix} 1 & 4 & 3 \\ 0 & 1 & 2 \\ 2 & 5 & 0 \end{bmatrix}
> \sim
> \begin{bmatrix} 1 & 4 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}.
> $$
>
> Only two pivots, so the columns are dependent: column $3$ equals $2$ times column
> $2$ minus $5$ times column $1$. The matching relation for the polynomials is
> $$
> 3 + 2t = 2(4 + t + 5t^2) - 5(1 + 2t^2).
> $$

### A plane isomorphic to $\mathbb{R}^2$

Coordinates also work inside a subspace.

> **Worked example.** Let $v_1 = (3, 6, 2)$, $v_2 = (-1, 0, 1)$, and
> $H = \operatorname{Span}\{v_1, v_2\}$, a plane through the origin in $\mathbb{R}^3$.
> Is $x = (3, 12, 7)$ in $H$, and if so what is $[x]_{\mathcal{B}}$ for
> $\mathcal{B} = \{v_1, v_2\}$?
>
> Solve $c_1 v_1 + c_2 v_2 = x$ by row reducing $[\,v_1 \;\; v_2 \;\; x\,]$:
> $$
> \left[\begin{array}{cc|c} 3 & -1 & 3 \\ 6 & 0 & 12 \\ 2 & 1 & 7 \end{array}\right]
> \sim
> \left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{array}\right].
> $$
>
> The system is consistent, so $x$ is in $H$ with $c_1 = 2$, $c_2 = 3$; thus
> $[x]_{\mathcal{B}} = (2, 3)$. Though $H$ sits in $\mathbb{R}^3$, the pair
> $\{v_1, v_2\}$ is a basis, so $H$ is isomorphic to $\mathbb{R}^2$ and carries a
> genuine two-dimensional coordinate system.

$$
% caption: The plane $H = \operatorname{Span}\{v_1, v_2\}$ in $\mathbb{R}^3$ carries
% its own coordinate grid; $x = 2v_1 + 3v_2$ has $\mathcal{B}$-coordinates $(2, 3)$
% even though it lives in three-dimensional space.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=0.85]
\definecolor{acc}{HTML}{2F6DB5}
\definecolor{red}{HTML}{B23A48}
\foreach \i in {0,1,2,3} {
  \draw[black] ({\i*0.5},{\i*1.0}) -- ({\i*0.5+2.6},{\i*1.0+0.4});
}
\foreach \j in {0,1,2} {
  \draw[black] ({\j*1.3},{\j*0.2}) -- ({\j*1.3+1.5},{\j*0.2+3.0});
}
\draw[->, acc, very thick] (0,0) -- (1.3,0.2) node[below right] {$v_1$};
\draw[->, acc, very thick] (0,0) -- (0.5,1.0) node[above left] {$v_2$};
\coordinate (xp) at ({2*1.3+3*0.5},{2*0.2+3*1.0});
\fill[red] (xp) circle (2.4pt);
\node[red, anchor=south east] at (xp) {$x = 2v_1 + 3v_2$};
\fill[black] (0,0) circle (1.4pt);
\node[anchor=north east] at (0,0) {$0$};
\end{tikzpicture}
$$

Whether a _different_ basis for $H$ produces the same conclusion — that $H$ is
isomorphic to $\mathbb{R}^2$ rather than some other $\mathbb{R}^k$ — is settled by
[dimension](/linear-algebra/vector-spaces/dimension-and-rank), which shows the
number of basis vectors does not depend on the basis chosen.

[^lay-coord]: **Lay**, _Linear Algebra and Its Applications_, §4.4 — Coordinate Systems: Theorem 7 (unique representation), the coordinate vector and change-of-coordinates matrix $P_{\mathcal{B}}$, and Theorem 8 (the coordinate mapping as an isomorphism onto $\mathbb{R}^n$).