Linear Algebra

Finding the matrix of a transformation

A printmaker turns or flips a design about a fixed pin. Where does every point of the design land, and which matrix does the moving?

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Why this exists

Nenad cuts a woodcut block and pins it to the press bed through a hole at the origin. Between pulls he turns the block about the pin, or flips it face down across a guide line through the pin. Libuse plans each sheet on graph paper and wants, for each move, one rule saying where every point lands.

Linear transformation proved that a linear rule T: \mathbb{R}^n \to \mathbb{R}^m is \mathbf{x} \mapsto A\mathbf{x}, where column j of A is T(\mathbf{e}_j). So Libuse only has to follow \mathbf{e}_1 and \mathbf{e}_2 through the move. This lesson does that geometry for rotations, reflections and projections that fix the origin, and for turns of space about a coordinate axis.

Definition

Let L be a line through the origin with a unit direction vector \mathbf{u}, so \lVert \mathbf{u} \rVert = 1.

The rotation by \theta, written R_\theta, turns every vector counterclockwise about the origin through the angle \theta, keeping its length (Rotations, with the center at the origin). A negative \theta turns clockwise.
The projection onto L, written P, sends \mathbf{x} to the foot of the perpendicular from \mathbf{x} to L, the point of L closest to \mathbf{x}.
The reflection across L, written F, sends \mathbf{x} to its mirror image: L is the perpendicular bisector of the segment from \mathbf{x} to F(\mathbf{x}), and F(\mathbf{x}) = \mathbf{x} for \mathbf{x} on L.

Each of these is linear (proved below), so each has a standard matrix \big[\,T(\mathbf{e}_1)\ \ T(\mathbf{e}_2)\,\big], called the matrix of the transformation. The same letter names the move and its matrix.

Intuition

The vector \mathbf{e}_1 is the point of the The unit circle at angle 0. Turned through \theta, it becomes the point at angle \theta, which is (\cos\theta, \sin\theta) by the definition of cosine and sine. The vector \mathbf{e}_2 is the point at 90^\circ, and it lands at \theta + 90^\circ:

\begin{aligned} &\big(\cos(\theta + 90^\circ),\ \sin(\theta + 90^\circ)\big) \\ &\quad = (-\sin\theta,\ \cos\theta). \end{aligned}

Those are the two columns. The figure shows Nenad’s third of a turn, \theta = 120^\circ.

Figure 1Turning by 120^\circ carries \mathbf{e}_1 to the point of the unit circle at 120^\circ and \mathbf{e}_2 to the point at 210^\circ. Those two images are the columns of R_{120^\circ}.

For a projection or a reflection the tool changes from the unit circle to the dot product: dropping a perpendicular onto L is a dot product with \mathbf{u}, and reflecting means walking to the foot of that perpendicular and the same distance again.

Results

Rotation of the plane.

R_\theta = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}.

It keeps lengths: \lVert R_\theta\mathbf{x} \rVert = \lVert \mathbf{x} \rVert for every \mathbf{x}.

Projection onto a line. If L has unit direction \mathbf{u} = (a, b), then P\mathbf{x} = (\mathbf{x} \cdot \mathbf{u})\,\mathbf{u}, and

P = \begin{bmatrix} a^2 & ab \\ ab & b^2 \end{bmatrix}.

P fixes every vector on L and sends every vector perpendicular to L to \mathbf{0}.

Reflection across a line. With the same \mathbf{u} = (a, b), F\mathbf{x} = 2P\mathbf{x} - \mathbf{x}, and

F = \begin{bmatrix} a^2 - b^2 & 2ab \\ 2ab & b^2 - a^2 \end{bmatrix}.

If L makes the angle \varphi with the positive x-axis, this is

F = \begin{bmatrix} \cos 2\varphi & \sin 2\varphi \\ \sin 2\varphi & -\cos 2\varphi \end{bmatrix}.

Reflecting twice returns every vector: F(F\mathbf{x}) = \mathbf{x}.

Rotation of space about a coordinate axis. Write c = \cos\theta and s = \sin\theta. A positive turn follows the right-hand rule (thumb along the positive axis, fingers curling in the direction of the turn): about the z-axis it carries x toward y, about the x-axis y toward z, and about the y-axis z toward x.

\begin{aligned} R_z &= \begin{bmatrix} c & -s & 0 \\ s & c & 0 \\ 0 & 0 & 1 \end{bmatrix}, \\ R_x &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & c & -s \\ 0 & s & c \end{bmatrix}, \\ R_y &= \begin{bmatrix} c & 0 & s \\ 0 & 1 & 0 \\ -s & 0 & c \end{bmatrix}. \end{aligned}

Which 2 \times 2 matrices these are. The columns of every rotation and reflection matrix are perpendicular unit vectors. Conversely, a 2 \times 2 matrix with perpendicular unit columns is either \begin{bmatrix} c & -s \\ s & c \end{bmatrix}, a rotation, or \begin{bmatrix} c & s \\ s & -c \end{bmatrix}, a reflection, where c^2 + s^2 = 1.

Why it is true

Rotations are linear. A turn about the origin is rigid: it carries the parallelogram on \mathbf{u} and \mathbf{v}, diagonal included, to the parallelogram on R\mathbf{u} and R\mathbf{v}, so R(\mathbf{u} + \mathbf{v}) = R\mathbf{u} + R\mathbf{v}; and it carries c\mathbf{u}, which is \mathbf{u} stretched by |c| and reversed if c < 0, to R\mathbf{u} stretched and reversed the same way, so R(c\mathbf{u}) = c\,R\mathbf{u}. The columns come from the Intuition section.

Lengths. For \mathbf{x} = (x, y),

\begin{aligned} \lVert R_\theta\mathbf{x} \rVert^2 &= (x\cos\theta - y\sin\theta)^2 \\ &\quad + (x\sin\theta + y\cos\theta)^2 \\ &= (x^2 + y^2)(\cos^2\theta + \sin^2\theta) \\ &= x^2 + y^2, \end{aligned}

since the cross terms \mp 2xy\sin\theta\cos\theta cancel.

Projection. The foot is t\mathbf{u} for some scalar t, and \mathbf{x} - t\mathbf{u} is perpendicular to \mathbf{u}:

(\mathbf{x} - t\mathbf{u}) \cdot \mathbf{u} = \mathbf{x} \cdot \mathbf{u} - t\,(\mathbf{u} \cdot \mathbf{u}) = 0.

Because \mathbf{u} \cdot \mathbf{u} = 1, t = \mathbf{x} \cdot \mathbf{u}. The dot product distributes over sums and lets scalars out, so \mathbf{x} \mapsto (\mathbf{x} \cdot \mathbf{u})\mathbf{u} is linear. Since \mathbf{e}_1 \cdot \mathbf{u} = a and \mathbf{e}_2 \cdot \mathbf{u} = b, the columns are a\mathbf{u} = (a^2, ab) and b\mathbf{u} = (ab, b^2). So P(k\mathbf{u}) = k\mathbf{u}, and P\mathbf{x} = \mathbf{0} when \mathbf{x} \cdot \mathbf{u} = 0.

Reflection. L is the perpendicular bisector of the segment from \mathbf{x} to F\mathbf{x}, so the midpoint of that segment is the foot P\mathbf{x}: \tfrac12(\mathbf{x} + F\mathbf{x}) = P\mathbf{x}, which gives F\mathbf{x} = 2P\mathbf{x} - \mathbf{x}. That is a combination of linear maps, so it is linear. Its first column is 2(a^2, ab) - (1, 0) = (2a^2 - 1, 2ab), and 2a^2 - 1 = a^2 - b^2 because a^2 + b^2 = 1; the second column works the same way. With a = \cos\varphi and b = \sin\varphi, the double-angle formulas turn a^2 - b^2 into \cos 2\varphi and 2ab into \sin 2\varphi.

Reflecting twice. P fixes every vector on L, and P\mathbf{x} is on L, so P(P\mathbf{x}) = P\mathbf{x}. Then by linearity P(F\mathbf{x}) = 2P(P\mathbf{x}) - P\mathbf{x} = P\mathbf{x}, and

\begin{aligned} F(F\mathbf{x}) &= 2P(F\mathbf{x}) - F\mathbf{x} \\ &= 2P\mathbf{x} - (2P\mathbf{x} - \mathbf{x}) = \mathbf{x}. \end{aligned}

Space. A turn about the z-axis fixes \mathbf{e}_3, on the axis, and turns \mathbf{e}_1 and \mathbf{e}_2 as in the plane. About the x-axis, \mathbf{e}_2 and \mathbf{e}_3 play those parts. About the y-axis the turn carries z toward x, so \mathbf{e}_3 plays the part of \mathbf{e}_1 and \mathbf{e}_1 that of \mathbf{e}_2: \mathbf{e}_3 \mapsto c\,\mathbf{e}_3 + s\,\mathbf{e}_1 and \mathbf{e}_1 \mapsto c\,\mathbf{e}_1 - s\,\mathbf{e}_3, which puts -s below the diagonal.

Which matrices. The columns of R_\theta and F are unit vectors with dot product 0. Conversely, a unit first column is (c, s) with c^2 + s^2 = 1, and the unit vectors perpendicular to it are (-s, c), giving R_\theta with \cos\theta = c, \sin\theta = s, and (s, -c), giving F with 2\varphi = \theta.

Worked examples

Example 1

A third of a turn

Nenad turns the block 120^\circ counterclockwise. Find the matrix, and where the corner at (4, 2) lands.

Solution

From the unit circle, \cos 120^\circ = -\tfrac12 and \displaystyle \sin 120^\circ = \tfrac{\sqrt3}{2}. So \mathbf{e}_1 lands at \displaystyle (-\tfrac12, \tfrac{\sqrt3}{2}) and \mathbf{e}_2 at \displaystyle (-\sin 120^\circ, \cos 120^\circ) = (-\tfrac{\sqrt3}{2}, -\tfrac12), as the figure in the Intuition section shows. These are the columns:

R_{120^\circ} = \begin{bmatrix} -\tfrac12 & -\tfrac{\sqrt3}{2} \\ \tfrac{\sqrt3}{2} & -\tfrac12 \end{bmatrix}.

For the corner, take 4 of the first column and 2 of the second:

R_{120^\circ}\begin{bmatrix} 4 \\ 2 \end{bmatrix} = \begin{bmatrix} -2 - \sqrt3 \\ 2\sqrt3 - 1 \end{bmatrix}.

Check: \displaystyle (2 + \sqrt3)^2 = 7 + 4\sqrt3 and \displaystyle (2\sqrt3 - 1)^2 = 13 - 4\sqrt3, which add to 20 = 4^2 + 2^2.

Example 2

A clockwise turn

Libuse turns a second copy 135^\circ clockwise. Find the matrix and the image of \displaystyle (\sqrt2, 3\sqrt2).

Solution

Clockwise means \theta = -135^\circ. The point of the unit circle at -135^\circ is \displaystyle (-\tfrac{\sqrt2}{2}, -\tfrac{\sqrt2}{2}), so \displaystyle \cos\theta = \sin\theta = -\tfrac{\sqrt2}{2}, and

R_{-135^\circ} = \begin{bmatrix} -\tfrac{\sqrt2}{2} & \tfrac{\sqrt2}{2} \\ -\tfrac{\sqrt2}{2} & -\tfrac{\sqrt2}{2} \end{bmatrix}.

In the second column, \displaystyle -\sin\theta = \tfrac{\sqrt2}{2}. Then

R_{-135^\circ}\begin{bmatrix} \sqrt2 \\ 3\sqrt2 \end{bmatrix} = \begin{bmatrix} -1 + 3 \\ -1 - 3 \end{bmatrix} = \begin{bmatrix} 2 \\ -4 \end{bmatrix},

since \displaystyle \tfrac{\sqrt2}{2} \cdot \sqrt2 = 1. Both vectors have squared length 20.

Example 3

Dropping the design onto a guide line

Libuse slides every point of the design straight onto the guide line y = \tfrac52 x, along the perpendicular. Find the matrix of this projection, and check it on (2, 5), (-5, 2) and (7, 3).

Solution

The line has direction (2, 5), of length \displaystyle \sqrt{29}, so \displaystyle \mathbf{u} = \tfrac{1}{\sqrt{29}}(2, 5). Then \displaystyle \mathbf{e}_1 \cdot \mathbf{u} = \tfrac{2}{\sqrt{29}} and \displaystyle \mathbf{e}_2 \cdot \mathbf{u} = \tfrac{5}{\sqrt{29}}, so

\begin{aligned} P\mathbf{e}_1 &= \tfrac{2}{\sqrt{29}}\mathbf{u} = (\tfrac{4}{29}, \tfrac{10}{29}), \\ P\mathbf{e}_2 &= \tfrac{5}{\sqrt{29}}\mathbf{u} = (\tfrac{10}{29}, \tfrac{25}{29}). \end{aligned}

Figure 2Projecting onto y = \tfrac52 x drops each standard vector straight onto the line. The feet, (\tfrac{4}{29}, \tfrac{10}{29}) and (\tfrac{10}{29}, \tfrac{25}{29}), are the columns of P.

The matrix is

P = \tfrac{1}{29}\begin{bmatrix} 4 & 10 \\ 10 & 25 \end{bmatrix}.

Checks: P(2, 5) = (2, 5), so a point on the line stays put, and P(-5, 2) = \mathbf{0}, as it must since (-5, 2) \cdot (2, 5) = 0. And P(7, 3) = \tfrac{1}{29}(58, 145) = (2, 5), which is right because (7, 3) - (2, 5) = (5, -2) is perpendicular to the line.

Example 4

Flipping across the same line

Nenad lays the block face down across the guide line y = \tfrac52 x. Find the matrix of the reflection, and confirm that flipping twice returns the design.

Solution

From F\mathbf{x} = 2P\mathbf{x} - \mathbf{x}, column by column, using Example 3:

\begin{aligned} F\mathbf{e}_1 &= 2(\tfrac{4}{29}, \tfrac{10}{29}) - (1, 0) \\ &= (-\tfrac{21}{29}, \tfrac{20}{29}), \\ F\mathbf{e}_2 &= 2(\tfrac{10}{29}, \tfrac{25}{29}) - (0, 1) \\ &= (\tfrac{20}{29}, \tfrac{21}{29}). \end{aligned}

F = \tfrac{1}{29}\begin{bmatrix} -21 & 20 \\ 20 & 21 \end{bmatrix}.

Figure 3Reflecting across y = \tfrac52 x sends \mathbf{e}_1 to (-\tfrac{21}{29}, \tfrac{20}{29}) and \mathbf{e}_2 to (\tfrac{20}{29}, \tfrac{21}{29}). The line crosses each dotted segment at its midpoint and at a right angle.

Check, as in the figure: the midpoint of \mathbf{e}_1 and F\mathbf{e}_1 is (\tfrac{4}{29}, \tfrac{10}{29}), on the line, and F\mathbf{e}_1 - \mathbf{e}_1 = (-\tfrac{50}{29}, \tfrac{20}{29}) is perpendicular to (2, 5). Flipping twice: F applied to its own columns gives

\begin{aligned} F(-\tfrac{21}{29}, \tfrac{20}{29}) &= \tfrac{1}{841}(441 + 400,\ 0) \\ &= (1, 0), \\ F(\tfrac{20}{29}, \tfrac{21}{29}) &= \tfrac{1}{841}(0,\ 400 + 441) \\ &= (0, 1). \end{aligned}

The second flip returns \mathbf{e}_1 and \mathbf{e}_2, and since it preserves combinations x_1\mathbf{e}_1 + x_2\mathbf{e}_2, it returns every \mathbf{x}.

Example 5

Reading the move off a matrix

Ruaridh finds two matrices in Nenad’s notes, M = \tfrac{1}{13}\begin{bmatrix} 5 & -12 \\ 12 & 5 \end{bmatrix} and N = \tfrac{1}{13}\begin{bmatrix} 5 & 12 \\ 12 & -5 \end{bmatrix}. Say what move each one is.

Solution

Both have unit columns, since 5^2 + 12^2 = 13^2, with dot product 0. So each is a rotation or a reflection; the sign pattern decides.

M has the pattern \begin{bmatrix} c & -s \\ s & c \end{bmatrix}: the rotation by the angle \theta with \cos\theta = \tfrac{5}{13}, \sin\theta = \tfrac{12}{13}, about 67.4^\circ.

N has the pattern \begin{bmatrix} c & s \\ s & -c \end{bmatrix}: a reflection, with \cos 2\varphi = \tfrac{5}{13} and \sin 2\varphi = \tfrac{12}{13}. By the half-angle formula, the line’s slope is

\tan\varphi = \frac{\sin 2\varphi}{1 + \cos 2\varphi} = \frac{12/13}{18/13} = \frac23,

so the line is y = \tfrac23 x. Check: N(3, 2) = \tfrac{1}{13}(39, 26) = (3, 2), fixed, and N(-2, 3) = \tfrac{1}{13}(26, -39) = (2, -3), sent to its negative, as a vector perpendicular to the mirror must be.

Example 6

A turn of space about the y-axis

Nenad’s press bed sits in the xz-plane of \mathbb{R}^3 and the pin runs along the y-axis. Find the matrix that turns space 150^\circ about the y-axis, and the image of (2, 1, 0).

Solution

Here \displaystyle c = \cos 150^\circ = -\tfrac{\sqrt3}{2} and s = \sin 150^\circ = \tfrac12. The axis vector \mathbf{e}_2 stays put. The positive turn carries z toward x, so \displaystyle \mathbf{e}_3 \mapsto c\,\mathbf{e}_3 + s\,\mathbf{e}_1 = (\tfrac12, 0, -\tfrac{\sqrt3}{2}) and \displaystyle \mathbf{e}_1 \mapsto c\,\mathbf{e}_1 - s\,\mathbf{e}_3 = (-\tfrac{\sqrt3}{2}, 0, -\tfrac12). Then

R_y = \begin{bmatrix} -\tfrac{\sqrt3}{2} & 0 & \tfrac12 \\ 0 & 1 & 0 \\ -\tfrac12 & 0 & -\tfrac{\sqrt3}{2} \end{bmatrix}.

Taking 2 of the first column and 1 of the second, \displaystyle R_y(2, 1, 0) = (-\sqrt3, 1, -1). Check: 3 + 1 + 1 = 5 = 2^2 + 1^2.

Example 7

Turning one mark onto another

Nenad wants the registration mark at (4, 3) to land on (0, 5). Find the rotation that does it, and where the mark at (1, -2) goes.

Solution

Both marks lie 5 from the pin, so a turn can work. Write c = \cos\theta, s = \sin\theta. Then R_\theta(4, 3) = (4c - 3s,\ 4s + 3c), and setting this equal to (0, 5) gives a linear system in c and s:

\begin{aligned} 4c - 3s &= 0, \\ 3c + 4s &= 5. \end{aligned}

Four times the second equation minus three times the first gives 25s = 20, so s = \tfrac45 and c = \tfrac35. As c^2 + s^2 = 1, these are the cosine and sine of an angle, about 53.1^\circ. So

R_\theta = \tfrac15\begin{bmatrix} 3 & -4 \\ 4 & 3 \end{bmatrix},

and R_\theta(1, -2) = \tfrac15(3 + 8,\ 4 - 6) = (\tfrac{11}{5}, -\tfrac25). Check: \tfrac{121 + 4}{25} = 5 = 1^2 + (-2)^2.

Boundaries

The pin must be the origin. A turn about another point, or a flip across a line missing the origin, moves \mathbf{0}, so it is not linear and has no 2 \times 2 matrix.
The direction must be a unit vector in P\mathbf{x} = (\mathbf{x} \cdot \mathbf{u})\mathbf{u}. For any nonzero direction \mathbf{d}, P\mathbf{x} = \dfrac{\mathbf{x} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}}\,\mathbf{d}.
A projection cannot be undone, since every vector perpendicular to L goes to \mathbf{0}. Rotations and reflections can; see When a matrix or transformation can be undone.
Unit columns alone are not enough; they must also be perpendicular (Exercise 18).

Where this goes next

One move after another is Composing transformations as a matrix product. The vectors a projection sends to \mathbf{0} and the line it lands on are its kernel and image. Rotations and reflections keep area while a projection flattens it, which the determinant measures. The line a reflection fixes and the line it reverses are early eigenvectors, and projection onto planes and larger subspaces is Orthogonal bases and projection onto a subspace.

Exercises

Rotations

Find the matrix of the rotation by 150^\circ counterclockwise, and the image of \displaystyle (2\sqrt3, 2).
Find the matrix of the rotation by 120^\circ clockwise.
Find the image of (0, 4) under the rotation by 210^\circ counterclockwise.
An angle \theta has \cos\theta = -\tfrac35, with \theta between 90^\circ and 180^\circ. Find R_\theta and the image of (5, 5).
Find the matrix of the rotation that sends (5, 12) to (13, 0).

Projections and reflections

Find the matrix of the projection onto the line y = -3x.
Find the matrix of the projection onto the line y = -\tfrac12 x, and the projection of (3, 4).
Find the matrix of the projection onto the line through the origin at 30^\circ to the positive x-axis.
Find the matrix of the reflection across the line y = -3x.
Find the matrix of the reflection across the line through the origin at 75^\circ to the positive x-axis.
Find the matrix of the reflection across y = -\tfrac12 x and the image of (1, 3). Check that the midpoint of (1, 3) and its image lies on the line.

In space

Find the matrix of the turn of \mathbb{R}^3 by 315^\circ about the z-axis, and the image of (2, 0, 5).
Find the matrix of the turn of \mathbb{R}^3 by 30^\circ about the x-axis, and the image of (1, 0, 2).
Find the matrices of the projection of \mathbb{R}^3 onto the xy-plane and of the reflection of \mathbb{R}^3 across the xy-plane.

Reading a matrix

Show that \tfrac{1}{17}\begin{bmatrix} 8 & -15 \\ 15 & 8 \end{bmatrix} is a rotation, and give \cos\theta and \sin\theta.
Show that \tfrac{1}{17}\begin{bmatrix} 8 & 15 \\ 15 & -8 \end{bmatrix} is a reflection, and find its line.
Show that \tfrac{1}{10}\begin{bmatrix} 9 & 3 \\ 3 & 1 \end{bmatrix} is the projection onto the line y = \tfrac13 x.
The matrix \begin{bmatrix} 0.6 & 0.8 \\ 0.8 & 0.6 \end{bmatrix} has unit columns. Show that it is neither a rotation nor a reflection by finding a vector whose length it changes.

Reasoning

Prove that R_\theta\mathbf{x} \cdot R_\theta\mathbf{y} = \mathbf{x} \cdot \mathbf{y} for all \mathbf{x}, \mathbf{y} in \mathbb{R}^2, and explain why this means a rotation keeps angles.
Explain why P(P\mathbf{x}) = P\mathbf{x} for a projection, and why no transformation can undo P.
Show that for every angle \varphi, the reflection matrix F sends its own first column to \mathbf{e}_1 and its second column to \mathbf{e}_2.
Libuse moves the pin to (1, 0) and turns the block about it. Explain why this move is not a linear transformation.
Explain why no rotation of the plane sends (1, 2) to (3, 0).

Answers

\displaystyle R_{150^\circ} = \begin{bmatrix} -\tfrac{\sqrt3}{2} & -\tfrac12 \\ \tfrac12 & -\tfrac{\sqrt3}{2} \end{bmatrix}, and the image is \displaystyle (-3 - 1,\ \sqrt3 - \sqrt3) = (-4, 0).
\theta = -120^\circ, \cos\theta = -\tfrac12, \displaystyle \sin\theta = -\tfrac{\sqrt3}{2}: \displaystyle \begin{bmatrix} -\tfrac12 & \tfrac{\sqrt3}{2} \\ -\tfrac{\sqrt3}{2} & -\tfrac12 \end{bmatrix}.
\displaystyle (2, -2\sqrt3): it is 4 times the second column \displaystyle (-\sin 210^\circ, \cos 210^\circ) = (\tfrac12, -\tfrac{\sqrt3}{2}).
\sin\theta = \tfrac45 (positive in that range), R_\theta = \tfrac15\begin{bmatrix} -3 & -4 \\ 4 & -3 \end{bmatrix}, and (5, 5) \mapsto (-7, 1). Both have squared length 50.
\tfrac{1}{13}\begin{bmatrix} 5 & 12 \\ -12 & 5 \end{bmatrix}, from solving 5c - 12s = 13, 12c + 5s = 0.
\displaystyle \mathbf{u} = \tfrac{1}{\sqrt{10}}(1, -3), so P = \tfrac{1}{10}\begin{bmatrix} 1 & -3 \\ -3 & 9 \end{bmatrix}.
\displaystyle \mathbf{u} = \tfrac{1}{\sqrt5}(2, -1), P = \tfrac15\begin{bmatrix} 4 & -2 \\ -2 & 1 \end{bmatrix}, and P(3, 4) = (\tfrac45, -\tfrac25).
\displaystyle \mathbf{u} = (\tfrac{\sqrt3}{2}, \tfrac12), so \displaystyle P = \begin{bmatrix} \tfrac34 & \tfrac{\sqrt3}{4} \\ \tfrac{\sqrt3}{4} & \tfrac14 \end{bmatrix}.
F = 2P - I with P from Exercise 6: \tfrac15\begin{bmatrix} -4 & -3 \\ -3 & 4 \end{bmatrix}.
2\varphi = 150^\circ: \displaystyle \begin{bmatrix} -\tfrac{\sqrt3}{2} & \tfrac12 \\ \tfrac12 & \tfrac{\sqrt3}{2} \end{bmatrix}.
F = \tfrac15\begin{bmatrix} 3 & -4 \\ -4 & -3 \end{bmatrix} and F(1, 3) = (-\tfrac95, -\tfrac{13}{5}). The midpoint is (-\tfrac25, \tfrac15), and -\tfrac12 \cdot (-\tfrac25) = \tfrac15.
\displaystyle \cos 315^\circ = \tfrac{\sqrt2}{2}, \displaystyle \sin 315^\circ = -\tfrac{\sqrt2}{2}: \displaystyle \begin{bmatrix} \tfrac{\sqrt2}{2} & \tfrac{\sqrt2}{2} & 0 \\ -\tfrac{\sqrt2}{2} & \tfrac{\sqrt2}{2} & 0 \\ 0 & 0 & 1 \end{bmatrix}, and \displaystyle (2, 0, 5) \mapsto (\sqrt2, -\sqrt2, 5).
\displaystyle \begin{bmatrix} 1 & 0 & 0 \\ 0 & \tfrac{\sqrt3}{2} & -\tfrac12 \\ 0 & \tfrac12 & \tfrac{\sqrt3}{2} \end{bmatrix}, and \displaystyle (1, 0, 2) \mapsto (1, -1, \sqrt3).
Projection: \mathbf{e}_3 \mapsto \mathbf{0}, giving \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}. Reflection: \mathbf{e}_3 \mapsto -\mathbf{e}_3, giving \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}.
8^2 + 15^2 = 17^2 gives unit columns, their dot product is 0, and the pattern is \begin{bmatrix} c & -s \\ s & c \end{bmatrix}: \cos\theta = \tfrac{8}{17}, \sin\theta = \tfrac{15}{17}.
Unit perpendicular columns, pattern \begin{bmatrix} c & s \\ s & -c \end{bmatrix}; \tan\varphi = \tfrac{15/17}{25/17} = \tfrac35, the line y = \tfrac35 x. Check: (5, 3) \mapsto (5, 3).
With \displaystyle \mathbf{u} = \tfrac{1}{\sqrt{10}}(3, 1), the formula gives \tfrac{1}{10}\begin{bmatrix} 9 & 3 \\ 3 & 1 \end{bmatrix}. Directly: (3, 1) \mapsto (3, 1) and (-1, 3) \mapsto (0, 0).
(1, -1) \mapsto (-0.2, 0.2), of length \displaystyle \sqrt{0.08} \approx 0.28, while \displaystyle \lVert (1, -1) \rVert = \sqrt2. (The columns have dot product 0.96, not 0.)
Expanding with \mathbf{x} = (x_1, x_2), \mathbf{y} = (y_1, y_2), the x_1y_1 and x_2y_2 terms each carry \cos^2\theta + \sin^2\theta = 1 and the cross terms cancel, leaving x_1y_1 + x_2y_2. Lengths are kept too, so the cosine of the angle, \dfrac{\mathbf{x} \cdot \mathbf{y}}{\lVert \mathbf{x} \rVert\,\lVert \mathbf{y} \rVert}, is unchanged.
P\mathbf{x} lies on L, where P fixes everything. For a nonzero \mathbf{v} perpendicular to L, P\mathbf{v} = \mathbf{0} = P\mathbf{0}, so an undoing map would have to send \mathbf{0} to both \mathbf{v} and \mathbf{0}.
Write c = \cos 2\varphi, s = \sin 2\varphi. Then F(c, s) = (c^2 + s^2,\ sc - cs) = (1, 0) and F(s, -c) = (0, 1).
A turn by \theta about (1, 0), for \theta not a multiple of 360^\circ, moves every point except (1, 0), so it moves \mathbf{0}. A linear transformation sends \mathbf{0} to \mathbf{0}.
A rotation keeps lengths, and \displaystyle \lVert (1, 2) \rVert = \sqrt5 while \lVert (3, 0) \rVert = 3.