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The SO(3) group consists of real three-by-three orthogonal matrices with determinant one. It acts by rotation in three dimensions. Its group manifold is ${\mathbb{RP}}^3$, since the Adjoint double cover from SU(2) to SO(3) identifies antipodal points of the three-sphere.

The compact group of real seven-by-seven orthogonal matrices with determinant one. Its Lie algebra has B3 root system. Its simply connected double cover is the Spin group $\mathrm{Spin}(7)$; the eight-dimensional spin representation belongs to that cover and does not descend to SO(7).

Every real symmetric matrix has an orthonormal basis of real eigenvectors. Equivalently, it admits an orthogonal diagonalization $A=O^{T}DO$ with $D$ real diagonal and $O$ an orthogonal matrix.

This is the finite-dimensional real-symmetric case of the spectral theorem.

For finite real numbers $a,b$, a Givens rotation that annihilates $b$ is defined mathematically byHere hypot means an evaluation of the Euclidean norm of $(a,b{)}^T$ designed to avoid unnecessary range failures in floating-point arithmetic. The zero case selects the identity matrix, avoiding division by zero. If $a=0,b\ne 0$, then $c=0,s=\mathrm{sign}(b)$; if $b=0,a\ne 0$, then $c=\mathrm{sign}(a),s=0$. Thus the retained component is nonnegative, including when $a<0$. In exact arithmetic, $ca+sb=r$ and $-sa+cb=0$.

Computing the square root of a*a + b*b directly can fail even when the final norm is representable. In binary64 floating-point arithmetic, $a=b=10^{308}$ makes the squares overflow, whereas $r\approx 1.4142\times 10^{308}$ is finite. With $a=b=10^{-300}$ the squares round to zero, although $r\approx 1.4142\times 10^{-300}$ is representable. A range-safe hypot avoids these intermediate failures.

The simple divisions $a/r,b/r$ still need care when $r$ is a subnormal number: rounding $r$ can cause large relative errors. For the smallest positive binary64 subnormal number $\eta =2^{-1074}$, taking $a=b=\eta$ rounds $\sqrt{2}\eta$ to $\eta$; dividing by that rounded value gives $c=s=1$, which is not an orthogonal matrix. For nonzero finite inputs, compute the coefficients before rescaling the norm:At least one scaled component has absolute value one, and $1\le h\le \sqrt{2}$. This construction keeps the coefficients meaningful even when rounding $r$ is severe. It cannot make an unrepresentably large $r$ finite, preserve an arbitrarily tiny $s$ against underflow, or repair inputs already rounded to zero. The row updates themselves can also overflow if their exact results exceed the available range.

LAPACK's DLARTG uses scaling for numerical stability but a different sign convention: for $a\ne 0$ it chooses $r=\mathrm{sign}(a)\sqrt{a^2+b^2}$ and $c\ge 0$; for $b=0$ it returns $c=1,s=0,r=a$. Its block still has the lower-left minus sign. Do not combine its returned coefficients with an assumed nonnegative $r$.^[ref]

If real matrices satisfy $A^T=A\le 0$ and $B^T=-B$, then $e^{tA}$ is a contraction and $e^{tB}$ is an orthogonal matrix for $t\ge 0$. Hence $\Vert e^{kA/2}{e}^{kB}{e}^{kA/2}{\Vert }_2\le 1$ for every $k\ge 0$. The bound is independent of commutativity and gives unconditional stability for a centered convection-diffusion semidiscretization with homogeneous Dirichlet boundary conditions. It does not itself bound the mesh-dependent commutators that enter a time-error estimate.

For a real skew-symmetric matrix $B$, its Cayley transform of a real skew-symmetric matrix with the opposite sign convention has inverse $C^{-1}=(I+B)(I-B{)}^{-1}$. Since $C$ is an orthogonal matrix, symmetry means $C=C^{-1}$, or $(I-B{)}^2=(I+B{)}^2$. This forces $B=0$. Skew-symmetry means $C=-C^{-1}$, or $(I-B{)}^2=-(I+B{)}^2$, which is equivalent to $B^2=-I$. In that case $C=-B$. Such examples exist in even dimension by taking two-dimensional right-angle rotation matrices in diagonal blocks; in odd real dimension the determinant equation $(\det B{)}^2=(-1{)}^n$ rules them out.

For a real orthogonal matrix $A$, $A^{\mathsf{T}}=A^{-1}$. If $Av=\lambda v$ for a nonzero complex eigenvector, then $A^{\mathsf{T}}v={\lambda }^{-1}v$. Therefore the same eigenvector has eigenvalue $(\lambda +{\lambda }^{-1})/2$ for the symmetric part of a matrix $(A+A^{\mathsf{T}})/2$. Orthogonality implies $|\lambda |=1$, so writing $\lambda =e^{i\theta }$ gives $\cos \theta$. The complete complex orthonormal eigenbasis of an orthogonal matrix shows that all the eigenvalues of its symmetric part lie in $[-1,1]$.

A real orthogonal matrix $T$ of size three with $\det T=-1$ has an eigenvector $n$ of eigenvalue $-1$. Its orthogonal complement is invariant, and the restriction there is an orientation-preserving plane rotation. Thus $T$ is a rotation about the axis $\mathbb{R}n$ composed with a reflection across the plane perpendicular to $n$. Writing the plane angle as $\varphi$, $\mathrm{tr}T=-1+2\cos \varphi$. A pure plane reflection is the case $\varphi =0$; if $\det (T-I)\ne 0$, it cannot be a pure plane reflection.