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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 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.

An identity element for addition: $0+a=a+0=a$. In a vector space, it is the zero vector; among matrices of fixed dimensions, it is the zero matrix.

Take $V={\mathbb{C}}^k$, $W={\mathbb{C}}^N$, $B_1,B_2\in \mathrm{End}(V)$, $I:W\to V$, and $J:V\to W$, satisfyingRegular data modulo $U(k)$ construct framed Anti-self-dual Yang-Mills equations solutions of charge magnitude $k$. The ADHM factorization identity gives an explicit gauge curvature proof. The two complex matrices and the two rectangular maps have $4k^2+4kN$ real parameters; three real matrix constraints remove $3k^2$, and quotient by $U(k)$ removes another $k^2$, giving dimension $4kN$ at regular points. Degenerate data can describe zero-size singular limits rather than smooth instantons.

Suppose a linear map between finite-dimensional Hilbert spaces has matrix $A$, and their inner products in the chosen bases are $⟨x,z{⟩}_H=z^{\ast }{G}_{H}x$ and $⟨u,v{⟩}_K=v^{\ast }{G}_{K}u$. The Gram matrices $G_H,G_K$ are positive-definite matrices. Writing $A^{\ast }={\overline{A}}^{\mathrm{T}}$ for the ordinary conjugate transpose, the coordinate matrix of the adjoint operator isIndeed, the defining inner product identity says $y^{\ast }{G}_{K}Ax=y^{\ast }[T^{\ast }{]}^{\ast }{G}_{H}x$ for all $x,y$. Thus simply transposing and conjugating is valid in orthonormal bases, where both Gram matrices are the identity matrix.

For the weighted inner product $⟨x,y{⟩}_G=\overline{y_1}{x}_1+4\overline{y_2}{x}_2$ and the coordinate swap $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, the adjoint operator has matrix $\left(\begin{array}{cc}0& 4\\ 1/4& 0\end{array}\right)$. With $x=(1,0{)}^{\mathrm{T}}$ and $y=(0,1{)}^{\mathrm{T}}$, both inner products equal $4$. The coordinate swap is its own ordinary transpose, but is not self-adjoint in this weighted inner product.

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.

Algebraic K-theory studies invariants built from projective modules and their automorphisms over rings. Its lowest groups record the additive projective-module classes and stable invertible matrices, respectively. It differs from Topological K-theory, although vector-bundle clutching and module patching have parallel forms.

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.

A periodic lossless line can consist of equal-length sections with impedances $Z_1,Z_2$. Its semi-infinite input is a fixed point of the two-section impedance transformation. The physical solution selects a forward-energy or decaying Bloch wave. At special frequencies where the cell matrix is the identity, the fixed-point equation alone loses uniqueness; limiting absorption or a termination specifies the response.

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.

Suppose the matrix $B$ is holomorphic at zero and $0$ is an eigenvalue of $R$. Let $m$ be the largest nonnegative integer for which $-m$ is an eigenvalue of $R$. Seek $F=z^m{\sum }_{n\ge 0}{h}_n{z}^n$, with $h_0\in \ker (mI+R)$ nonzero. The Frobenius method gives $((m+n)I+R)h_n=-{\sum }_{j=0}^{n-1}{B}_j{h}_{n-1-j}$ for $n\ge 1$, where scalar terms multiply the identity. Every left-hand matrix is invertible by the choice of $m$. Its inverse is $O(1/n)$, so a scalar convergent majorant proves convergence of the series. This yields a nonzero analytic solution; it is necessarily nonconstant when $m>0$, but can be constant when $m=0$. For the displayed sign convention, indicial exponents are eigenvalues of $-R$.

The unitary matrix relating the anyon fusion trees $((ab{)}_{u}c{)}_t$ and $(a(bc{)}_v{)}_t$. Fix the convention $|((ab{)}_{u}c{)}_t⟩={\sum }_v[F_t^{abc}{]}_{uv}|(a(bc{)}_v{)}_t⟩$. An inverse change of parentheses uses the adjoint matrix, so its entries are complex conjugates with exchanged indices.

A matrix encoding multiplication by charge $b$ in the anyon fusion rule. These matrices commute in a braided monoidal category. On a sphere, the dimension of the anyon fusion space with puncture charges $a_1,\dots ,a_n$ and total vacuum anyon charge is $(N_{a_1}\cdots N_{a_n}{)}_{11}$.

For dimension-compatible matrices, $(AB)C=A(BC)$. In coordinates both sides have $(i,j)$ entry ${\sum }_{k,l}{A}_{ik}{B}_{kl}{C}_{lj}$; distributing and regrouping the finite sums proves equality. This permits accumulating a product left to right even when a recursive algorithm groups the same transformations from the right. It does not permit changing factor order, because matrix multiplication is generally not commutative.

For high-risk group fraction $p$, fixed total contact rates and reciprocal cross-group contacts, let $C=Bp/(1-p)$. The displayed SIS model describes within-group infectious fractions. In fractions the transmission matrix is $M=\left(\begin{array}{cc}A& B\\ C& D-C\end{array}\right)$; in infected-person counts it is $\mathrm{diag}(p,1-p)M\mathrm{diag}(p,1-p{)}^{-1}$. A common recovery rate gives basic reproduction number $\rho (M)/\gamma$. The mean population susceptible fraction does not generally equal its reciprocal.

The augmented matrix of a linear system $Ax=b$ appends the right-hand side as one extra column. Elementary row operations preserve its solution set. The linear system is consistent exactly when the coefficient matrix rank equals the augmented matrix rank, because this is equivalent to $b$ lying in the span of the columns of $A$. A larger augmented matrix rank produces a zero coefficient row with a nonzero right-hand side during Gaussian elimination.

The full symmetry group of a cube acts as all signed permutation matrices. The stabilizer subgroups of $(1,0,0)$, $(1,1,0)$ and $(1,1,1)$ have orders eight, four and six, respectively, and their fixed-point subspaces of a group action are the lines through those vectors. The three group orbits have sizes six, twelve and eight. They therefore give distinct axial types for the equivariant branching lemma. More generally the fixed-space dimension of a vector stabilizer subgroup equals the number of distinct nonzero absolute coordinate values, so these are precisely its one-dimensional types.

// Destination: dynamical-systems.bigb

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 matrix whose columns are the coordinate vectors of an ordered basis in a fixed reference basis. It is an invertible matrix, and multiplication by it converts coordinates in the column basis to the reference coordinates: $v=B[v{]}_B$. For a real inner product space with orthonormal reference basis, its Gram matrix is $B^{T}B$, and its reciprocal basis has basis matrix $B^{-T}$.

For a bipartite graph with ordered parts $X,Y$, its biadjacency matrix has rows indexed by $X$ and columns indexed by $Y$, with entry one for an edge and zero otherwise. It has size $|X|$ by $|Y|$, rather than the square full bipartite adjacency matrix on $X\bigsqcup Y$, which is the block matrix $\left(\begin{array}{cc}0& M\\ M^{\mathsf{T}}& 0\end{array}\right)$. Its singular values describe the normalized adjacency operator of a bipartite graph after division by $\sqrt{|X||Y|}$.