Exercises — Orthogonal matrices — properties, det = ±1

$\det (Q^{\mathsf{T}})=\det Q$ Transposing a matrix does not change its determinant. Quick reason: the determinant is a sum over permutations of products of entries $a_{1\sigma (1)}\cdots a_{n\sigma (n)}$; transposing just relabels each such product (swap row and column indices), giving the same set of products with the same signs. So $\det (A^{\mathsf{T}})=\det A$ for every square $A$. We use this in Level 5.

Test each: are both columns unit length, and perpendicular to each other?

$A$: columns $c_1=\left(\genfrac{}{}{0ex}{}{0}{1}\right)$, $c_2=\left(\genfrac{}{}{0ex}{}{-1}{0}\right)$. Lengths $\sqrt{0+1}=1$ and $\sqrt{1+0}=1$ ✓. Dot product $c_1\cdot c_2=0\cdot (-1)+1\cdot 0=0$ ✓. Orthogonal.

$B$: columns $\left(\genfrac{}{}{0ex}{}{2}{0}\right)$, $\left(\genfrac{}{}{0ex}{}{0}{1}\right)$. First column length $=2\ne 1$ ✗. NOT orthogonal (it stretches the $x$-axis by $2$).

$C$: columns $\frac{1}{\sqrt{2}}\left(\genfrac{}{}{0ex}{}{1}{1}\right)$, $\frac{1}{\sqrt{2}}\left(\genfrac{}{}{0ex}{}{1}{-1}\right)$. Length${}^2=\frac{1}{2}(1+1)=1$ each ✓. Dot $=\frac{1}{2}(1\cdot 1+1\cdot (-1))=\frac{1}{2}(0)=0$ ✓. Orthogonal.

Answer: $A$ and $C$ are orthogonal; $B$ is not.

$\det A=0\cdot 0-(-1)(1)=1$ → rotation (it's a $90^{\circ }$ turn, orientation kept). $\det C=\frac{1}{2}(1\cdot (-1)-1\cdot 1)=\frac{1}{2}(-2)=-1$ → reflection (handedness flips).

The first column $c_1=\left(\genfrac{}{}{0ex}{}{3/5}{4/5}\right)$ already has length $1$ ($\frac{9}{25}+\frac{16}{25}=1$ ✓; see the yellow arrow on the unit circle). We need $c_2$ unit-length AND perpendicular to $c_1$.

Why rotate by $90^{\circ }$? In ${\mathbb{R}}^2$, the vector perpendicular to $\left(\genfrac{}{}{0ex}{}{a}{b}\right)$ is $\left(\genfrac{}{}{0ex}{}{-b}{a}\right)$ (turn left — the blue arrow) or $\left(\genfrac{}{}{0ex}{}{b}{-a}\right)$ (turn right) — both unit if the original is. This is the fastest way to guarantee $c_1\cdot c_2=0$ (the pink right-angle mark in the figure).

Left turn: $c_2=\left(\genfrac{}{}{0ex}{}{-4/5}{3/5}\right)$. Check $\det$: $\frac{3}{5}\cdot \frac{3}{5}-(-\frac{4}{5})\cdot \frac{4}{5}=\frac{9}{25}+\frac{16}{25}=1$ ✓ ($+1$, so this is the right choice).

$Q=\frac{1}{5}\left(\begin{array}{cc}3& -4\\ 4& 3\end{array}\right).$ (The other perpendicular $\left(\genfrac{}{}{0ex}{}{4/5}{-3/5}\right)$ — the right turn — gives $\det =-1$, a reflection — rejected.)

$Q^{\mathsf{T}}Q=\frac{1}{25}\left(\begin{array}{cc}3& 4\\ -4& 3\end{array}\right)\left(\begin{array}{cc}3& -4\\ 4& 3\end{array}\right)=\frac{1}{25}\left(\begin{array}{cc}9+16& -12+12\\ -12+12& 16+9\end{array}\right)=\frac{1}{25}\left(\begin{array}{cc}25& 0\\ 0& 25\end{array}\right)=I.✓$

What we use: eigenvalues of orthogonal $Q$ satisfy $|\lambda |=1$ (they sit on the unit circle in the figure); complex eigenvalues come in conjugate pairs $e^{\pm i\theta }$ (whose product is $e^{i\theta }{e}^{-i\theta }=1$); and $\det Q$ = product of all eigenvalues.

For $n=3$, the eigenvalues (with multiplicity) are either three reals from $\{+1,-1\}$, or one real ${\lambda }_1\in \{+1,-1\}$ plus a conjugate pair $e^{\pm i\theta }$.

Case A (one real + a pair): product $={\lambda }_1\cdot e^{i\theta }{e}^{-i\theta }={\lambda }_1\cdot 1={\lambda }_1$. This equals $\det Q=-1$, so ${\lambda }_1=-1$. ✓ eigenvalue $-1$ present.

Case B (three reals from $\{\pm 1\}$): their product is $-1$, so the count of $-1$'s is odd — hence at least one $-1$. ✓

Conclusion: in either case $-1$ is an eigenvalue. Geometrically, an odd-dimensional orientation-flip must reverse some axis — that reversed axis is the $-1$ eigenvector. See Eigenvalues and eigenvectors.

Column 1 is a unit vector in ${\mathbb{R}}^2$, so it lies on the unit circle: write it as $\left(\genfrac{}{}{0ex}{}{\cos \theta }{\sin \theta }\right)$ for some angle $\theta$. (Why? Every unit vector in the plane is $(\cos \theta ,\sin \theta )$ — that's the definition of the unit circle.)

Column 2 is a unit vector perpendicular to column 1, so it is $\pm \left(\genfrac{}{}{0ex}{}{-\sin \theta }{\cos \theta }\right)$ (the two $90^{\circ }$ turns of column 1).

• Choosing $+\left(\genfrac{}{}{0ex}{}{-\sin \theta }{\cos \theta }\right)$ gives $\det =\cos \theta \cos \theta -(-\sin \theta )\sin \theta ={\cos }^2\theta +{\sin }^2\theta =+1$ ✓.
• Choosing $-\left(\genfrac{}{}{0ex}{}{-\sin \theta }{\cos \theta }\right)$ gives $\det =-1$ ✗ (rejected).

So $\det =+1$ forces the rotation form. See Rotations and reflections in $\mathbb{R}^2$ and $\mathbb{R}^3$.

Step 1 — first column $c_1=\frac{1}{3}(2,2,1{)}^{\mathsf{T}}$. Check length: $\frac{1}{9}(4+4+1)=1$ ✓.

Step 2 — pick $c_2\perp c_1$, then normalize. Choose raw $v=(1,-1,0{)}^{\mathsf{T}}$; check $c_1\cdot v=\frac{1}{3}(2-2+0)=0$ ✓. Length $\sqrt{2}$, so $c_2=\frac{1}{\sqrt{2}}(1,-1,0{)}^{\mathsf{T}}$.

Step 3 — $c_3\perp c_1,c_2$. The clean way is the cross product $c_3=c_1\times c_2$ (guaranteed perpendicular to both): $c_1\times c_2=\frac{1}{3}\cdot \frac{1}{\sqrt{2}}\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 2& 1\\ 1& -1& 0\end{array}\right|=\frac{1}{3\sqrt{2}}(2\cdot 0-1\cdot (-1),1\cdot 1-2\cdot 0,2\cdot (-1)-2\cdot 1).$ $=\frac{1}{3\sqrt{2}}(1,1,-4{)}^{\mathsf{T}}.$

Why is $c_3$ automatically unit length? The magnitude formula for a cross product is $\Vert a\times b\Vert =\Vert a\Vert \Vert b\Vert \sin \varphi$, where $\varphi$ is the angle between $a$ and $b$. Here $a=c_1$ and $b=c_2$ are orthonormal: $\Vert a\Vert =\Vert b\Vert =1$ and $\varphi =90^{\circ }$, so $\sin \varphi =1$. Hence $\Vert c_3\Vert =1\cdot 1\cdot 1=1$ — no extra normalization needed. Direct check: $\frac{1}{18}(1+1+16)=\frac{18}{18}=1$ ✓.

Result: $Q=\left(\begin{array}{ccc}\frac{2}{3}& \frac{1}{\sqrt{2}}& \frac{1}{3\sqrt{2}}\\ \frac{2}{3}& -\frac{1}{\sqrt{2}}& \frac{1}{3\sqrt{2}}\\ \frac{1}{3}& 0& -\frac{4}{3\sqrt{2}}\end{array}\right).$ Because $c_3=c_1\times c_2$ follows the right-hand rule, this is a right-handed frame: $\det Q=+1$. (Swapping $c_2\leftrightarrow c_3$ would give $-1$.)

Claim: $+1$ is an eigenvalue, and its eigenvector is the rotation axis. For $n=3$, eigenvalues are one real ${\lambda }_1\in \{\pm 1\}$ plus either two more reals from $\{\pm 1\}$ or a pair $e^{\pm i\theta }$. Product $=\det Q=+1$.

• If a conjugate pair is present: ${\lambda }_1\cdot 1=+1\Rightarrow {\lambda }_1=+1$. ✓
• If all three real: their product is $+1$, so the number of $-1$'s is even (0 or 2), leaving at least one $+1$. ✓

So there is a nonzero $v$ with $Qv=v$: the axis. Vectors perpendicular to $v$ are rotated by the angle $\theta$ from the $e^{\pm i\theta }$ pair. This is exactly a rotation about $v$. Links to Spectral theorem and Eigenvalues and eigenvectors.

Determinant part. Using $\det (AB)=\det A\det B$ and $\det (Q^{\mathsf{T}})=\det Q$ (the reminder at the top of the page): $\det (Q^{\mathsf{T}}MQ)=\det (Q^{\mathsf{T}})\det (M)\det (Q)=(\det Q{)}^2\det M.$ Since $Q$ orthogonal $\Rightarrow (\det Q{)}^2=1$: $\det (Q^{\mathsf{T}}MQ)=\det M.✓$

Eigenvalue part. Let $A=Q^{\mathsf{T}}MQ$. Then for any scalar $\lambda$: $A-\lambda I=Q^{\mathsf{T}}MQ-\lambda Q^{\mathsf{T}}Q=Q^{\mathsf{T}}(M-\lambda I)Q,$ using $I=Q^{\mathsf{T}}Q$. Taking determinants and applying the fact above with $M\to M-\lambda I$: $\det (A-\lambda I)=\det (M-\lambda I).$ The two characteristic polynomials are identical, so $A$ and $M$ have the same eigenvalues. This is the backbone of the Spectral theorem and QR decomposition (orthogonal similarity preserves spectrum). See also Inner product spaces.

Use $Q^2=I$ to pin the eigenvalues. If $Qx=\lambda x$ then $Q^{2}x={\lambda }^{2}x=x$, so ${\lambda }^2=1$, i.e. $\lambda \in \{+1,-1\}$ — no genuine complex pair survives (a pair $e^{\pm i\theta }$ would need $e^{2i\theta }=1$, forcing $\theta \in \{0,\pi \}$, which is again real $\pm 1$).

Now impose $\det Q=-1$. With $n=3$ eigenvalues each $\pm 1$ and their product $=\det Q=-1$, the number of $-1$'s must be odd: either one $-1$ or three $-1$'s. Two possibilities:

• Eigenvalues $\{+1,+1,-1\}$ → a reflection across a plane: the two $+1$ directions span the mirror plane (fixed), the single $-1$ direction is flipped.
• Eigenvalues $\{-1,-1,-1\}$ → the point inversion $Q=-I$: every direction is flipped through the origin.

Explicit example (reflection across the $xy$-plane): $Q=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right),Q^2=I,\det Q=-1.✓$ Here the mirror plane is $z=0$ (the $\{+1,+1\}$ eigenspace) and the flipped axis is $z$ (the $-1$ eigenvector). The other case, $Q=-I$, also satisfies $Q^2=I$ and $\det (-I)=(-1{)}^3=-1$.