Worked examples — Orthogonal matrices — properties, det = ±1

Before we start, three reminders in plain words, so no symbol appears un-earned.

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The scenario matrix

Every orthogonal-matrix problem is one (or a mix) of the cells below. The examples that follow are tagged with the cell they hit, so you can see the whole territory is covered.

Cell	Case class	What's tricky about it	Covered by
A	Pure rotation ($2\times 2$), $\det =+1$	reading the angle off the matrix	Ex 1
B	Pure reflection ($2\times 2$), $\det =-1$	axis of the flip, self-inverse	Ex 2
C	The impostor: $\det =\pm 1$ but not orthogonal	necessary vs sufficient	Ex 3
D	Degenerate / zero-angle & identity	limiting case $\theta \to 0$	Ex 4
E	Build one from a single vector (Gram–Schmidt feel)	normalising, choosing $\perp$ partner, sign choice	Ex 5
F	$3\times 3$ rotation about an axis, $\det =+1$	one fixed eigenvector, block structure	Ex 6
G	$3\times 3$ reflection, $\det =-1$	orientation flip in 3D	Ex 7
H	Word problem (robot / camera)	translating physical rigid motion to $Q$	Ex 8
I	Exam twist: product & "guess the det"	closure, $\det$ of a product	Ex 9
J	Eigenvalues on the unit circle	complex eigenvalues of a rotation	Ex 10

The figure below sums up the whole table in one picture: the left panel is the archetype of every "$+1$" cell (A, D, F, H, I-rotation — a spin that keeps orientation), and the right panel is the archetype of every "$-1$" cell (B, G, I-reflection — a flip across a mirror line). As you work each example, place it mentally on the correct side of this picture.

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Example 1 — Cell A: pure rotation, read the angle

Steps.

1. Compare to the rotation template $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$. Matching: $\cos \theta =\frac{1}{2}$, $\sin \theta =\frac{\sqrt{3}}{2}$. Why this step? The parent's rotation matrix has a fixed shape; if $Q$ fits it, the angle is read off directly.
2. Solve for $\theta$: $\cos \theta =\frac{1}{2}$ and $\sin \theta =\frac{\sqrt{3}}{2}$ both hold at $\theta =60^{\circ }=\frac{\pi }{3}$. Why this step? We need one angle satisfying both equations — one equation alone is ambiguous.
3. Confirm orthonormal columns. Column 1 is $\frac{1}{2}(1,\sqrt{3})$, length${}^2=\frac{1}{4}(1+3)=1$ ✓. Column 2 is $\frac{1}{2}(-\sqrt{3},1)$, length${}^2=\frac{1}{4}(3+1)=1$ ✓. Dot product $=\frac{1}{4}(-\sqrt{3}+\sqrt{3})=0$ ✓. Why this step? This is the test $Q^{\mathsf{T}}Q=I$ done column-by-column.
4. Determinant: $\det Q=\frac{1}{4}(1\cdot 1-(-\sqrt{3})(\sqrt{3}))=\frac{1}{4}(1+3)=1$. Why this step? $\det =+1$ certifies it's a rotation (a spin), consistent with orientation preserved.

Worked example Verify

${\cos }^{2}60^{\circ }+{\sin }^{2}60^{\circ }=\frac{1}{4}+\frac{3}{4}=1$: unit columns ✓. Apply $Q$ to $e_1=(1,0)$: you get $\frac{1}{2}(1,\sqrt{3})$, an arrow at exactly $60^{\circ }$ above the $x$-axis — matches the claimed angle.

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Example 2 — Cell B: pure reflection, find the mirror line

Steps.

1. Orthonormal columns. $c_1=\frac{1}{5}(3,4)$: length${}^2=\frac{9+16}{25}=1$ ✓. $c_2=\frac{1}{5}(4,-3)$: length${}^2=\frac{16+9}{25}=1$ ✓. Dot $=\frac{1}{25}(12-12)=0$ ✓. Why this step? Verifies $R^{\mathsf{T}}R=I$, i.e. it is orthogonal.
2. Determinant: $\det R=\left(\frac{3}{5}\right)(-\frac{3}{5})-\left(\frac{4}{5}\right)\left(\frac{4}{5}\right)=-\frac{9}{25}-\frac{16}{25}=-1$. Why this step? $\det =-1$ means it is a flip, not a spin.
3. Find the fixed direction (mirror line). A reflection fixes vectors on the mirror: solve $Rv=v$, i.e. $(R-I)v=0$. Row 1: $(\frac{3}{5}-1)x+\frac{4}{5}y=0\Rightarrow -\frac{2}{5}x+\frac{4}{5}y=0\Rightarrow y=\frac{1}{2}x$. So the mirror is the line $y=\frac{1}{2}x$, direction $(2,1)$. Why this step? A reflection has eigenvalue $+1$ along the mirror and $-1$ perpendicular to it — the parent's eigenvalue fact.

Worked example Verify

Plug $v=(2,1)$: $Rv=\frac{1}{5}(3\cdot 2+4\cdot 1,4\cdot 2-3\cdot 1)=\frac{1}{5}(10,5)=(2,1)=v$ ✓ (fixed). Plug $w=(1,-2)$ (perpendicular to mirror): $Rw=\frac{1}{5}(3-8,4+6)=\frac{1}{5}(-5,10)=(-1,2)=-w$ ✓ (flipped).

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Example 3 — Cell C: the impostor ($\det =1$ but not orthogonal)

Steps.

1. Don't trust $\det$ alone. $\det =\pm 1$ is necessary but not sufficient (parent's mistake box). Run the real test. Why this step? Orientation-preserving maps include shears, which stretch — we must check lengths.
2. Test the columns. $c_1=(2,1)$, length${}^2=4+1=5\ne 1$. Why this step? Orthogonal columns must be unit length; $5\ne 1$ already kills it.
3. Compute $A^{\mathsf{T}}A$ to be sure. $A^{\mathsf{T}}A=\left(\begin{array}{cc}2& 1\\ 3& 2\end{array}\right)\left(\begin{array}{cc}2& 3\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}5& 8\\ 8& 13\end{array}\right)\ne I$. Why this step? The definitive certificate: $A^{\mathsf{T}}A\ne I$ ⇒ not orthogonal.

Worked example Verify

$\Vert A(1,0)\Vert =\Vert (2,1)\Vert =\sqrt{5}\ne 1$, so $A$ changes lengths — an isometry never does. Not orthogonal ✓.

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Example 4 — Cell D: degenerate / zero-angle limit

Steps.

1. Plug $\theta =0$: $R(0)=\left(\begin{array}{cc}\cos 0& -\sin 0\\ \sin 0& \cos 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$. Why this step? The identity is the "do-nothing" rotation — the limiting/degenerate case.
2. Check: $I^{\mathsf{T}}I=I$ ✓, $\det I=1$. It is orthogonal, in $SO(2)$ (the $\det =+1$ rotations). Why this step? Confirms the identity sits inside the orthogonal group (it's the group's neutral element).
3. Plug $\theta =180^{\circ }$: $R(\pi )=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)=-I$. $\det (-I)=(-1)(-1)=1$. Why this step? This is a subtle case: $-I$ flips both axes, yet $\det =+1$ — it is a rotation (by $180^{\circ }$), not a reflection, because flipping both axes is two flips = a spin.

Worked example Verify

$\det (-I_2)=1$ (even dimension), $\det (-I_3)=-1$ (odd dimension). Parity of $n$ decides the sign of $\det (-I_n)$.

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Example 5 — Cell E: build one from a single vector

Steps.

1. Normalize. $\Vert (1,1)\Vert =\sqrt{2}$, so $u_1=\frac{1}{\sqrt{2}}(1,1)$. Why this step? Columns of $Q$ must be unit vectors (Gram–Schmidt step 1).
2. Choose a perpendicular unit partner. Rotating $u_1$ by $+90^{\circ }$ gives $u_2=\frac{1}{\sqrt{2}}(-1,1)$; by $-90^{\circ }$ gives $u_2^{\prime }=\frac{1}{\sqrt{2}}(1,-1)$. Why this step? In the plane exactly two unit vectors are perpendicular to $u_1$ (opposite directions) — that's the sign freedom.
3. Assemble and pick the rotation. $Q_{+}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$, $\det Q_{+}=\frac{1}{2}(1+1)=1$ → rotation. $Q_{-}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$, $\det Q_{-}=\frac{1}{2}(-1-1)=-1$ → reflection. Why this step? The sign of the partner controls whether orientation is kept ($+1$) or flipped ($-1$).

Worked example Verify

$Q_{+}^{\mathsf{T}}{Q}_{+}=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)=\frac{1}{2}\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)=I$ ✓. $\det Q_{+}=1$, $\det Q_{-}=-1$ ✓.

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Example 6 — Cell F: a $3\times 3$ rotation about an axis

Steps.

1. Insert $\theta =90^{\circ }$: $\cos 90^{\circ }=0,\sin 90^{\circ }=1$, so $Q=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right)$. Why this step? Concrete numbers make the checks unambiguous.
2. Orthonormal columns: $c_1=(1,0,0)=e_1$, $c_2=(0,0,1)=e_3$, $c_3=(0,-1,0)=-e_2$: each unit length, pairwise dot products $0$ ✓. Why this step? Direct verification of $Q^{\mathsf{T}}Q=I$ in 3D.
3. Determinant via block structure: the top-left $1$ multiplies the $2\times 2$ rotation block whose det is ${\cos }^2\theta +{\sin }^2\theta =1$. So $\det Q=1\cdot 1=1$. Why this step? $\det =+1$ ⇒ a genuine 3D rotation, so $Q\in SO(3)$ (the $\det =+1$ rotations in space).
4. Axis = fixed direction: $Q{e}_1=(1,0,0)=e_1$, so the $x$-axis is fixed; the rotation happens in the $y$–$z$ plane. Why this step? A 3D rotation always fixes a line (its eigenvector with eigenvalue $+1$) — that line is the axis.

Worked example Verify

$Q(0,1,0)=(0,0,1)$ and $Q(0,0,1)=(0,-1,0)$: the $y$–$z$ plane vectors rotate $90^{\circ }$, while $e_1$ stays put ✓. $\det =1$ ✓.

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Example 7 — Cell G: a $3\times 3$ reflection

Steps.

1. Orthogonal check: columns are $e_1,e_2,-e_3$ — orthonormal ✓, and $S^{\mathsf{T}}S=S^2=I$ (flip twice returns home). Why this step? Confirms $S\in O(3)$.
2. Determinant: diagonal matrix ⇒ $\det =1\cdot 1\cdot (-1)=-1$. Why this step? $\det =-1$ ⇒ reflection (orientation flips), so $S$ is not in $SO(3)$ — exactly one axis flipped is an odd count.

Worked example Verify

Take the right-handed triple $(e_1,e_2,e_3)$. After $S$: $(e_1,e_2,-e_3)$ — a left-handed triple. Handedness flipped, matching $\det =-1$ ✓.

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Example 8 — Cell H: word problem (camera gimbal)

Steps.

1. Rotation about $z$ leaves $z$ fixed, acts on $x,y$: $Q=\left(\begin{array}{ccc}\cos 30^{\circ }& -\sin 30^{\circ }& 0\\ \sin 30^{\circ }& \cos 30^{\circ }& 0\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}\frac{\sqrt{3}}{2}& -\frac{1}{2}& 0\\ \frac{1}{2}& \frac{\sqrt{3}}{2}& 0\\ 0& 0& 1\end{array}\right)$. Why this step? Spinning about the vertical axis is the $z$-axis rotation block, mirroring Example 6.
2. New forward vector: $Q(1,0,0)=(\frac{\sqrt{3}}{2},\frac{1}{2},0)\approx (0.866,0.5,0)$. Why this step? This is the physical answer — where the lens now points.
3. No scaling check: $\Vert Q(1,0,0)\Vert =\sqrt{\frac{3}{4}+\frac{1}{4}}=1$, same as $\Vert (1,0,0)\Vert =1$. Why this step? An orthogonal map is an isometry — the drone's "zoom" (length) can't change from a pure rotation. Units: a unit direction stays a unit direction.

Worked example Verify

$\det Q=1\cdot ({\cos }^{2}30^{\circ }+{\sin }^{2}30^{\circ })=1$ ✓. $0.866^2+0.5^2=0.75+0.25=1$ ✓.

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Example 9 — Cell I: exam twist (product & guess the det)

Steps.

1. Orthogonality by closure: $(Q_1{Q}_2{)}^{\mathsf{T}}(Q_1{Q}_2)=Q_2^{\mathsf{T}}{Q}_1^{\mathsf{T}}{Q}_1{Q}_2=Q_2^{\mathsf{T}}I{Q}_2=I$. So $Q_1{Q}_2$ is orthogonal. Why this step? The orthogonal group is closed under products — products stay orthogonal (parent's closure box).
2. Determinant is multiplicative: $\det (Q_1{Q}_2)=\det Q_1\cdot \det Q_2=(+1)(-1)=-1$. Why this step? $\det (AB)=\det A\det B$ — no multiplication of the matrices needed.
3. Interpret: $\det =-1$ ⇒ the combined map is a reflection (it leaves $SO(2)$). A rotation followed by a reflection is a reflection (odd total number of flips). Why this step? The sign of $\det$ is the "flip parity"; even = rotation, odd = reflection.

Worked example Verify

Concretely: $Q_1=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ ($\det 1$), $Q_2=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ ($\det -1$). $Q_1{Q}_2=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, $\det =-1$ ✓ — the swap/reflection matrix from the parent.

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Example 10 — Cell J: eigenvalues on the unit circle

Steps.

1. Characteristic polynomial: $\det (Q-\lambda I)=\det \left(\begin{array}{cc}-\lambda & -1\\ 1& -\lambda \end{array}\right)={\lambda }^2+1=0$. Why this step? Eigenvalues are roots of $\det (Q-\lambda I)=0$ (see Eigenvalues and eigenvectors).
2. Solve: ${\lambda }^2=-1\Rightarrow \lambda =\pm i$. Why this step? A rotation by $90^{\circ }$ has no real eigenvector (nothing is left pointing the same way), so the eigenvalues are complex — exactly as the parent predicted, $e^{\pm i\theta }$ with $\theta =90^{\circ }$.
3. Check magnitude: $|\pm i|=\sqrt{0^2+1^2}=1$. Why this step? The parent proved every eigenvalue of an orthogonal matrix has $|\lambda |=1$; here $\lambda =e^{\pm i\pi /2}$, on the unit circle.

Worked example Verify

Product of eigenvalues $=(i)(-i)=1=\det Q$ ✓; sum $=i+(-i)=0=\text{trace}=\cos \theta +\cos \theta$ with $\theta =90^{\circ }$ ✓.

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Recall One-line summary of every cell

Spin = $\det +1$; flip = $\det -1$; impostor = $\det$ right but $Q^{\mathsf{T}}Q\ne I$; identity/$-I_2$ are rotations; $-I_3$ is a reflection; product's det multiplies; eigenvalues live on the unit circle.

Active recall