Coordinates Relative to a Basis

When you write a vector such as $v = \begin{bmatrix} 5 \\ 1 \end{bmatrix}$ in $\mathbb{R}^2$, you are silently reading off its components in the standard basis $\mathcal{E} = \{e_1, e_2\}$, where $e_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $e_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$. The entries $5$ and $1$ are the amounts of $e_1$ and $e_2$ needed to build $v$: $v = 5e_1 + 1e_2$.

But the standard basis is just one choice of "measuring sticks." A basis $B = \{b_1, \dots, b_n\}$ of $\mathbb{R}^n$ is any set of $n$ linearly independent vectors that span the space. Relative to a different basis $B$, the same point in space gets a different list of numbers — its coordinates.

• $v$ — the vector itself, an actual point/arrow in space. It does not change.
• $b_1, \dots, b_n$ — the basis vectors, our chosen directions.
• $c_1, \dots, c_n$ — the coordinates: how much of each $b_i$ we stack up to reach $v$.

Think of the basis as a coordinate grid tilted and stretched to taste. The vector stays put; the coordinates are just the grid-readings along that particular grid.

The whole idea only works because a basis gives exactly one recipe for each vector.

Unique Representation Theorem. If $B = \{b_1, \dots, b_n\}$ is a basis of a vector space $V$, then for every $v \in V$ there exist unique scalars $c_1, \dots, c_n$ with $v = c_1 b_1 + c_2 b_2 + \cdots + c_n b_n.$ Uniqueness is what makes coordinates well-defined: it comes directly from the basis being linearly independent. (If two recipes gave the same $v$, subtracting them would be a nontrivial dependence among the $b_i$.)

We collect those unique scalars into the coordinate vector of $v$ relative to $B$: $[v]_B = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}.$

Two cautions:

• $B$ is an ordered basis. Swapping $b_1$ and $b_2$ swaps $c_1$ and $c_2$, so the order is part of the data.
• Relative to the standard basis $\mathcal{E}$, coordinates and components coincide: $[v]_{\mathcal{E}} = v$.

To find $[v]_B$, read the defining equation $v = c_1 b_1 + \cdots + c_n b_n$ as a matrix equation. Stack the basis vectors as columns to form the change-of-coordinates matrix $P_B = B = [\,b_1\ \ b_2\ \cdots\ b_n\,].$ Then $c_1 b_1 + \cdots + c_n b_n = Bc$, so the coordinates $c = [v]_B$ are the solution of $Bc = v.$ Because the columns of $B$ are a basis, $B$ is invertible, which guarantees the unique solution $[v]_B = B^{-1} v.$ In practice you rarely invert the matrix by hand — you just solve the system $Bc = v$ by elimination or substitution. Each symbol:

• $B$ (equivalently $P_B$): $n \times n$ matrix whose $i$-th column is $b_i$.
• $c = [v]_B$: the unknown coordinate vector you are solving for.
• $v$: the known right-hand side (the vector's standard components).

Direction of the map: $B$ turns coordinates into the vector ($v = B[v]_B$), while $B^{-1}$ turns a vector into its coordinates ($[v]_B = B^{-1}v$).

Going the other way is pure arithmetic — no system to solve. Given a coordinate vector $[v]_B = \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix}$ and the basis $B$, rebuild the actual vector with a linear combination: $v = c_1 b_1 + c_2 b_2 + \cdots + c_n b_n = B[v]_B.$ This is the ideal sanity check for any coordinate computation: after you solve $Bc = v$, plug the $c_i$ back into $c_1 b_1 + \cdots + c_n b_n$ and confirm you recover the original $v$. If it does not match, a coordinate is wrong.

The two operations are perfect inverses — a round trip. Starting from $v$, computing $[v]_B = B^{-1}v$, and reconstructing $B[v]_B = B B^{-1} v = v$ returns you exactly where you started.