2.7 Coordinates in a New Basis

If a subspace has infinitely many bases and no reason to prefer one, then naming a point of it is a two-part act: you choose a basis, and only then can you say which combination of it you mean. That second part is what coordinates are, and the previous section quietly established that they always exist and are always unique — spanning gives existence, independence gives uniqueness.

Make the choice deliberately awkward, so that nothing can be read off by habit. In the plane $V : z = 0.5 x + 0.5 y$ , take

$𝐛_{1} = (2, 1, 1.5), 𝐛_{2} = (1, 3, 2)$

Both lie in $V$ — check the third coordinates against $0.5 x plus 0.5 y$ : $0.5 (2) + 0.5 (1) = 1.5$ and $0.5 (1) + 0.5 (3) = 2$ . Neither is a multiple of the other, so by the previous section's counting argument they are a basis of $V$ . They are also at an oblique angle and of different lengths, which is the point: a basis owes you nothing but spanning and independence.

Given coefficients $s$ and $a$ , the point they name is

$s 𝐛_{1} + a 𝐛_{2} = (2 s + a, s + 3 a, 1.5 s + 2 a)$

and we write $[𝐩]_{ℬ} = (s, a)$ to say that $𝐩$ is the point with those coordinates in the basis $ℬ equals open brace 𝐛 sub 1, 𝐛 sub 2 close brace$ . The figure turns that formula into a pair of dials.

A large semi-transparent tilted plane through the origin. A short red arrow and a short blue arrow lie in it, meeting at the origin at an oblique angle rather than a right angle - the skewed basis. Two amber lines, one parallel to the red arrow and one parallel to the blue, also lie in the plane and cross at a single point; two sliders slide them across the sheet, so their crossing is the point with the chosen pair of coordinates. A black dot further out on the sheet marks the target (5, 5, 5); dialling the sliders to 2 and 1 slides the crossing exactly onto it, which is what it means for that point to have coordinates 2 and 1 in this basis. — The plane $V$ with the skewed basis drawn from the origin: red $𝐛 sub 1 equals open parenthesis 2, 1, 1.5 close parenthesis$ and blue $𝐛 sub 2 equals open parenthesis 1, 3, 2 close parenthesis$ . The two amber lines are the coordinate lines of that basis, one running in each basis direction, and they cross at the point named by the current sliders $s$ and $a$ . The black dot further out on the sheet is the target $open parenthesis 5, 5, 5 close parenthesis$ .

Explore the figure

At the defaults $s equals 1$ , $a equals 0$ the crossing sits on the tip of the red arrow. Of course it does — coordinates $open parenthesis 1, 0 close parenthesis$ mean one $𝐛 sub 1$ and no $𝐛 sub 2$ , so the basis vectors are the points with coordinates $open parenthesis 1, 0 close parenthesis$ and $open parenthesis 0, 1 close parenthesis$ , exactly as $𝐞 sub 1$ and $𝐞 sub 2$ are in the standard grid.
Set $s equals 0$ , $a equals 0$ . The crossing collapses onto the origin. Every basis puts its origin in the same place, because $0 𝐛_{1} + 0 𝐛_{2} = 0$ whatever the basis is.
Now hunt the black dot. Drag $s$ and $a$ until the crossing lands on $open parenthesis 5, 5, 5 close parenthesis$ ; you should find it at $s equals 2$ , $a equals 1$ . Those two numbers are the target's coordinates in this basis, and reading them off the sliders is reading them off the figure.
Watch the amber lines rather than the crossing as you drag one slider. They stay parallel to the basis directions and slide, never rotating. The two families of amber lines are a grid on the plane — a skewed one, but a perfectly good one.

Finding coordinates is solving a system

Step 3 was a search. Doing it properly means solving for $s$ and $a$ , and the equation is $s 𝐛_{1} + a 𝐛_{2} = 𝐩$ . Since a point of $V$ is determined by its first two coordinates — the third is forced by the plane — only two equations carry information:

$2 s + a = 5, s + 3 a = 5$

Solve however you like: the first gives $a = 5 - 2 s$ , and substituting into the second gives $s + 15 - 6 s = 5$ , so $- 5 s = - 10$ and $s equals 2$ , whence $a equals 1$ . The coordinates of $open parenthesis 5, 5, 5 close parenthesis$ are $open bracket open parenthesis 5, 5, 5 close parenthesis close bracket sub ℬ equals open parenthesis 2, 1 close parenthesis$ , agreeing with the sliders. Confirm it forwards: $2 (2, 1, 1.5) + (1, 3, 2) = (4, 2, 3) + (1, 3, 2) = (5, 5, 5)$ .

Doing that from scratch for every point is wasteful, because the left-hand side never changes. Collect the basis vectors' first two coordinates into the coordinate matrix

$C equals open parenthesis 2, 1; 1, 3 close parenthesis$

whose columns are $𝐛 sub 1$ and $𝐛 sub 2$ as seen in the standard grid. Then $C open parenthesis s, a close parenthesis equals open parenthesis p sub 1, p sub 2 close parenthesis$ , and the coordinates come from inverting once and reusing forever.

[\begin{matrix} \frac{3}{5} & - \frac{1}{5} \\ - \frac{1}{5} & \frac{2}{5} \end{matrix}]

The change-of-coordinates matrix ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The inverse of the coordinate matrix, computed symbolically. Multiplying a point's standard coordinates by this matrix returns its $ℬ$ -coordinates in one step. On the target: $\frac{3}{5} (5) - \frac{1}{5} (5) = 3 - 1 = 2$ and $- \frac{1}{5} (5) + \frac{2}{5} (5) = - 1 + 2 = 1$ , which is the pair $open parenthesis 2, 1 close parenthesis$ the sliders found. An inverse exists here for the same reason the basis is a basis — the columns are independent — and the next chapter makes that equivalence official.

The two matrices are a translation pair. $C$ takes coordinates and returns the point; $C^{- 1}$ takes the point and returns its coordinates. Neither direction is more fundamental than the other, which is worth insisting on, because the standard grid feels like the truth and a skewed basis feels like a description of it. It is not so: $𝐞 sub 1, 𝐞 sub 2, 𝐞 sub 3$ are themselves a basis chosen for convenience, and the entries of any vector you have ever written down were already coordinates with respect to it.

Example 1 A point, its coordinates, and the round trip

The point $𝐪 equals open parenthesis 4, 7, 5.5 close parenthesis$ lies in $V$ . Find $[𝐪]_{ℬ}$ using the inverse, then reconstruct $𝐪$ from the coordinates as a check.

First confirm $𝐪$ is in $V$ : $0.5 (4) + 0.5 (7) = 5.5$ , matching the third coordinate, so the question is legitimate — a point outside $V$ would have no coordinates in a basis of $V$ at all.

Apply $C^{- 1}$ to the first two standard coordinates $open parenthesis 4, 7 close parenthesis$ :

$s = \frac{3}{5} (4) - \frac{1}{5} (7) = \frac{12 - 7}{5} = 1, a = - \frac{1}{5} (4) + \frac{2}{5} (7) = \frac{- 4 + 14}{5} = 2$

so $open bracket 𝐪 close bracket sub ℬ equals open parenthesis 1, 2 close parenthesis$ .

Now the round trip. $1 𝐛_{1} + 2 𝐛_{2} = (2, 1, 1.5) + (2, 6, 4) = (4, 7, 5.5) = 𝐪$ . Note that the third coordinate came out right without being asked for: the reconstruction used only the two-dimensional system, and the plane supplied the rest. That is the sense in which a two-dimensional subspace really only has room for two numbers, however many coordinates its points appear to carry.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.