13. Normal vectors

Suppose you have drawn a vector $q$ perpendicular to another vector $p$, and now you transform $p$ by a matrix $\textcolor{#1d4ed8}{M}$ into $\textcolor{#1d4ed8}{M}p$. You would like the perpendicular vector to follow along and stay perpendicular. The catch is that $\textcolor{#1d4ed8}{M}q$ is usually not perpendicular to $\textcolor{#1d4ed8}{M}p$. So the question is: by what matrix $\textcolor{#be185d}{N}$ should we transform $q$ so that $\textcolor{#be185d}{N}q$ stays perpendicular to $\textcolor{#1d4ed8}{M}p$?

13.1 The problem

import numpy as np

pie.clear()
pie.scalar(110, out='unit')
GRAY = '#71717a'
PAIR_COLORS = ['#dc2626', '#d97706', '#16a34a', '#2563eb']
p0 = pie.point(0, 0, out='p0', hidden=True)
pie.circle(p0, 1, out='circ', stroke=GRAY)
# Four points p on the unit circle, and a normal q = p turned by 90° at the tip of each
angles = np.radians([0, 90, 180, 270])
px, py = np.cos(angles), np.sin(angles)
pvecs = pie.la_vec2d(px.tolist(), py.tolist(), out='pvecs')
pie.set_stroke(pvecs, PAIR_COLORS)
qvecs = pie.la_vec2d((-py).tolist(), px.tolist(), px.tolist(), py.tolist(), out='qvecs')
pie.set_stroke(qvecs, PAIR_COLORS)
lab_p = pie.text('p', 1.25, 0, 15, out='labP', stroke=PAIR_COLORS[0])
lab_q = pie.text('q', 1.25, 1.25, 15, out='labQ', stroke=PAIR_COLORS[0])

Take several points $p$ evenly spaced on a circle and attach to each a normal vector $q$, meaning a vector perpendicular to $p$. If we transform the points by a matrix

for node in [pvecs, qvecs, lab_p, lab_q]:
    pie.hide(node)
# M starts as the identity, row-major
M = [pie.scalar(v, out=f'm{k}') for k, v in enumerate([1, 0, 0, 1])]
# Row `row` of a 2×2 matrix times the vectors (xs[i], ys[i])
def apply_row(m11, m12, m21, m22, xs, ys, row):
    xs, ys = np.array(xs), np.array(ys)
    return m11 * xs + m12 * ys if row == 0 else m21 * xs + m22 * ys

apply = pie.lift(apply_row, pie.Array[pie.Scalar])
mpx, mpy = apply(*M, px.tolist(), py.tolist(), 0), apply(*M, px.tolist(), py.tolist(), 1)
mqx, mqy = apply(*M, (-py).tolist(), px.tolist(), 0), apply(*M, (-py).tolist(), px.tolist(), 1)
mpvecs = pie.la_vec2d(mpx, mpy, out='mpvecs')
pie.set_stroke(mpvecs, PAIR_COLORS)
mqvecs = pie.la_vec2d(mqx, mqy, mpx, mpy, out='mqvecs')
pie.set_stroke(mqvecs, PAIR_COLORS)
pie.la_mat_to_ellipse([[M[0], M[1]], [M[2], M[3]]], p0, out='ellipse', stroke=GRAY)

# Labels and the angle mark for the first pair
scaled = pie.lift(lambda k, *vs: k * sum(vs), pie.Scalar)
pie.text('Mp', scaled(1.25, mpx[0]), scaled(1.25, mpy[0]), 15, out='labMP', stroke=PAIR_COLORS[0])
pie.text('Mq', scaled(1.25, mpx[0], mqx[0]), scaled(1.25, mpy[0], mqy[0]), 15, out='labMQ', stroke=PAIR_COLORS[0])
mp_tip = pie.point(mpx[0], mpy[0], out='mptipA', hidden=True)
ext_p = pie.point(scaled(2, mpx[0]), scaled(2, mpy[0]), out='extPA', hidden=True)
mq_tip = pie.point(scaled(1, mpx[0], mqx[0]), scaled(1, mpy[0], mqy[0]), out='mqtipA', hidden=True)
line_mp = pie.line(mp_tip, ext_p, out='lineMpA', hidden=True)
line_mq = pie.line(mp_tip, mq_tip, out='lineMqA', hidden=True)
pie.annotate_angle_mark(line_mp, line_mq, out='angleA')
pie.text('Transforming p and its normal q by the same M breaks the right angle: Mq no longer meets Mp at 90 degrees.', -0.5, -2.5, 14, 3.5, -0.5, out='sumBoxA', stroke=GRAY)

and transform their normals the same way, the transformed pairs are no longer at right angles.

pie.translate_scalar_to(M, [1.5, -1, 0, -1], animate=True)

13.2 Working out the solution

We know $\langle p, q \rangle = 0$ and we want $\langle \textcolor{#1d4ed8}{M}p, \textcolor{#be185d}{N}q \rangle = 0$, where $\textcolor{#be185d}{N}$ is the unknown matrix to apply to the normal. Writing the dot product as a matrix product $\langle x, y \rangle = x^T y$ and using $(\textcolor{#1d4ed8}{M}x)^T = x^T \textcolor{#1d4ed8}{M}^T$:

Compare this last line with the fact we started from, $\langle p, q \rangle = p^T q = 0$. If we could make the middle factor $\textcolor{#1d4ed8}{M}^T \textcolor{#be185d}{N}$ disappear, the two expressions would match and $\langle \textcolor{#1d4ed8}{M}p, \textcolor{#be185d}{N}q \rangle$ would be forced to zero exactly when $\langle p, q \rangle$ is. That happens precisely when

So $\textcolor{#be185d}{N}$ is the inverse of the transpose of $\textcolor{#1d4ed8}{M}$. This matrix is important enough that it has a name: it is called the inverse-transpose (or the cotangent / normal matrix in graphics, where it is exactly how surface normals are transformed).

13.3 The solution

For our matrix

transposing gives

and inverting that gives

import numpy as np

pie.clear()
pie.scalar(110, out='unit')
GRAY = '#71717a'
PAIR_COLORS = ['#dc2626', '#d97706', '#16a34a', '#2563eb']
p0 = pie.point(0, 0, out='p0', hidden=True)
pie.circle(p0, 1, out='circ', stroke=GRAY)
angles = np.radians([0, 90, 180, 270])
px, py = np.cos(angles), np.sin(angles)
# M and N start as the identity, row-major
M = [pie.scalar(v, out=f'm{k}') for k, v in enumerate([1, 0, 0, 1])]
N = [pie.scalar(v, out=f'n{k}') for k, v in enumerate([1, 0, 0, 1])]
# Row `row` of a 2×2 matrix times the vectors (xs[i], ys[i])
def apply_row(m11, m12, m21, m22, xs, ys, row):
    xs, ys = np.array(xs), np.array(ys)
    return m11 * xs + m12 * ys if row == 0 else m21 * xs + m22 * ys

apply = pie.lift(apply_row, pie.Array[pie.Scalar])
mpx, mpy = apply(*M, px.tolist(), py.tolist(), 0), apply(*M, px.tolist(), py.tolist(), 1)
pie.la_mat_to_ellipse([[M[0], M[1]], [M[2], M[3]]], p0, out='ellipse', stroke=GRAY)
mpvecs = pie.la_vec2d(mpx, mpy, out='mpvecs')
pie.set_stroke(mpvecs, PAIR_COLORS)
scaled = pie.lift(lambda k, *vs: k * sum(vs), pie.Scalar)
pie.text('Mp', scaled(1.25, mpx[0]), scaled(1.25, mpy[0]), 15, out='labMP', stroke=PAIR_COLORS[0])

Let's redo the experiment on the same four pairs, but this time transform each $p$ by $M$

pie.translate_scalar_to(M, [1.5, -1, 0, -1], animate=True)

and its normal $q$ by $N$ instead of $M$.

nqx, nqy = apply(*N, (-py).tolist(), px.tolist(), 0), apply(*N, (-py).tolist(), px.tolist(), 1)
nqvecs = pie.la_vec2d(nqx, nqy, mpx, mpy, out='nqvecs')
pie.set_stroke(nqvecs, PAIR_COLORS)
pie.text('Nq', scaled(1.25, mpx[0], nqx[0]), scaled(1.25, mpy[0], nqy[0]), 15, out='labNQ', stroke=PAIR_COLORS[0])
mp_tip = pie.point(mpx[0], mpy[0], out='mptipB', hidden=True)
ext_p = pie.point(scaled(2, mpx[0]), scaled(2, mpy[0]), out='extPB', hidden=True)
nq_tip = pie.point(scaled(1, mpx[0], nqx[0]), scaled(1, mpy[0], nqy[0]), out='nqtipB', hidden=True)
line_mp = pie.line(mp_tip, ext_p, out='lineMpB', hidden=True)
line_nq = pie.line(mp_tip, nq_tip, out='lineNqB', hidden=True)
pie.annotate_angle_mark(line_mp, line_nq, out='angleB')
pie.text('Transforming the normal by N restores every right angle: Nq meets Mp at 90 degrees again.', -0.5, -2.5, 14, 3.5, -0.5, out='sumBoxB', stroke=GRAY)
pie.translate_scalar_to(N, [0.6667, 0, -0.6667, -1], animate=True)

Now every normal meets its $Mp$ at a right angle again, all the way around the circle.

This is the whole point of the inverse-transpose. Whenever you transform geometry by a matrix $\textcolor{#1d4ed8}{M}$, its normals must be transformed by $(\textcolor{#1d4ed8}{M}^T)^{-1}$ to keep pointing perpendicular to the surface.

Why not just reuse $M$?

It is tempting to think a normal is "just another vector" and transform it by the same $\textcolor{#1d4ed8}{M}$. Section 13.1 shows why that fails: $\textcolor{#1d4ed8}{M}$ stretches and shears the space, and a direction that was perpendicular before the shear is tilted afterwards. Only the inverse-transpose undoes exactly the part of $\textcolor{#1d4ed8}{M}$ that would have spoiled the right angle.

In the next chapter we turn to eigenvectors and eigenvalues, the special directions a matrix leaves unrotated, scaling them by just a number.

← 12. Rotations · 14. Eigenvectors and eigenvalues →