Math Core

The standard basis is convenient, but it is often not the natural one. A crystal lattice, a rotated coordinate frame or a polynomial expanded around $t = 1$ each comes with its own preferred basis. Once a basis is fixed, every vector gets a unique list of coordinates, and a change-of-basis matrix translates coordinates from one basis to another.

Coordinates relative to a basis

The key fact is that a basis gives each vector exactly one address.

Unique representation

If $\mathcal{B} = \lbrace \mathbf{b}_1, \dots, \mathbf{b}_n \rbrace$ is a basis for $V$ , then every $\mathbf{x}$ in $V$ can be written in exactly one way as

\mathbf{x} = c_1\mathbf{b}_1 + \cdots + c_n\mathbf{b}_n.

Such weights exist because $\mathcal{B}$ spans $V$ . They are unique because if $\mathbf{x}$ also equals $d_1\mathbf{b}_1 + \cdots + d_n\mathbf{b}_n$ , subtracting gives $(c_1 - d_1)\mathbf{b}_1 + \cdots + (c_n - d_n)\mathbf{b}_n = \mathbf{0}$ , and independence forces every $c_i - d_i = 0$ .

Definition

Coordinate vector

The weights $c_1, \dots, c_n$ are the coordinates of $\mathbf{x}$ relative to $\mathcal{B}$ , and the vector

[\mathbf{x}]_{\mathcal{B}} = \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix}

in $\mathbb{R}^n$ is the $\mathcal{B}$ -coordinate vector of $\mathbf{x}$ .

In $\mathbb{R}^n$ , collect the basis vectors into the change-of-coordinates matrix $P_{\mathcal{B}} = [\,\mathbf{b}_1 \;\cdots\; \mathbf{b}_n\,]$ . The vector equation above says exactly

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

Its columns are a basis, so $P_{\mathcal{B}}$ is invertible and $[\mathbf{x}]_{\mathcal{B}} = P_{\mathcal{B}}^{-1}\mathbf{x}$ . In practice you usually solve the system rather than invert.

Worked example: Finding and using coordinates

Let $\mathcal{B} = \lbrace (2, 1), (-1, 1) \rbrace$ , a basis for $\mathbb{R}^2$ .

(a) Find $[\mathbf{x}]_{\mathcal{B}}$ for $\mathbf{x} = (1, 5)$ . Solve $c_1(2, 1) + c_2(-1, 1) = (1, 5)$ :

2c_1 - c_2 = 1, \qquad c_1 + c_2 = 5.

Adding the equations gives $3c_1 = 6$ , so $c_1 = 2$ and $c_2 = 3$ . Thus $[\mathbf{x}]_{\mathcal{B}} = (2, 3)$ . Check: $2(2, 1) + 3(-1, 1) = (1, 5)$ .

(b) If $[\mathbf{y}]_{\mathcal{B}} = (-1, 4)$ , then $\mathbf{y} = -1(2, 1) + 4(-1, 1) = (-6, 3)$ .

Picture it this way: the basis vectors $(2, 1)$ and $(-1, 1)$ lay down a slanted grid over the plane, and $[\mathbf{x}]_{\mathcal{B}} = (2, 3)$ says "walk $2$ steps along $\mathbf{b}_1$ and $3$ steps along $\mathbf{b}_2$ ."

(2, 1)(-1, 1)(1, 5)y = x/2y = -xOpen in grapher →

Coordinates in other vector spaces

Coordinates work in any vector space with a finite basis, and they turn abstract problems into problems in $\mathbb{R}^n$ . The map $\mathbf{x} \mapsto [\mathbf{x}]_{\mathcal{B}}$ is linear and one-to-one onto $\mathbb{R}^n$ (an isomorphism), so questions about spans and independence in $V$ can be answered by the same questions about coordinate vectors.

Worked example: Coordinates of a polynomial

Let $\mathcal{B} = \lbrace 1,\; 1 + t,\; 1 + t + t^2 \rbrace$ , a basis for $\mathbb{P}_2$ . Find $[p]_{\mathcal{B}}$ for $p(t) = 4 - t + 2t^2$ .

Write $c_1 + c_2(1 + t) + c_3(1 + t + t^2) = (c_1 + c_2 + c_3) + (c_2 + c_3)t + c_3 t^2$ and match coefficients:

c_3 = 2, \qquad c_2 + c_3 = -1, \qquad c_1 + c_2 + c_3 = 4.

So $c_3 = 2$ , $c_2 = -3$ and $c_1 = 5$ , giving $[p]_{\mathcal{B}} = (5, -3, 2)$ .

Changing from one basis to another

Now suppose $V$ has two bases, $\mathcal{B} = \lbrace \mathbf{b}_1, \dots, \mathbf{b}_n \rbrace$ and $\mathcal{C} = \lbrace \mathbf{c}_1, \dots, \mathbf{c}_n \rbrace$ . If you know $[\mathbf{x}]_{\mathcal{B}} = (x_1, \dots, x_n)$ , then $\mathbf{x} = x_1\mathbf{b}_1 + \cdots + x_n\mathbf{b}_n$ . Taking $\mathcal{C}$ -coordinates, which is a linear operation,

[\mathbf{x}]_{\mathcal{C}} = x_1[\mathbf{b}_1]_{\mathcal{C}} + \cdots + x_n[\mathbf{b}_n]_{\mathcal{C}}.

That is a matrix-vector product.

The change-of-basis matrix

There is a unique $n \times n$ matrix $\underset{\mathcal{C} \leftarrow \mathcal{B}}{P}$ with

[\mathbf{x}]_{\mathcal{C}} = \underset{\mathcal{C} \leftarrow \mathcal{B}}{P}\,[\mathbf{x}]_{\mathcal{B}} \quad \text{for every } \mathbf{x} \text{ in } V.

Its columns are the $\mathcal{C}$ -coordinate vectors of the $\mathcal{B}$ basis vectors:

\underset{\mathcal{C} \leftarrow \mathcal{B}}{P} = \Big[\, [\mathbf{b}_1]_{\mathcal{C}} \;\cdots\; [\mathbf{b}_n]_{\mathcal{C}} \,\Big].

It is invertible, and its inverse is $\underset{\mathcal{B} \leftarrow \mathcal{C}}{P}$ .

In $\mathbb{R}^n$ you find all the columns at once. Finding $[\mathbf{b}_j]_{\mathcal{C}}$ means solving $[\,\mathbf{c}_1 \;\cdots\; \mathbf{c}_n\,]\mathbf{y} = \mathbf{b}_j$ , and these systems share a coefficient matrix. So row reduce

[\,\mathbf{c}_1 \;\cdots\; \mathbf{c}_n \mid \mathbf{b}_1 \;\cdots\; \mathbf{b}_n\,] \;\sim\; [\, I \mid \underset{\mathcal{C} \leftarrow \mathcal{B}}{P} \,].

Worked example: Building a change-of-basis matrix

In $\mathbb{R}^2$ , let $\mathbf{b}_1 = (1, 1)$ , $\mathbf{b}_2 = (1, -1)$ , $\mathbf{c}_1 = (1, 0)$ , $\mathbf{c}_2 = (1, 1)$ . Find $\underset{\mathcal{C} \leftarrow \mathcal{B}}{P}$ , then find $[\mathbf{x}]_{\mathcal{C}}$ when $[\mathbf{x}]_{\mathcal{B}} = (3, 1)$ .

Row reduce $[\,\mathbf{c}_1 \; \mathbf{c}_2 \mid \mathbf{b}_1 \; \mathbf{b}_2\,]$ with $R_1 - R_2$ :

\left[\begin{array}{cc|cc} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & -1 \end{array}\right] \sim \left[\begin{array}{cc|cc} 1 & 0 & 0 & 2 \\ 0 & 1 & 1 & -1 \end{array}\right], \qquad \underset{\mathcal{C} \leftarrow \mathcal{B}}{P} = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix}.

(Check a column: $[\mathbf{b}_2]_{\mathcal{C}} = (2, -1)$ means $2(1, 0) - (1, 1) = (1, -1)$ . Correct.)

Then

[\mathbf{x}]_{\mathcal{C}} = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \end{bmatrix}.

Both answers describe the same vector: $3(1, 1) + 1(1, -1) = (4, 2)$ and $2(1, 0) + 2(1, 1) = (4, 2)$ .

Common mistake

The columns of $\underset{\mathcal{C} \leftarrow \mathcal{B}}{P}$ are the $\mathcal{B}$ vectors written in $\mathcal{C}$ -coordinates, not the other way around. The notation reads right to left: it accepts $\mathcal{B}$ -coordinates on the right and produces $\mathcal{C}$ -coordinates on the left. In the augmented matrix, the target basis $\mathcal{C}$ goes on the left of the bar.

Tip

When $\mathcal{C}$ is the standard basis, $[\mathbf{b}_j]_{\mathcal{C}} = \mathbf{b}_j$ , so $\underset{\mathcal{C} \leftarrow \mathcal{B}}{P}$ is just $P_{\mathcal{B}}$ . In general, $\underset{\mathcal{C} \leftarrow \mathcal{B}}{P} = P_{\mathcal{C}}^{-1}P_{\mathcal{B}}$ : convert from $\mathcal{B}$ to standard, then from standard to $\mathcal{C}$ .

Practice

Practice 1

Let $\mathcal{B} = \lbrace (1, 1), (1, -1) \rbrace$ . Find $[\mathbf{x}]_{\mathcal{B}}$ for $\mathbf{x} = (5, 1)$ . Enter it as $(c_1, c_2)$ .

Enter a point like (2, -3)

Practice 2

Let $\mathcal{B} = \lbrace (2, 0, 1), (0, 1, 0), (1, 0, 1) \rbrace$ and $[\mathbf{x}]_{\mathcal{B}} = (1, -2, 3)$ . Find $\mathbf{x}$ .

Enter a point like (2, -3)

Practice 3

Let $\mathcal{B} = \lbrace (1, 0, 0), (1, 1, 0), (1, 1, 1) \rbrace$ . Find $[\mathbf{x}]_{\mathcal{B}}$ for $\mathbf{x} = (3, -1, 2)$ .

Enter a point like (2, -3)

Practice 4

Let $\mathcal{B} = \lbrace 1,\; t - 1,\; (t - 1)^2 \rbrace$ , a basis for $\mathbb{P}_2$ . Find $[p]_{\mathcal{B}}$ for $p(t) = t^2$ .

Enter a point like (2, -3)

Practice 5

$\mathcal{B}$ and $\mathcal{C}$ are bases of a vector space $V$ . What are the columns of $\underset{\mathcal{C} \leftarrow \mathcal{B}}{P}$ ?

Practice 6

Bases $\mathcal{B} = \lbrace \mathbf{b}_1, \mathbf{b}_2 \rbrace$ and $\mathcal{C} = \lbrace \mathbf{c}_1, \mathbf{c}_2 \rbrace$ of a vector space satisfy $\mathbf{b}_1 = 2\mathbf{c}_1 + \mathbf{c}_2$ and $\mathbf{b}_2 = -\mathbf{c}_1 + 3\mathbf{c}_2$ . If $[\mathbf{x}]_{\mathcal{B}} = (1, 2)$ , find $[\mathbf{x}]_{\mathcal{C}}$ .

Enter a point like (2, -3)

Practice 7

In $\mathbb{R}^2$ , let $\mathcal{B} = \lbrace (1, 2), (2, 5) \rbrace$ and $\mathcal{C} = \lbrace (1, 1), (1, 2) \rbrace$ . Find $\underset{\mathcal{C} \leftarrow \mathcal{B}}{P}$ , and use it to find $[\mathbf{x}]_{\mathcal{C}}$ when $[\mathbf{x}]_{\mathcal{B}} = (2, -1)$ .

Enter a point like (2, -3)