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All the transformations below fix the origin. A planar rotation preserves lengths and angles and turns every vector through one common oriented angle. Its rotation matrix isAn orthogonal reflection in a line fixes that line and reverses its perpendicular direction; reflection in the horizontal axis has reflection matrix $\mathrm{diag}(1,-1)$. A uniform dilation multiplies every vector by the same positive scale $s$, represented by $sI$; for example $2I$ doubles lengths. A shear mapping fixes one line pointwise and shifts vectors parallel to that line by an amount proportional to their transverse coordinate. A horizontal shear mapping has matrixwith $S_1$ a nontrivial example.

The first given matrix is the clockwise planar rotation through $\pi /4$, namely $A=R_{-\pi /4}$. Direct multiplication givesThus $C$ is a horizontal shear mapping of strength $2$. Because $A^{-1}=R_{\pi /4}$,So $B$ is the composition of that shear mapping followed by an anticlockwise planar rotation through $\pi /4$; order matters. It is not individually one of the four elementary types under these definitions. In particular, its determinant is $1$, but it is not an orthogonal matrix, excluding both a planar rotation and an orthogonal reflection. It is not a scalar matrix, excluding a uniform dilation; its eigenvalues $\sqrt{2}+1$ and $\sqrt{2}-1$ exclude a pure shear mapping, whose eigenvalues are both $1$. As a symmetric positive-definite matrix, $B$ can also be described as stretching two perpendicular principal directions by these reciprocal factors, preserving area.

The quadratic expression is ${\mathbf{x}}^{T}A\mathbf{x}$. Normalize the preceding eigenvectors to obtain an orthonormal basisWith the orthogonal matrix $U=({\mathbf{u}}_1{\mathbf{u}}_2{\mathbf{u}}_3)$ and coordinates $\mathbf{y}=U^T\mathbf{x}$, the surface equation becomes $y_1^2+2y_2^2+4y_3^2=1$. All coefficients are positive, so it is an ellipsoid, with principal semiaxes $1,1/\sqrt{2},1/2$ along the respective eigenvectors. This is principal-axis reduction of a quadric.

A map to the unit sphere is $\mathbf{x}\mapsto \mathrm{diag}(1,\sqrt{2},2)U^T\mathbf{x}$. One explicit matrix isIndeed $B^{T}B=A$, so $|B\mathbf{x}{|}^2={\mathbf{x}}^{T}A\mathbf{x}$. Since $B$ is invertible, every point on the unit sphere has a preimage on the ellipsoid; the map is onto, not merely into. The requested $B$ is not unique, since an additional orthogonal matrix on the left preserves the unit sphere.

An eigenvalue can be complex even though $P$ is real, so use the complex Hermitian inner product. For a nonzero complex eigenvector $\mathbf{v}$, the real orthogonal matrix also satisfies $P^{\ast }P=P^{T}P=I$. Henceand $\boxed{|\lambda |=1}$. Taking complex conjugates in $P\mathbf{v}=\lambda \mathbf{v}$ gives $P\overline{\mathbf{v}}=\overline{\lambda }\overline{\mathbf{v}}$ because $P$ is real. The conjugate vector is nonzero, so $\boxed{{\lambda }^{\ast }=\overline{\lambda }\text{ is also an eigenvalue}}$.

For the unheaded geometrical conclusion, the third eigenvalue of $Q$ must be real: nonreal eigenvalues of a real matrix occur in conjugate pairs. Its modulus is one, so it is $1$ or $-1$. Exact algebraic multiplicity two for $1$ excludes a third $1$, leaving $-1$. By the supplied diagonalisability fact, the $1$-eigenspace is a two-dimensional plane and the $-1$-eigenspace is a line. They are orthogonal: preservation of the inner product between a fixed vector and a negated vector gives $\mathbf{u}\cdot \mathbf{v}=-\mathbf{u}\cdot \mathbf{v}$. Therefore $Q$ is the orthogonal reflection in its fixed plane. It fixes the two tangential directions and reverses the normal direction; with unit normal $\mathbf{n}$, $Q=I-2\mathbf{n}{\mathbf{n}}^T$.

For a real orthogonal matrix, $P^{T}P=I$. ConsequentlyThus it preserves every real inner product, and in particular lengths and angles.

Each plane rotation satisfies $R_j^T{R}_j=I$. Hence $U=R_3{R}_2{R}_1$ is an orthogonal matrix and $D=UA$ is an upper triangular matrix. Multiplication by $U^T$ gives the QR decompositionThe zero-pair cases already handled make this construction valid for every real $3\times 3$ matrix, including rank-deficient matrices. Orthogonality follows from the product identity, not from any invertibility assumption on $A$.

Use the orthonormal right-handed basisDirect multiplication gives $B(r)u=u$, $B(r)v=rv+w/\sqrt{2}$ and $B(r)w=-v/\sqrt{2}+rw$. Thus the matrix in this basis isThe last two columns are orthogonal and both have squared norm $r^2+1/2$. Therefore $B(r)$ is an orthogonal matrix precisely when $r^2+1/2=1$, giving the unique positive valueIts determinant is $r_0^2+1/2=1$, so it is a proper rotation matrix, not a reflection. The transverse block has cosine and sine both $1/\sqrt{2}$. ConsequentlyWith the oriented axis $u$, the angle is positive by the right-hand rule, since $u\times v=w$. The axis also follows from the unit-eigenvalue eigenspace. As a check, $\mathrm{tr}B(r_0)=1+\sqrt{2}=1+2\cos (\pi /4)$.

For a Cartesian change of basis represented by an orthogonal matrix $Q$, vectors transform as $E_i^{\prime }=Q_{ia}{E}_a$ and $B_i^{\prime }=Q_{ia}{B}_a$. Their squared norms are invariant and $Q_{ia}{Q}_{jb}{\delta }_{ab}={\delta }_{ij}$. Substitution into the given expression therefore givesthe transformation law of a Cartesian second-rank tensor. The magnetic field's extra axial sign under an improper physical reflection, if included, appears twice and cancels in its quadratic contribution.

Write the Maxwell stress tensor as $T=\mathbf{EE}+\mathbf{BB}-\frac{1}{2}(E^2+B^2)I$. Its divergence isThe identity $(\mathbf{A}\cdot \nabla )\mathbf{A}-\frac{1}{2}\nabla A^2=-\mathbf{A}\times (\nabla \times \mathbf{A})$ follows by contracting two Levi-Civita symbols, or directly by differentiating components. Using Maxwell's equations consequently yieldsHence the local conservation of electromagnetic momentum isThe two subtracted terms are the Lorentz force density, while $\mathbf{E}\times \mathbf{B}$ is the electromagnetic momentum density in the units used here.

For fixed finite matrices, the matrix exponential and a Taylor expansion showThis agrees through degree two with $e^{h(A+B)}$, even when $A$ and $B$ do not commute. On a fixed time interval, $\Vert S(h)\Vert \le e^{h(\Vert A\Vert +\Vert B\Vert )}$ and the telescoping identityturns the $O(h^3)$ one-step defect into $O(N{h}^3)=O(h^2)$ global error for $Nh\le T$. Hence Strang splitting is second order in general; for commuting matrices it is exact.

For the spatial discretization, use $d=2/(M+1)$, $x_m=-1+md$ and the interior vector $U=(u_1,\dots ,u_M{)}^T$, with $u_0=u_{M+1}=0$. The central finite differences giveHere $\mathrm{tridiag}$ lists lower diagonal, main diagonal and upper diagonal, in that order. Thus $A^T=A$ and $B^T=-B$. Discrete summation by parts givesThe matrix exponential $e^{hA/2}$ is consequently a contraction in the Euclidean norm, while $e^{hB}$ is an orthogonal matrix for real $\alpha$. ThereforeRepeated split steps are contractive in the mesh-weighted norm $d\sum |u_m{|}^2$, uniformly in the spatial mesh and without a Courant–Friedrichs–Lewy condition. The split semidiscretization is unconditionally stable. No commutativity or shared eigenvectors are needed for this Strang splitting contraction for symmetric diffusion and skew advection. Its time-order proof is for fixed spatial matrices; as $d\to 0$, commutators can grow, so this stability result alone is not a uniform-in-mesh second-order error estimate.

Denote the four displayed matrices, from left to right, by $T_1,T_2,T_3,T_4$. An orthogonal matrix has an orthonormal set of columns, equivalently $T^{T}T=I$. Direct column dot products give $T_j^T{T}_j=I$ for $j=1,2,4$. The first column of $T_3$ has squared Euclidean norm $(1+6+1)/6=4/3$, so the third matrix is the unique nonorthogonal matrix.

For $T_1$, the determinant is $1$. A nonidentity orthogonal transformation of ${\mathbb{R}}^3$ with determinant $1$ is a rotation: it has a fixed axis and restricts to a plane rotation on its orthogonal complement. Consequently $T_1$ is the rotation.

For $T_2$, calculation givesA plane reflection fixes its entire plane, and therefore has eigenvalue $1$. Since $T_2-I$ is invertible, $T_2$ is not a plane reflection. Thus $T_2$ is the combination of a rotation and a reflection. More explicitly, $(1,-2,0{)}^T$ is an eigenvector with eigenvalue $-1$. On the perpendicular plane, $T_2$ is a rotation with $\cos \varphi =2/3$, because $\mathrm{tr}{T}_2=-1+2\cos \varphi =1/3$. This gives a three-dimensional improper orthogonal transformation.

For $T_4$, set $n=(1,2,2{)}^T/3$. ThenThis reflection matrix reverses the normal $n$ and fixes every vector perpendicular to $n$. Therefore $T_4$ is reflection in the plane $x+2y+2z=0$. In particular, $\det T_4=-1$ and $\det (T_4-I)=0$. Finally $T_3$ represents none of the three listed orthogonal transformations, since each preserves the Euclidean norm.

True. The determinant gives a group homomorphismFor an orthogonal matrix, $A^{T}A=I$ implies $(\det A{)}^2=1$, and a reflection has determinant $-1$, so the target has order two and the map is onto. Its kernel of a group homomorphism is exactlyThis is the defining special orthogonal group inside the orthogonal group.

For a differentiable curve $R(t)$ in the SO(3) group with $R(0)=I$, differentiating $R(t{)}^{T}R(t)=I$ gives $X^T+X=0$, where $X=R^{\prime }(0)$. Conversely, if $X$ is a real skew-symmetric matrix, $e^{tX}$ is an orthogonal matrix and has determinant $e^{t\mathrm{tr}X}=1$. HenceThe displayed generators of the SO(3) Lie algebra areThey are linearly independent and span the three independent entries of a real skew-symmetric matrix. Equivalently, $T_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The vector triple-product identity then givesThus the structure constants of a Lie algebra in this basis are

Use a real, antisymmetric convention throughout the SO(3) vector Higgs model. With gauge coupling $g$, the gauge covariant derivative and gauge field strength areThere is no factor of $i$ in these formulas because $T_a$ are real antisymmetric generators. For a local non-Abelian gauge transformation ${\Phi }^{\prime }=R\Phi$, with $R(x)$ in the SO(3) group, the compatible transformation of the gauge potential isIt gives $D_{\mu }^{\prime }{\Phi }^{\prime }=R{D}_{\mu }\Phi$ and $F_{\mu \nu }^{\prime }=R{F}_{\mu \nu }{R}^{-1}$. These covariance laws fix the relative signs in the gauge covariant derivative and gauge field strength.

An SO(3) invariant quartic scalar potential isSince $(R\Phi {)}^T(R\Phi )={\Phi }^T\Phi$, this scalar potential is gauge-invariant. Its minimum is the sphere ${\Phi }^T\Phi =v^2$, rather than a preferred direction in the internal vector space.

Writing ${\mathbf{A}}_{\mu }=(A_{\mu 1},A_{\mu 2},A_{\mu 3})$, the cross product form of the gauge covariant derivative is $D_{\mu }\Phi ={\partial }_{\mu }\Phi +g{\mathbf{A}}_{\mu }\times \Phi$. In components this isCorrespondingly, the gauge field strength has components

A covariantly constant vector Higgs field has constant norm, becauseThe last equality uses antisymmetry of the gauge potential. On a connected region write this norm as $\varphi \ge 0$. If $\varphi =0$, the vector Higgs field is identically zero. If $\varphi >0$, the SO(3) group acts transitively on unit vectors, so choose a smooth local non-Abelian gauge transformation rotating $\Phi /\varphi$ to ${\mathbf{e}}_3$. In this unitary gauge, $\Phi =(0,0,\varphi {)}^T$ with $\varphi$ constant. This is a local construction, and is global on a trivial contractible region; a nontrivial bundle can require more than one gauge patch.

For nonzero $\varphi$, substituting the constant aligned vector Higgs field into $D_{\mu }\Phi =0$ gives $g\varphi (A_{\mu 2},-A_{\mu 1},0{)}^T=0$. For nonzero gauge coupling, the most general compatible gauge potential and gauge field strength arewhere $a_{\mu }$ is an arbitrary real one-form. Thus the surviving gauge group is the $SO(2)$ stabilizer subgroup of ${\mathbf{e}}_3$. A covariantly constant vector Higgs field does not force the gauge field strength to vanish: it only requires the curvature to annihilate that field.

For the Higgs mechanism, choose a vacuum of the SO(3) invariant quartic scalar potential, so the constant norm is $v$. The canonically normalized kinetic term then containsThe two broken-direction gauge bosons have mass $gv$ (or $|g|v$ if the sign of $g$ is unrestricted), while the $T_3$ gauge boson remains massless. The two angular Goldstone bosons supply the massive vectors' longitudinal gauge-boson polarizations. The radial fluctuation remains a physical scalar field, with $m_h^2=2\lambda v^2$ for this normalization. The condition $D_{\mu }\Phi =0$ selects backgrounds without broken-direction gauge potentials; fluctuations about such a vacuum exhibit the two massive modes. Constancy of the norm alone does not require it to minimize the scalar potential.