Linear Algebra

Coordinates in a basis and change of basis

A point on a slanted tile grid can be named by its screen position or by its tile steps. How do you turn one description into the other, and between two different grids?

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Finding an inverse by row reducing A next to ILinear Algebra Basis and dimensionLinear Algebra

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DiagonalizationLinear Algebra

Where you are 8 in Vectors in n dimensions

Linear algebra6

Vectors in n dimensions8

▴ 3 above Span Subspaces Basis and dimension

Coordinates in a basis and change of basis

Dot product, length and orthogonality in n dimensions

Why this exists

Pehuen is drawing a pixel-art game on a slanted tile grid, in the spirit of isometric art. The engine stores every position as a screen vector, so many units right and so many units up. Amancay, who designs the levels, never thinks that way: a gem sits “two tiles along one edge, three along the other”. One tile edge is \mathbf{b}_1 = (2, 1) on screen and the other is \mathbf{b}_2 = (1, 3), so her gem is at

2\mathbf{b}_1 + 3\mathbf{b}_2 = (4, 2) + (3, 9) = (7, 11).

The game needs both directions of translation: tile steps to screen position for drawing, and screen position to tile steps when a player clicks. Then Kuyen brings in a second tile set with different edges, and every level has to be converted from one grid to the other. All three jobs are the same idea. A basis is a coordinate grid, the coordinates of a vector are its steps on that grid, and an invertible matrix carries one set of coordinates to another.

Definition

Let V be a subspace of \mathbb{R}^n and \mathcal{B} = \{\mathbf{b}_1, \dots, \mathbf{b}_k\} a basis of V, taken in this order (Basis and dimension). For \mathbf{x} in V, the coordinate vector of \mathbf{x} relative to \mathcal{B} is the column of weights

[\mathbf{x}]_{\mathcal{B}} = \begin{bmatrix} c_1 \\ \vdots \\ c_k \end{bmatrix} \quad \text{where} \quad \mathbf{x} = c_1\mathbf{b}_1 + \dots + c_k\mathbf{b}_k.

The rule \mathbf{x} \mapsto [\mathbf{x}]_{\mathcal{B}}, from V to \mathbb{R}^k, is the coordinate map. In prose a column is written as a list, [\mathbf{x}]_{\mathcal{B}} = (2, 3).

When \mathcal{B} is a basis of all of \mathbb{R}^n, its vectors as columns form the n \times n matrix P_{\mathcal{B}} = \begin{bmatrix} \mathbf{b}_1 & \cdots & \mathbf{b}_n \end{bmatrix}.

If \mathcal{C} is a second basis of the same V, the change-of-coordinates matrix from \mathcal{B} to \mathcal{C} is the k \times k matrix whose columns are the \mathcal{C}-coordinates of the \mathcal{B} vectors:

P_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} [\mathbf{b}_1]_{\mathcal{C}} & \cdots & [\mathbf{b}_k]_{\mathcal{C}} \end{bmatrix}.

Intuition

The standard basis draws the square grid, and on it a vector’s coordinates are its entries. Any other basis draws a slanted grid: lines through the multiples of \mathbf{b}_1 and \mathbf{b}_2 and their sums, tiling the plane with copies of one parallelogram. Coordinates are walking instructions on that grid.

Figure 1Two steps along \mathbf{b}_1 and three along \mathbf{b}_2 reach \mathbf{x} = (7, 11), so [\mathbf{x}]_{\mathcal{B}} = (2, 3).

The matrix P_{\mathcal{B}} is a translator. Fed tile steps (c_1, c_2), it returns c_1\mathbf{b}_1 + c_2\mathbf{b}_2, the screen position, because multiplying a matrix by a vector combines its columns with those weights (Multiplying a matrix by a vector). Going back, from screen to tiles, undoes P_{\mathcal{B}}, so it is multiplication by P_{\mathcal{B}}^{-1}. Changing grids is the two trips in a row: from \mathcal{B}-steps out to the screen, then from the screen into \mathcal{C}-steps. The vector never moves; only its description changes.

Results

1. Coordinates are unique. Relative to a basis of V, each \mathbf{x} in V has exactly one coordinate vector.

2. From coordinates to the vector and back. If \mathcal{B} is a basis of \mathbb{R}^n, then P_{\mathcal{B}} is invertible and, for every \mathbf{x} in \mathbb{R}^n,

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

3. The coordinate map is linear and one-to-one onto. For \mathbf{x}, \mathbf{y} in V and a scalar c, [\mathbf{x} + \mathbf{y}]_{\mathcal{B}} = [\mathbf{x}]_{\mathcal{B}} + [\mathbf{y}]_{\mathcal{B}} and [c\mathbf{x}]_{\mathcal{B}} = c[\mathbf{x}]_{\mathcal{B}}. Different vectors have different coordinates, and every list in \mathbb{R}^k is the coordinate vector of some \mathbf{x} in V. So a list in V is independent exactly when its coordinate vectors are (Exercise 7).

4. Converting between bases. For bases \mathcal{B}, \mathcal{C} of V, P_{\mathcal{C} \leftarrow \mathcal{B}} is the only matrix with [\mathbf{x}]_{\mathcal{C}} = P_{\mathcal{C} \leftarrow \mathcal{B}}[\mathbf{x}]_{\mathcal{B}} for every \mathbf{x} in V. When V = \mathbb{R}^n,

P_{\mathcal{C} \leftarrow \mathcal{B}} = P_{\mathcal{C}}^{-1}P_{\mathcal{B}},

and row reducing \begin{bmatrix} P_{\mathcal{C}} & P_{\mathcal{B}} \end{bmatrix} to \begin{bmatrix} I & M \end{bmatrix} produces M = P_{\mathcal{C} \leftarrow \mathcal{B}}.

5. The converting matrix is invertible. P_{\mathcal{C} \leftarrow \mathcal{B}} is invertible with inverse P_{\mathcal{B} \leftarrow \mathcal{C}}, and for three bases P_{\mathcal{D} \leftarrow \mathcal{C}}P_{\mathcal{C} \leftarrow \mathcal{B}} = P_{\mathcal{D} \leftarrow \mathcal{B}}. Conversely, every invertible k \times k matrix is P_{\mathcal{C} \leftarrow \mathcal{D}} for some basis \mathcal{D} (Example 7).

Why it is true

Result 1. Basis and dimension states this; here is the proof. A basis spans V, so some weights exist. If \mathbf{x} = \sum c_i\mathbf{b}_i = \sum d_i\mathbf{b}_i, subtracting gives \sum (c_i - d_i)\mathbf{b}_i = \mathbf{0}. The \mathbf{b}_i are independent, so every c_i - d_i = 0. Spanning gives existence; independence gives uniqueness.

Result 2. By the column view of a product, P_{\mathcal{B}}\mathbf{c} = c_1\mathbf{b}_1 + \dots + c_n\mathbf{b}_n, so P_{\mathcal{B}}[\mathbf{x}]_{\mathcal{B}} = \mathbf{x} is the definition of coordinates read as a matrix equation. The n columns of the square matrix P_{\mathcal{B}} are independent, so it has a pivot in every column, hence in every row, and it reduces to I; such a matrix is invertible (Finding an inverse by row reducing A next to I). Multiply P_{\mathcal{B}}[\mathbf{x}]_{\mathcal{B}} = \mathbf{x} on the left by P_{\mathcal{B}}^{-1}.

Result 3. If \mathbf{x} = \sum c_i\mathbf{b}_i and \mathbf{y} = \sum d_i\mathbf{b}_i, then \mathbf{x} + \mathbf{y} = \sum (c_i + d_i)\mathbf{b}_i and c\mathbf{x} = \sum (cc_i)\mathbf{b}_i, and by Result 1 these weights are the coordinates. Equal coordinates name the same combination, hence the same vector. Any (c_1, \dots, c_k) is the coordinate vector of \sum c_i\mathbf{b}_i, which lies in V because a subspace is closed under combinations.

Result 4. Let \mathbf{x} = \sum c_j\mathbf{b}_j. By Result 3, [\mathbf{x}]_{\mathcal{C}} = \sum c_j[\mathbf{b}_j]_{\mathcal{C}}, which is P_{\mathcal{C} \leftarrow \mathcal{B}} times (c_1, \dots, c_k) = [\mathbf{x}]_{\mathcal{B}}. Only one matrix does this: taking \mathbf{x} = \mathbf{b}_j, whose \mathcal{B}-coordinates are \mathbf{e}_j, shows column j of any such matrix must be [\mathbf{b}_j]_{\mathcal{C}}. In \mathbb{R}^n, Result 2 gives [\mathbf{x}]_{\mathcal{C}} = P_{\mathcal{C}}^{-1}\mathbf{x} = P_{\mathcal{C}}^{-1}P_{\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}. The row operations that turn P_{\mathcal{C}} into I amount to multiplying on the left by P_{\mathcal{C}}^{-1} (Finding an inverse by row reducing A next to I), so done to the whole block they turn P_{\mathcal{B}} into P_{\mathcal{C}}^{-1}P_{\mathcal{B}}.

Result 5. In \mathbb{R}^n, P_{\mathcal{B} \leftarrow \mathcal{C}}P_{\mathcal{C} \leftarrow \mathcal{B}} = P_{\mathcal{B}}^{-1}P_{\mathcal{C}}P_{\mathcal{C}}^{-1}P_{\mathcal{B}} = I, and the product in the other order is I the same way. For three bases, P_{\mathcal{D}}^{-1}P_{\mathcal{C}}P_{\mathcal{C}}^{-1}P_{\mathcal{B}} = P_{\mathcal{D}}^{-1}P_{\mathcal{B}}. For a basis of a smaller subspace, P_{\mathcal{B}} is not square and the same facts are proved from Result 4 alone (Exercise 18). The meaning is simple: converting to \mathcal{C} and back changes nothing, so the second trip undoes the first.

Worked examples

Examples 1 to 3 belong to the first day; Examples 4 to 7 to the second.

Example 1

Day 1: tile steps and screen positions

Pehuen’s grid has \mathcal{B} = \{(2, 1), (1, 3)\}. (a) A sprite has [\mathbf{x}]_{\mathcal{B}} = (2, 3). Where is it on screen? (b) A player clicks the screen point \mathbf{y} = (1, 8). Find [\mathbf{y}]_{\mathcal{B}}. (c) Find the coordinates of \mathbf{z} = (1, 0) and say what they mean on the grid.

Solution

(a) By Result 2, \mathbf{x} = P_{\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}:

\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}\begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 7 \\ 11 \end{bmatrix}.

That is the walk in the first figure.

(b) The two vectors are not multiples, so \mathcal{B} is a basis of \mathbb{R}^2. By the formula for a 2 \times 2 inverse (Inverse of a 2 by 2 matrix, and solving with it), with 2 \cdot 3 - 1 \cdot 1 = 5,

\begin{aligned} P_{\mathcal{B}}^{-1} &= \frac{1}{5}\begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}, \\ P_{\mathcal{B}}^{-1}\begin{bmatrix} 1 \\ 8 \end{bmatrix} &= \frac{1}{5}\begin{bmatrix} -5 \\ 15 \end{bmatrix} = \begin{bmatrix} -1 \\ 3 \end{bmatrix}. \end{aligned}

So [\mathbf{y}]_{\mathcal{B}} = (-1, 3). Check: -(2, 1) + 3(1, 3) = (1, 8).

(c) P_{\mathcal{B}}^{-1}(1, 0) = \left(\tfrac{3}{5}, -\tfrac{1}{5}\right), the first column of P_{\mathcal{B}}^{-1}. Every screen point has \mathcal{B}-coordinates, but the tile corners are exactly the points whose coordinates are integers, and \mathbf{z} sits inside a tile.

Example 2

Day 1: a basis of \mathbb{R}^3 from the columns of a matrix

Let \mathcal{U} = \{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\} be the columns of

P = \begin{bmatrix} 1 & 0 & 2 \\ 1 & 1 & 3 \\ 0 & 1 & 2 \end{bmatrix}.

Show \mathcal{U} is a basis of \mathbb{R}^3, find P^{-1}, and find [\mathbf{x}]_{\mathcal{U}} for \mathbf{x} = (5, 9, 7).

Solution

Row reduce P with I beside it. R_2 - R_1, then R_3 - R_2:

\left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0 \\ 0 & 1 & 1 & -1 & 1 & 0 \\ 0 & 0 & 1 & 1 & -1 & 1 \end{array}\right].

Three pivots, so the columns are independent and, three of them in \mathbb{R}^3, a basis. Clearing above the last pivot with R_2 - R_3 and R_1 - 2R_3 leaves

P^{-1} = \begin{bmatrix} -1 & 2 & -2 \\ -2 & 2 & -1 \\ 1 & -1 & 1 \end{bmatrix}.

Then

P^{-1}\begin{bmatrix} 5 \\ 9 \\ 7 \end{bmatrix} = \begin{bmatrix} -5 + 18 - 14 \\ -10 + 18 - 7 \\ 5 - 9 + 7 \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \\ 3 \end{bmatrix},

so [\mathbf{x}]_{\mathcal{U}} = (-1, 1, 3). Check: -(1, 1, 0) + (0, 1, 1) + 3(2, 3, 2) = (5, 9, 7). Once P^{-1} is known, every further vector costs one product; its columns are [\mathbf{e}_1]_{\mathcal{U}}, [\mathbf{e}_2]_{\mathcal{U}}, [\mathbf{e}_3]_{\mathcal{U}}.

Example 3

Day 1: coordinates on a plane in \mathbb{R}^3

Kuyen tilts a level into the plane V: x - y + z = 0 in \mathbb{R}^3, with basis \mathcal{B} = \{(1, 1, 0), (1, 0, -1)\}. Find [\mathbf{x}]_{\mathcal{B}} for \mathbf{x} = (4, 1, -3), and explain why (1, 0, 0) has no coordinates.

Solution

Both vectors satisfy the equation and are not multiples, and the plane has dimension 2, so \mathcal{B} is a basis of V. A combination is

c_1(1, 1, 0) + c_2(1, 0, -1) = (c_1 + c_2,\ c_1,\ -c_2).

Matching (4, 1, -3): c_1 = 1, c_2 = 3, and the first entry checks, 1 + 3 = 4. So [\mathbf{x}]_{\mathcal{B}} = (1, 3), a list of two numbers for a vector with three entries: the coordinate map sends V to \mathbb{R}^2, and by Result 3 it turns the plane into a copy of \mathbb{R}^2. Here P_{\mathcal{B}} is 3 \times 2 and has no inverse, so coordinates come from solving.

For (1, 0, 0) the entries force c_1 = 0 and c_2 = 0, and then the first entry is 0, not 1. No weights work, because 1 - 0 + 0 \neq 0: the vector is not in V, and coordinates relative to \mathcal{B} exist only for vectors of V.

Example 4

Day 2: converting to a second tile set

Amancay’s second grid uses \mathcal{C} = \{(1, 1), (1, 2)\}. Find P_{\mathcal{C} \leftarrow \mathcal{B}} for Pehuen’s \mathcal{B} = \{(2, 1), (1, 3)\}, and convert the sprite with [\mathbf{x}]_{\mathcal{B}} = (2, 3).

Solution

By Result 4, P_{\mathcal{C} \leftarrow \mathcal{B}} = P_{\mathcal{C}}^{-1}P_{\mathcal{B}}. Here 1 \cdot 2 - 1 \cdot 1 = 1, so

\begin{aligned} P_{\mathcal{C}}^{-1} &= \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}, \\ P_{\mathcal{C} \leftarrow \mathcal{B}} &= \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} \\ &= \begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}. \end{aligned}

The columns say \mathbf{b}_1 = 3\mathbf{c}_1 - \mathbf{c}_2 and \mathbf{b}_2 = -\mathbf{c}_1 + 2\mathbf{c}_2; for instance 3(1, 1) - (1, 2) = (2, 1). Then

[\mathbf{x}]_{\mathcal{C}} = \begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \end{bmatrix},

and 3(1, 1) + 4(1, 2) = (7, 11), the same screen point as before.

Figure 2The same point \mathbf{x} = (7, 11) on the second grid is four steps along \mathbf{c}_2 and three along \mathbf{c}_1, so [\mathbf{x}]_{\mathcal{C}} = (3, 4). The order of the steps does not matter.

Example 5

Day 2: going back

Find P_{\mathcal{B} \leftarrow \mathcal{C}} for the bases of Example 4 in two ways. Then find the \mathcal{B}-coordinates of the points with [\mathbf{y}]_{\mathcal{C}} = (1, 3) and [\mathbf{w}]_{\mathcal{C}} = (1, 0).

Solution

By Result 5 it is the inverse of P_{\mathcal{C} \leftarrow \mathcal{B}}, whose entries give 3 \cdot 2 - (-1)(-1) = 5:

P_{\mathcal{B} \leftarrow \mathcal{C}} = \frac{1}{5}\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}.

Directly, P_{\mathcal{B}}^{-1}P_{\mathcal{C}} = \tfrac{1}{5}\begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} gives the same matrix. Then [\mathbf{y}]_{\mathcal{B}} = \tfrac{1}{5}(2 + 3, 1 + 9) = (1, 2); check, \mathbf{y} = (1, 1) + 3(1, 2) = (4, 7) = (2, 1) + 2(1, 3). And [\mathbf{w}]_{\mathcal{B}} = \left(\tfrac{2}{5}, \tfrac{1}{5}\right).

So P_{\mathcal{C} \leftarrow \mathcal{B}} has integer entries and its inverse does not. On the grids: every corner of a \mathcal{B} tile is a corner of a \mathcal{C} tile, but not the other way. A \mathcal{B} tile has area 5 and a \mathcal{C} tile area 1, so the coarse grid’s corners are a fifth of the fine grid’s.

Example 6

Day 2: two bases of \mathbb{R}^3 by one row reduction

With \mathcal{U} the columns of P from Example 2, let \mathcal{W} = \{(1, 1, 1), (1, 0, 1), (1, 2, 0)\}. Find P_{\mathcal{U} \leftarrow \mathcal{W}}, and convert [\mathbf{x}]_{\mathcal{W}} = (1, 1, 1).

Solution

By Result 4, row reduce \begin{bmatrix} P & P_{\mathcal{W}} \end{bmatrix}:

\left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 1 & 1 \\ 1 & 1 & 3 & 1 & 0 & 2 \\ 0 & 1 & 2 & 1 & 1 & 0 \end{array}\right].

The same four operations as in Example 2 (R_2 - R_1, R_3 - R_2, R_2 - R_3, R_1 - 2R_3) turn P into I. On the right they give

P_{\mathcal{U} \leftarrow \mathcal{W}} = \begin{bmatrix} -1 & -3 & 3 \\ -1 & -3 & 2 \\ 1 & 2 & -1 \end{bmatrix},

which is P^{-1}P_{\mathcal{W}}; with P^{-1} already known, that product is a second route. (Reducing P_{\mathcal{W}} alone gives three pivots, so \mathcal{W} is a basis.) Check the second column: -3(1, 1, 0) - 3(0, 1, 1) + 2(2, 3, 2) = (1, 0, 1). Finally

[\mathbf{x}]_{\mathcal{U}} = P_{\mathcal{U} \leftarrow \mathcal{W}}(1, 1, 1) = (-1, -2, 2),

the row sums of the matrix. Indeed \mathbf{x} = (1, 1, 1) + (1, 0, 1) + (1, 2, 0) = (3, 3, 2), and -(1, 1, 0) - 2(0, 1, 1) + 2(2, 3, 2) = (3, 3, 2).

Example 7

Day 2: which matrices convert coordinates

Kuyen wants a third grid \mathcal{D} with P_{\mathcal{C} \leftarrow \mathcal{D}} = M = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}, with Amancay’s \mathcal{C}. Find \mathcal{D}. Then show no grid has P_{\mathcal{C} \leftarrow \mathcal{D}} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}.

Solution

The columns of P_{\mathcal{C} \leftarrow \mathcal{D}} are [\mathbf{d}_1]_{\mathcal{C}} and [\mathbf{d}_2]_{\mathcal{C}}, so

\begin{aligned} \mathbf{d}_1 &= 2\mathbf{c}_1 + \mathbf{c}_2 = (3, 4), \\ \mathbf{d}_2 &= \mathbf{c}_1 + \mathbf{c}_2 = (2, 3). \end{aligned}

In matrix form P_{\mathcal{D}} = P_{\mathcal{C}}M = \begin{bmatrix} 3 & 2 \\ 4 & 3 \end{bmatrix}, which has 3 \cdot 3 - 2 \cdot 4 = 1 \neq 0, so it is invertible and \mathcal{D} is a basis.

For the second matrix the same recipe gives \mathbf{d}_1 = \mathbf{c}_1 + 2\mathbf{c}_2 = (3, 5) and \mathbf{d}_2 = 2\mathbf{c}_1 + 4\mathbf{c}_2 = (6, 10) = 2\mathbf{d}_1, a dependent list. In general P_{\mathcal{D}} = P_{\mathcal{C}}M is invertible exactly when M is, so the invertible matrices are exactly the change-of-coordinates matrices, as Result 5 says.

Boundaries

Coordinates need a basis. A dependent spanning list gives some vectors many weight lists (Exercise 10); an independent list that does not span gives some vectors none.
A basis is an ordered list here. Reordering it reorders every coordinate vector (Exercise 9).
[\mathbf{x}]_{\mathcal{B}} exists only for \mathbf{x} in the space \mathcal{B} spans (Example 3).
[\mathbf{x}]_{\mathcal{B}} = P_{\mathcal{B}}^{-1}\mathbf{x} and P_{\mathcal{C} \leftarrow \mathcal{B}} = P_{\mathcal{C}}^{-1}P_{\mathcal{B}} need bases of all of \mathbb{R}^n. For a basis of a smaller subspace P_{\mathcal{B}} is n \times k with k < n and has no inverse; coordinates come from solving, and P_{\mathcal{C} \leftarrow \mathcal{B}} from its columns (Exercise 17).
P_{\mathcal{C} \leftarrow \mathcal{B}} needs \mathcal{B} and \mathcal{C} to be bases of the same space.

Where this goes next

A Linear transformation has a matrix in the standard basis, but nothing forces that choice. Described in the coordinates of another basis, the same transformation has the matrix P_{\mathcal{B}}^{-1}AP_{\mathcal{B}}, and choosing the basis well can make that matrix diagonal. That is Diagonalization, where a basis of eigenvectors turns a transformation into plain scaling along each direction. The coordinate map also gives a working rule used everywhere: any question about independence, span or dimension inside a subspace can be moved to \mathbb{R}^k and answered by row reduction there.

Exercises

Day 1: coordinates in a basis

Let \mathcal{B} = \{(1, 2), (3, 1)\}. (a) If [\mathbf{x}]_{\mathcal{B}} = (2, -1), find \mathbf{x}. (b) Find [\mathbf{y}]_{\mathcal{B}} for \mathbf{y} = (8, 1).
For \mathcal{B} of Exercise 1, write P_{\mathcal{B}} and P_{\mathcal{B}}^{-1}, and use them to find the coordinates of (5, 0) and (0, 5).
Show that (1, 0, 1), (1, 1, 0), (0, 1, 1) is a basis of \mathbb{R}^3 and find the coordinates of (3, 5, 4) relative to it.
Let Q = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 3 & 3 \end{bmatrix}. Find Q^{-1} by row reduction, then the coordinates of (2, 1, 4) and of \mathbf{e}_3 relative to the columns of Q.
Show that (1, 2, 0), (0, 1, 1) is a basis of the plane 2x - y + z = 0, find the coordinates of (3, 7, 1), and explain why (1, 1, 1) has none.
On Pehuen’s grid \mathcal{B} = \{(2, 1), (1, 3)\}, which of the screen points (0, 5), (3, 2), (2, -4) are tile corners?

Day 1: reasoning

Let \mathcal{B} be a basis of V. Prove that vectors \mathbf{x}_1, \dots, \mathbf{x}_m in V are independent exactly when [\mathbf{x}_1]_{\mathcal{B}}, \dots, [\mathbf{x}_m]_{\mathcal{B}} are.
Explain why [\mathbf{b}_j]_{\mathcal{B}} = \mathbf{e}_j, and why, for a basis of \mathbb{R}^n, column j of P_{\mathcal{B}}^{-1} is [\mathbf{e}_j]_{\mathcal{B}}.
Let \mathcal{B}' list Pehuen’s vectors in the other order, (1, 3) then (2, 1). Find [(7, 11)]_{\mathcal{B}'} and explain how [\mathbf{x}]_{\mathcal{B}'} relates to [\mathbf{x}]_{\mathcal{B}} for every \mathbf{x}.
The list (2, 1), (4, 2) is not a basis of \mathbb{R}^2. Show that (5, 5) is no combination of it and that (6, 3) is a combination in more than one way, and say which half of Result 1 fails each time.

Day 2: change of basis

Let \mathcal{B} = \{(1, 2), (3, 1)\} and \mathcal{C} = \{(1, 0), (1, 1)\}. Find P_{\mathcal{C} \leftarrow \mathcal{B}} and [\mathbf{x}]_{\mathcal{C}} when [\mathbf{x}]_{\mathcal{B}} = (2, -1).
For the bases of Exercise 11, find P_{\mathcal{B} \leftarrow \mathcal{C}} and the \mathcal{B}-coordinates of the vector with \mathcal{C}-coordinates (1, 3).
Let \mathcal{Q} be the columns of Q in Exercise 4 and \mathcal{W} the basis of Exercise 3. Find P_{\mathcal{Q} \leftarrow \mathcal{W}} by row reducing \begin{bmatrix} Q & P_{\mathcal{W}} \end{bmatrix}, and convert [\mathbf{x}]_{\mathcal{W}} = (1, 2, 3).
Kuyen’s grid is \mathcal{D} = \{(1, -1), (1, 1)\}. Find P_{\mathcal{D} \leftarrow \mathcal{B}} for Pehuen’s \mathcal{B}, and the \mathcal{D}-coordinates of the sprite with \mathcal{B}-coordinates (2, 3).
With \mathcal{C} = \{(1, 1), (1, 2)\} and \mathcal{D} of Exercise 14, find P_{\mathcal{D} \leftarrow \mathcal{C}} and check that P_{\mathcal{D} \leftarrow \mathcal{C}}P_{\mathcal{C} \leftarrow \mathcal{B}} equals your answer to Exercise 14.
With \mathcal{C} = \{(1, 0), (1, 1)\}, find the basis \mathcal{D} for which P_{\mathcal{C} \leftarrow \mathcal{D}} = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}.
On the plane 2x - y + z = 0 take \mathcal{B} = \{(1, 2, 0), (0, 1, 1)\} and \mathcal{C} = \{(1, 1, -1), (1, 3, 1)\}. Find the 2 \times 2 matrix P_{\mathcal{C} \leftarrow \mathcal{B}} and convert [\mathbf{x}]_{\mathcal{B}} = (3, 1).

Day 2: reasoning

Let \mathcal{B}, \mathcal{C}, \mathcal{D} be bases of a subspace V. Using only Result 4, prove P_{\mathcal{B} \leftarrow \mathcal{B}} = I and P_{\mathcal{D} \leftarrow \mathcal{C}}P_{\mathcal{C} \leftarrow \mathcal{B}} = P_{\mathcal{D} \leftarrow \mathcal{B}}, and deduce that P_{\mathcal{C} \leftarrow \mathcal{B}} is invertible.
Prove that if P_{\mathcal{C} \leftarrow \mathcal{B}} = I then \mathcal{B} = \mathcal{C}, and that a matrix that is not invertible is never a change-of-coordinates matrix.
Prove that [\mathbf{x}]_{\mathcal{B}} = \mathbf{x} for every \mathbf{x} in \mathbb{R}^n exactly when \mathcal{B} is the standard basis.

Answers

(a) \mathbf{x} = (-1, 3): 2(1, 2) - (3, 1). (b) [\mathbf{y}]_{\mathcal{B}} = (-1, 3): c_1 + 3c_2 = 8 and 2c_1 + c_2 = 1 give c_1 = -1, c_2 = 3.
P_{\mathcal{B}} = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}, \displaystyle P_{\mathcal{B}}^{-1} = \frac{1}{5}\begin{bmatrix} -1 & 3 \\ 2 & -1 \end{bmatrix} (here 1 - 6 = -5). Coordinates (-1, 2) and (3, -1): -(1, 2) + 2(3, 1) = (5, 0), 3(1, 2) - (3, 1) = (0, 5).
Coordinates (1, 2, 3). Adding the three equations c_1 + c_2 = 3, c_2 + c_3 = 5, c_1 + c_3 = 4 gives c_1 + c_2 + c_3 = 6, then c_3 = 3, c_1 = 1, c_2 = 2. Every vector is reached the same way, so the list spans, and three spanning vectors in \mathbb{R}^3 are a basis.
Q^{-1} = \begin{bmatrix} 0 & -3 & 1 \\ 1 & 2 & -1 \\ -1 & -1 & 1 \end{bmatrix}, from R_3 - R_1, R_3 - R_2, R_2 - R_3, R_1 - R_3, R_1 - 2R_2. Coordinates of (2, 1, 4): (1, 0, 1); of \mathbf{e}_3: (1, -1, 1), the third column of Q^{-1}.
Coordinates (3, 1). Both vectors satisfy the equation and are not multiples; the plane has dimension 2. Matching c_1(1, 2, 0) + c_2(0, 1, 1) = (c_1, 2c_1 + c_2, c_2) gives c_1 = 3, c_2 = 1, and 6 + 1 = 7. (1, 1, 1) gives 2 - 1 + 1 = 2 \neq 0, so it is not in the plane.
(0, 5) and (2, -4). With P_{\mathcal{B}}^{-1} from Example 1, the coordinates are (-1, 2), \left(\tfrac{7}{5}, \tfrac{1}{5}\right) and (2, -2); only integer coordinates are corners.
By Result 3, \sum a_i\mathbf{x}_i = \mathbf{0} exactly when \sum a_i[\mathbf{x}_i]_{\mathcal{B}} = [\mathbf{0}]_{\mathcal{B}} = \mathbf{0}, since the map is linear and only \mathbf{0} has zero coordinates. So the two lists have the same dependence relations.
\mathbf{b}_j = 0\mathbf{b}_1 + \dots + 1\mathbf{b}_j + \dots + 0\mathbf{b}_n, and by Result 1 these weights are the coordinates. Column j of P_{\mathcal{B}}^{-1} is P_{\mathcal{B}}^{-1}\mathbf{e}_j = [\mathbf{e}_j]_{\mathcal{B}} by Result 2.
[(7, 11)]_{\mathcal{B}'} = (3, 2). The same combination with the terms listed in the other order, so the two entries of [\mathbf{x}]_{\mathcal{B}} always swap.
a(2, 1) + b(4, 2) = (2a + 4b, a + 2b) always has first entry twice the second, and (5, 5) does not: existence fails, because the list does not span. (6, 3) = 3(2, 1) = 1(2, 1) + 1(4, 2): uniqueness fails, because the list is dependent.
P_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} -1 & 2 \\ 2 & 1 \end{bmatrix}, columns [(1, 2)]_{\mathcal{C}} and [(3, 1)]_{\mathcal{C}}, since (1, 2) = -(1, 0) + 2(1, 1) and (3, 1) = 2(1, 0) + (1, 1). [\mathbf{x}]_{\mathcal{C}} = (-4, 3), and -4(1, 0) + 3(1, 1) = (-1, 3) as in Exercise 1.
\displaystyle P_{\mathcal{B} \leftarrow \mathcal{C}} = \frac{1}{5}\begin{bmatrix} -1 & 2 \\ 2 & 1 \end{bmatrix}, the inverse of the matrix in Exercise 11 (-1 - 4 = -5). Coordinates (1, 1): the vector is (1, 0) + 3(1, 1) = (4, 3) = (1, 2) + (3, 1).
P_{\mathcal{Q} \leftarrow \mathcal{W}} = \begin{bmatrix} 1 & -3 & -2 \\ 0 & 3 & 1 \\ 0 & -2 & 0 \end{bmatrix}, by the operations of Exercise 4 on \begin{bmatrix} Q & P_{\mathcal{W}} \end{bmatrix}. [\mathbf{x}]_{\mathcal{Q}} = (-11, 9, -4); check, Q(-11, 9, -4) = (3, 5, 4).
\displaystyle P_{\mathcal{D} \leftarrow \mathcal{B}} = \frac{1}{2}\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}, from \displaystyle P_{\mathcal{D}}^{-1} = \frac{1}{2}\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} times P_{\mathcal{B}}. Coordinates (-2, 9): -2(1, -1) + 9(1, 1) = (7, 11).
\displaystyle P_{\mathcal{D} \leftarrow \mathcal{C}} = P_{\mathcal{D}}^{-1}P_{\mathcal{C}} = \frac{1}{2}\begin{bmatrix} 0 & -1 \\ 2 & 3 \end{bmatrix}. Times \begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix} it gives \displaystyle \frac{1}{2}\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}.
\mathcal{D} = \{(5, 3), (3, 2)\}: \mathbf{d}_1 = 2(1, 0) + 3(1, 1) and \mathbf{d}_2 = (1, 0) + 2(1, 1). The matrix has 4 - 3 = 1 \neq 0, so \mathcal{D} is a basis.
\displaystyle P_{\mathcal{C} \leftarrow \mathcal{B}} = \frac{1}{2}\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}: (1, 2, 0) = \tfrac{1}{2}(1, 1, -1) + \tfrac{1}{2}(1, 3, 1) and (0, 1, 1) = -\tfrac{1}{2}(1, 1, -1) + \tfrac{1}{2}(1, 3, 1). [\mathbf{x}]_{\mathcal{C}} = (1, 2): \mathbf{x} = (3, 7, 1) = (1, 1, -1) + 2(1, 3, 1).
I satisfies [\mathbf{x}]_{\mathcal{B}} = I[\mathbf{x}]_{\mathcal{B}}, and Result 4 says only one matrix does, so P_{\mathcal{B} \leftarrow \mathcal{B}} = I. The product P_{\mathcal{D} \leftarrow \mathcal{C}}P_{\mathcal{C} \leftarrow \mathcal{B}} sends [\mathbf{x}]_{\mathcal{B}} to P_{\mathcal{D} \leftarrow \mathcal{C}}[\mathbf{x}]_{\mathcal{C}} = [\mathbf{x}]_{\mathcal{D}} for every \mathbf{x}, so by uniqueness it is P_{\mathcal{D} \leftarrow \mathcal{B}}. Taking \mathcal{D} = \mathcal{B} gives P_{\mathcal{B} \leftarrow \mathcal{C}}P_{\mathcal{C} \leftarrow \mathcal{B}} = I, and swapping the roles gives the other order.
If P_{\mathcal{C} \leftarrow \mathcal{B}} = I, column j says [\mathbf{b}_j]_{\mathcal{C}} = \mathbf{e}_j, so \mathbf{b}_j = \mathbf{c}_j for each j. A change-of-coordinates matrix has an inverse by Result 5, so a matrix without one is not such a matrix.
[\mathbf{x}]_{\mathcal{B}} = \mathbf{x} for all \mathbf{x} means P_{\mathcal{B}}^{-1} = I (compare columns, using \mathbf{x} = \mathbf{e}_j), so P_{\mathcal{B}} = I and \mathbf{b}_j = \mathbf{e}_j. Conversely the standard basis has P_{\mathcal{B}} = I.