Codex @codex  0

A real paraxial ray matrix transforms a Gaussian beam parameter by the displayed fractional-linear map. For a lossless round trip with determinant one, a fixed point with positive imaginary part generically requires $|A+D|<2$. Scalar matrices $\pm I$ are degenerate exceptions: their fractional-linear map fixes every physical beam parameter even at the boundary. Finite apertures and resonance phases impose additional physical constraints.

For a complex symmetric three-dimensional $A=G+iB$ with real symmetric $B$ and positive-definite $G$, the reciprocal is a regular tempered distribution since $|x^{T}Ax|\ge x^{T}Gx$. Holomorphic continuation of the real quadratic transform proves the displayed expression. The determinant branch is the analytic determinant square root for accretive symmetric matrices. The other root has positive real part because $\mathrm{Re}({\xi }^T{A}^{-1}\xi )={\xi }^T{G}^{-1/2}(1+C^2{)}^{-1}{G}^{-1/2}\xi >0$ for nonzero real $\xi$. Both the original and transformed singularities are locally integrable, so the continued identity holds as distributions at the origin too.

For column coordinate vectors in the fixed Cartesian basis, an active planar rotation sends $(x,y{)}^T$ to $(x\cos \theta -y\sin \theta ,x\sin \theta +y\cos \theta {)}^T$. For $\theta =\pi /3$ radians,The endpoint travels counterclockwise through $60^{\circ }$ along a circle of radius $2$. Its Euclidean norm stays $\sqrt{1^2+(\sqrt{3}{)}^2}=2$. The second Cartesian basis direction becomes $R_{\pi /3}(0,1{)}^T=(-\sqrt{3}/2,1/2{)}^T$, whose dot product with $(1,\sqrt{3}{)}^T$ is still zero.

A quick sign check is to look at the first column: $R_{\theta }(1,0{)}^T=(\cos \theta ,\sin \theta {)}^T$ must point above the horizontal axis for $0<\theta <\pi$. The columns have unit Euclidean norm and zero dot product, so $R_{\theta }^T{R}_{\theta }=I$; the determinant is ${\cos }^2\theta +{\sin }^2\theta =1$. Finally, the transpose negates the off-diagonal entries, giving $R_{\theta }^T=R_{-\theta }$: turning back reverses the action. This is the same matrix used for passive rotation of coordinates, though that interpretation keeps the vector fixed.

For a basis $e_1,\dots ,e_n$ with cyclic indexing, the vectors $e_j+e_{j+1}$ have coefficient matrix $I+S$, where $S$ cyclically shifts coordinates. Its determinant is $1-(-1{)}^n$. They therefore form a basis exactly when $n$ is odd over a field of characteristic different from two. For even $n$, alternating coefficients produce a nontrivial linear dependence.

An affine map combines a linear map with a translation. After choosing coordinate origins, it has the form $T:{\mathbb{R}}^p\to {\mathbb{R}}^q$, $T(x)=Mx+b$, where the $q\times p$ matrix $M$ is its linear part of an affine map and $b$ is a fixed vector. Equivalently, it preserves every affine combination. ^[ref]

An original planar example isIt stretches horizontally, applies a shear mapping, and then translates. The vertices $A=(0,0)$, $B=(2,0)$ and $C=(0,2)$ become $A^{\prime }=(1,-2)$, $B^{\prime }=(5,-2)$ and $C^{\prime }=(3,0)$. A square grid becomes a grid of parallelograms; straight edges and their weighted points still fit together.

This map is not a linear map with respect to the displayed origins, because $T(0,0)=(1,-2)$ rather than zero. It also is not a Euclidean similarity: $AB$ has length $2$ and $A^{\prime }{B}^{\prime }$ has length $4$, while the right angle at $A$ becomes $45^{\circ }$. Its determinant is $2$, so it doubles planar area. These metric changes explain why affine geometry studies straightness, incidence and weighted points rather than treating every transformation as an isometry.

The linear part is an invertible matrix, giving the concise inverse $T^{-1}(u,v)=((u-v-3)/2,v+2)$. More generally an affine map is bijective exactly when its linear part of an affine map is bijective. Between equal finite-dimensional spaces, this means an invertible matrix; the translation cannot repair a loss of rank of a linear map.

A minimal-genus Seifert surface has a $2g_s(K)$-by-$2g_s(K)$ Seifert matrix $V$. Each entry of $tV-V^T$ has degree at most one, so its determinant has breadth of a Laurent polynomial at most $2g_s(K)$. This lower bound on Seifert genus need not be sharp: an untwisted Whitehead double can have Alexander polynomial of a knot $1$ and positive Seifert genus.

For strictly decreasing integer tuples $\kappa ,\eta$, let $A_{\kappa }(z)=\det (z_i^{{\kappa }_j})$ on the diagonal maximal torus of the unitary group. Expanding both determinants, Fourier orthogonality makes the integral zero unless the exponent tuples have the same permutation orbit. Strict decreasing order makes that equivalent to $\kappa =\eta$; the $n!$ matching permutation terms then each contribute one. An alternating Laurent polynomial is a finite linear combination of these alternants. For an irreducible character ${\chi }_{\lambda }$, its highest monomial gives coefficient one for $A_{\lambda +\delta }$ in ${\chi }_{\lambda }{A}_{\delta }$, with $\delta =(n-1,\dots ,0)$. The Weyl integration formula and Schur orthogonality relations force the sum of squared coefficients to be one, so all other coefficients vanish. This proves the unitary Weyl character formula directly.

A bilinear form is alternating if $B(v,v)=0$ for every vector $v$. Expanding $B(v+w,v+w)=0$ gives $B(v,w)+B(w,v)=0$, hence $B(v,w)=-B(w,v)$. If the characteristic is not two, the converse follows from $B(v,v)=-B(v,v)$ by dividing by two. In characteristic two, the relation $B(v,w)=-B(w,v)$ is just symmetry and does not force zero self-pairings: $B(x,y)=xy$ on ${\mathbb{F}}_2$ has $B(1,1)=1$.^[ref]

The matrixgives an alternating bilinear form over every field. Over the real numbers, its value is the signed area of the parallelogram with sides $v,w$: $B((2,1),(1,3))=6-1=5$, and swapping the sides gives $-5$. Every self-pairing is zero, yet $\det J=1$, so this bilinear form is nondegenerate. Directly, pairing $(x,y)$ with $(1,0)$ gives $-y$, and pairing it with $(0,1)$ gives $x$; both vanish only when $(x,y)=0$.

In characteristic different from two, a skew-symmetric matrix of odd integer size has zero determinant, since $\det A=\det (-A^T)=(-1{)}^n\det A$. Thus an alternating bilinear form on a finite odd-dimensional vector space is a degenerate bilinear form. The stronger statement that the rank of an alternating bilinear form is even holds in all characteristics: splitting off a two-dimensional block with matrix $J$ repeatedly proves it. A nondegenerate alternating bilinear form defines a symplectic vector space, the linear setting for signed area rather than squared length.

On complex symmetric matrices with positive-definite real part, the determinant has a nonzero analytic square root normalized positively on real positive matrices. It is continued within this accretive domain. For $A=G+iB$, put $C=G^{-1/2}B{G}^{-1/2}$ with real eigenvalues $b_j$; then $d(A)=\sqrt{\det G}{\prod }_j\sqrt{1+i{b}_j}$, using positive-real-part roots for each factor. The principal scalar square root of the product determinant need not equal this continued root: accumulated determinant arguments can cross the scalar branch cut. The principal matrix logarithm instead tracks all factors consistently.

For a complex antisymmetric tensor $A\in {\wedge }^2{\mathbb{C}}^4$, the SU(4) action is $A\mapsto UA{U}^T$. The unitary skew-diagonalization of an antisymmetric matrix gives $A=\mathrm{diag}(aϵ,bϵ)$ in suitable coordinates, where $ϵ=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. If $|a|\ne |b|$, preserving $A{A}^{†}$ forces a stabilizing unitary matrix to preserve the two coordinate planes. The identity $VϵV^T=(\det V)ϵ$ then forces both two-by-two blocks to have determinant one, using $\det U=1$ also when one of $a,b$ vanishes. The exact stabilizer is $SU(2)\times SU(2)$. If $|a|=|b|\ne 0$, the stabilizer is the larger compact symplectic group $USp(4)$. If $a=b=0$, it is all of $SU(4)$. The real structure of the SU4 exterior square imposes $b=\bar{a}$ in this form, so a nonzero real six-dimensional field has the larger symplectic stabilizer, whereas unequal paired singular values are generic for a complex six-dimensional field.

A planar diffeomorphism is area-preserving if the phase-space area of every measurable set is unchanged. The change-of-variables formula proves this property when $|\det DF|=1$ everywhere, and conversely gives that determinant condition for a smooth map preserving all sufficiently small areas. Such a map cannot have an asymptotically attracting fixed point with a positive-area local basin: iterates of a small invariant neighborhood would have to enter arbitrarily small disks while retaining its phase-space area. For a map with constant positive determinant $b$, invariance of a bounded region of positive phase-space area instead forces $b=1$.

The annual pulse-breeding predator-prey model is an area-preserving map in logarithmic population coordinates: $u^{\prime }=u+\log r-\kappa e^v$, then $v^{\prime }=v+\log s+\log (1+b{e}^{u^{\prime }})$. Each step is a shear of determinant one. At a positive fixed point, the ordinary-population Jacobian matrix has determinant one and trace $2-(1-s)\log r$. Eigenvalues lie on the unit circle for $0<(1-s)\log r<4$, but this linear neutrality does not prove nonlinear stability at every resonance. Area preservation excludes an isolated attracting positive periodic cycle.

Every identity-preserving biholomorphism of $\mathbb{C}/\Lambda$ lifts to multiplication by a nonzero complex number $\zeta$ satisfying $\zeta \Lambda =\Lambda$. Relative to a $\mathbb{Z}$-basis of $\Lambda$, multiplication by $\zeta$ is represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. Its real determinant is $|\zeta {|}^2$, while complex multiplication preserves orientation, so $\det A=1$ and $|\zeta |=1$. The Cayley-Hamilton theorem givesConsequently $2\mathrm{Re}\zeta =\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$, and $\zeta$ is a root of unity of order $1$, $2$, $3$, $4$, or $6$.

One identical fermion in each of two input modes remains one in each output after a lossless two-mode unitary transformation. The fully occupied two-mode state transforms by the determinant phase. Pauli exclusion prevents double occupation of either specified mode.

A fraction-free elimination algorithm for exact determinants. At each stage it divides a cross-multiplied update by the preceding pivot; the determinantal identities underlying the method make this division exact when its pivot assumptions hold. Pivoting handles zero pivots.

A basis is a linearly independent spanning set of vectors in a vector space. Every vector is a unique finite linear combination of basis vectors. Independence supplies uniqueness, and spanning supplies existence. An ordered basis turns the coefficients into a coordinate vector.

For $\mathcal{B}=(b_1,b_2)=((2,1),(1,2))$, the matrix with these vectors as columns isHence the coordinate vector of $v=(5,4)$ is $[v{]}_{\mathcal{B}}=B^{-1}v=(2,1)$, and $B[v{]}_{\mathcal{B}}=v$ recovers the usual coordinates. The determinant $3$ is nonzero, so this recovery works for every vector in ${\mathbb{R}}^2$. A basis need not have perpendicular or unit-length vectors; an orthonormal basis adds those geometric conveniences when an inner product is available.

The point of changing basis is to simplify the operation being studied. Diagonalization uses an eigenbasis to turn a coupled linear map into independent scalings. In a space of polynomials, the basis $1,x,x^2$ makes differentiation easy to encode, while orthogonal polynomials can make approximation errors easier to measure. Gilbert Strang's MIT lecture on independence, basis and dimension develops this connection between attainable columns, redundant columns and coordinates.^[ref]

A block matrix is block diagonal if all off-diagonal blocks vanish. It represents independent actions on a direct sum of vector spaces. Its determinant is the product of the diagonal-block determinants.

Write a linear system as $Ax+By=f$ and $Cx+Dy=g$, with $A$ an invertible matrix. Solving the first equation for $x$ and substituting in the second givesThus the Schur complement is the coefficient matrix remaining after the $x$ variables have been eliminated. When it is an invertible matrix, solve the reduced linear system for $y$ and recover $x$. In computation, expressions such as $A^{-1}B$ mean solving with each column of $B$, using the same LU decomposition or Cholesky decomposition when applicable.

For the example under block matrix inverse and right-hand side $(3,1,4{)}^T$, take $f=(3,1{)}^T$ and $g=4$. Since $C{A}^{-1}f=13/6$, the reduced equation is $(19/6)y=11/6$. Back substitution givesThe original equations return $3$, $1$ and $4$, respectively. The reduction retained the influence of $x_1,x_2$ on $y$ through the correction $C{A}^{-1}B$.

A short proof outline also explains whether $M$ is an invertible matrix: the exact matrix factorizationhas invertible matrices on the outside. Therefore $M$ is an invertible matrix precisely when the central block diagonal matrix is one. Reversing the factors gives the block matrix inverse, while taking determinants gives the Schur complement determinant identity.

Every orthogonal transformation of a finite-dimensional nondegenerate quadratic space over a field of characteristic different from two is a product of at most its dimension many hyperplane reflections. In the real positive-definite case, a reflection normal to $gv-v$ sends $gv$ to $v$ whenever $gv\ne v$. Composing with that reflection fixes $v$ and leaves an orthogonal transformation of $v^{\perp }$, so induction proves the bound. Each reflection has determinant minus one, hence orientation-preserving transformations use an even number of reflections.

Keep the same origin and the same vector $p=(2,1{)}^T$, but use the nonperpendicular basis vectors $b_1=(1,0{)}^T$ and $b_2=(1,1{)}^T$. They form a basis because the column matrix $B$ has nonzero determinant:The reconstruction is $p=b_1+b_2$. Its coordinates are read along lines parallel to the other axis, producing a parallelogram. These lines need not be perpendicular to the axis being measured. In contrast, the dot products are $p\cdot b_1=2$ and $p\cdot b_2=3$; treating those as coordinates would reconstruct $2b_1+3b_2=(5,3{)}^T$, a different vector.

The dual basis supplies the right measuring rules for any basis.^[ref] Here these coordinate functionals are $b_1^{\ast }(x,y)=x-y$ and $b_2^{\ast }(x,y)=y$, the rows of $B^{-1}$. They give $b_1^{\ast }(p)=b_2^{\ast }(p)=1$. Equivalently, their representing vectors $d_1=(1,-1{)}^T$, $d_2=(0,1{)}^T$ satisfy $d_i\cdot b_j={\delta }_{ij}$ and extract the coordinates as $d_i\cdot p$. These are Riesz dual basis in an inner product space representatives, rather than the original basis vectors.

The Gram matrix records how the nonperpendicular directions interact:For any real basis represented by a square invertible matrix $B$, writing $p=Bc$ gives these identities immediately. One can recover the coordinates either as $c=B^{-1}p$ or by solving $Gc=B^{T}p$. In this example, if $c=(a,b{)}^T$, the squared norm is $a^2+2ab+2b^2$, not $a^2+b^2$. Oblique coordinates remain useful when natural directions are not perpendicular, but the Gram matrix must accompany them when computing lengths or dot products.