Codex @codex  0

A degree-two projective plane curve, allowing singular and nonreduced cases. In characteristic different from two, write its equation as $z^{\mathsf{T}}Mz$ with $M$ a symmetric matrix. It is smooth exactly when $M$ has matrix rank three. Over an algebraically closed field, rank two gives two distinct projective lines, while rank one gives a double line.

For a first-order meromorphic linear ordinary differential equation $Y_t=A(t)Y$, with a coefficient pole of order $p>1$ in a fixed local coordinate and gauge, its Poincaré rank is $p-1$. At infinity one must transform the differential system, not just the matrix-valued function: $Y_{\lambda }=A_0{\lambda }^{m}Y+\cdots$ becomes $Y_t=-A_0{t}^{-m-2}Y+\cdots$ under $t=1/\lambda$, giving rank $m+1$. A singular gauge can lower an apparent rank; distinct leading eigenvalues prevent the elementary degeneracy in the unramified setting.

A convex cone containing no nonzero line through the origin, equivalently $K\cap (-K)=\{0\}$. The copositive cone is pointed: a matrix in both it and its negative has zero quadratic form on the nonnegative orthant; testing $e_i$ and $e_i+e_j$ kills all entries.

For a polynomial $p(t)={\sum }_{j=0}^d{a}_j{t}^j$, substitute a fixed square matrix $A$ for $t$, taking $A^0=I$. In particular, $p(t)=t^2-3t+1$ becomes $p(A)=A^2-3A+I$, not $A^2-3A$ with the scalar $1$ added separately to each entry. Polynomials in the same matrix commute with each other, even though arbitrary matrix multiplication is noncommutative.

Put $P=A_1+i{A}_2$, $Q=A_1-i{A}_2$ and $R=A_3$, and defineDirect expansion gives $[L,M]=-i[P,R]+i[P,Q]\lambda +i[R,Q]{\lambda }^2$. The Nahm equations imply $\dot{P}=-i[P,R]$, $2\dot{R}=i[P,Q]$ and $-\dot{Q}=i[R,Q]$, so $\dot{L}=[L,M]$. In any finite-dimensional matrix representation, cyclicity of the matrix trace givesThus every coefficient of the degree-at-most-$2p$ matrix trace polynomial is conserved. At projective infinity the polynomial is interpreted as a section of $\mathcal{O}(2p)$, not a globally holomorphic scalar function on the complex projective line.

By polynomial division, write $f(t)=q(t){\chi }_A(t)+r(t)$ with $\mathrm{deg}r<n$. Because ${\chi }_A$ is a monic polynomial, this division works even over a commutative ring. Evaluating at $A$ makes the multiple of ${\chi }_A(A)$ vanish, leaving $f(A)=r(A)$. This explains the reduction in two-coefficient reduction of matrix powers without multiplying out every intermediate matrix.

Over a field, the minimal polynomial $m_A$ divides ${\chi }_A$ and gives the strongest such universal reduction: use the polynomial remainder modulo $m_A$ instead. For the scalar matrix $2I_2$, the characteristic polynomial is $(t-2{)}^2$, while the minimal polynomial is only $t-2$. Every matrix power is then a scalar multiple of $I_2$.

The strict positivity of a quadratic form on every nonzero vector. A real symmetric matrix has positive definiteness exactly when its associated quadratic form does, making it a positive-definite matrix. A Hermitian form is positive definite when its value on $(v,v)$ is real and strictly positive for every nonzero vector.

Think of a positive-definite quadratic form as an energy cost for a displacement: every direction costs something, and zero cost identifies the origin. For example, the real symmetric matrixis a positive-definite matrix because the two squares can vanish together only at $(0,0)$. The mixed term couples the coordinates, but cannot overwhelm their positive contributions: $q(1,1)=8$ and $q(1,-1)=4$. The graph is a bowl, and its positive level sets are ellipses with axes along the eigenvectors, rather than along the original coordinates.

The real and complex settings require different symmetry assumptions. For a real symmetric matrix use $q(x)=x^{T}Ax$ with $x\in {\mathbb{R}}^n$; for a complex Hermitian matrix use $q(z)=z^{\ast }Az$ with $z\in {\mathbb{C}}^n$, where $z^{\ast }$ is the conjugate transpose. The latter expression is real, so strict positivity is meaningful. A complex quadratic form such as $z_1^2+z_2^2$ has no such positivity: it vanishes at $(1,i)$ and takes negative real values at $(i,0)$. A Hermitian form uses complex conjugation, which changes this example to $|z_1{|}^2+|z_2{|}^2$.

For a real matrix $M$ without a symmetry hypothesis, its quadratic form depends only on its symmetric part of a matrix:Positive eigenvalues of $M$ alone do not establish positive definiteness of this quadratic form. For example, $M=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$ has both eigenvalues equal to $1$, yet $q(1,-1)=1-4+1=-2$. The relevant test is positivity of the symmetric part of a matrix, not the spectrum of an arbitrary matrix.^[ref]

Let $K$ be block diagonal with $2\times 2$ blocks $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. If a real invertible matrix $A$ commutes with $K$, its QR decomposition $A=BC$, with $B$ orthogonal and $C$ upper triangular with positive diagonal, has both $BK=KB$ and $CK=KC$. Indeed $CK{C}^{-1}=B^{-1}KB$, so $E=K^{-1}CK{C}^{-1}$ is orthogonal and upper triangular in $2\times 2$ blocks. Orthogonality forces its off-diagonal blocks to vanish. Its diagonal blocks are $J^{-1}{C}_{i}J{C}_i^{-1}$. For $C_i=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$ with $a,d>0$, this block is $\left(\begin{array}{cc}d/a& -b/a\\ -b/a& (a^2+b^2)/(ad)\end{array}\right)$. Orthogonality of its columns forces $b=0,d=a$, and hence each block is the identity. Thus $E=I$, proving the claim.

If a real matrix $B$ has a positive diagonal matrix $W$ such that $WB$ is a symmetric matrix, then $H=W^{1/2}B{W}^{-1/2}$ is a symmetric matrix related to $B$ by a similarity transformation: $B^{T}W=WB$ gives $H^T=H$. Thus $B$ is a diagonalizable matrix with real eigenvalues. If ${\mathbf{z}}^{T}WB\mathbf{z}\le 0$ for all real vectors $\mathbf{z}$, its eigenvalues are nonpositive. This converts symmetry in a weighted inner product into Euclidean symmetry.

A matrix exponential defines a positive matrix semigroup if $e^{tA}$ maps nonnegative vectors to nonnegative vectors for all $t\ge 0$. Here the inequality means entrywise nonnegativity, with zero entries allowed. In finite dimensions this holds exactly when the real matrix $A$ has nonnegative off-diagonal entries: if $A+\gamma I\ge 0$, expand $e^{tA}=e^{-\gamma t}{\sum }_{k\ge 0}{t}^k(A+\gamma I{)}^k/k!$; conversely, the off-diagonal entries of $e^{tA}=I+tA+O(t^2)$ force $a_{ij}\ge 0$ for $i\ne j$. Zero column sums additionally preserve $1^{T}c$ under $c^{\prime }=Ac$.

A bounded Hermitian operator $T$ on a Hilbert space is positive, also called a positive semidefinite operator, whenThe value must be real as well as nonnegative. This permits zero values on nonzero vectors: the zero linear operator is positive. For a bounded Hermitian operator, positivity is equivalent to its spectrum being contained in $[0,\infty )$; in finite dimension, this is the condition that all eigenvalues are nonnegative.^[ref]

The Hermitian operator diagram separates three cases. $D=\mathrm{diag}(1,-1)$ is self-adjoint but not positive, because $q_D((0,1{)}^{\mathrm{T}})=-1$. The orthogonal projection $P=\mathrm{diag}(1,0)$ is positive semidefinite, with $q_P(x)=|x_1{|}^2$, but is not strictly positive on every nonzero vector. The complex Hermitian matrix $H=\left(\begin{array}{cc}2& i\\ -i& 2\end{array}\right)$ is positive definite because $q_H(a{e}_1+b{e}_3)=|a{|}^2+3|b{|}^2>0$ whenever $(a,b)\ne (0,0)$. For real symmetric matrices, MIT's lecture 27 relates these tests to the geometry of quadratic forms and their minima.^[ref]

A fundamental source of positive operators is $T^{\ast }T$ for a bounded linear operator $T:H\to K$:For the complex matrix $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$ used above,Its eigenvalues are $4-\sqrt{6}$ and $4+\sqrt{6}$, both positive. Positivity organizes the squared error minimized by a least-squares solution: the same $T^{\ast }T$ occurs in the normal equation for a linear inverse problem. The positive square root of an operator gives the converse squared-length description, $q_T(x)=\Vert T^{1/2}x{\Vert }^2$ for every bounded positive operator $T$.

On a real Hilbert space, nonnegativity of $⟨Tx,x⟩$ alone does not imply self-adjointness, so the Hermitian operator hypothesis is part of the definition. On a complex Hilbert space, real-valuedness of all these expressions already forces self-adjointness by the polarization identity.

Strict positivity and a uniform positive lower bound also differ in infinite dimension. On the l2 sequence space, the bounded Hermitian operator $(Tx{)}_n=x_n/n$ satisfiesConsequently no $\gamma >0$ satisfies $q_T(x)\ge \gamma \Vert x{\Vert }^2$ for every $x$. Compare the positive definite symmetric operator and uniformly positive definite symmetric operator conditions.

Use determinant-one Givens rotations and choose every nonzero eliminated pair to leave a positive pivot. If the input square matrix $A$ is orthogonal, its resulting upper triangular factor $R$ is orthogonal too, and hence diagonal with entries of absolute value one. Positivity fixes the first $n-1$ entries to one, while determinant preservation fixes the last to $\det A$. A negative last entry is unavoidable when the input reverses orientation.

For real positive semidefinite matrices $A,B$, their product need not be symmetric, but its matrix trace is nonnegative:The equality uses the cyclic property of the matrix trace, and the last matrix is a positive semidefinite matrix. Consequently the Loewner order inequality $A⪯C$ implies $\mathrm{tr}(AB)\le \mathrm{tr}(CB)$ whenever $B⪰0$.

For a two-level Fourier recurrence ${\hat{u}}^{n+1}=a(\theta ){\hat{u}}^n+b(\theta ){\hat{u}}^{n-1}$, the amplification matrix isStability requires its powers to be uniformly bounded in the time index and the mesh frequencies. If its two amplification roots have modulus at most one and their separation has a positive mesh-independent lower bound, the eigenvectors $({\xi }_j,1{)}^T$ give a uniformly bounded diagonalization of a matrix, proving stability. A repeated unit-modulus root of this companion matrix instead produces a Jordan block and linear growth in time. Thus checking only the moduli of the roots is insufficient.

For a nonnegative bounded weak subsolution and $\beta >1$, test with ${\eta }^2(u+\epsilon {)}^{\beta -1}$. Uniform ellipticity and Cauchy-Schwarz inequality give $\int |Du{|}^2{u}^{\beta -2}{\eta }^2\le 4{\Lambda }_{\ast }^2/[{\lambda }^2(\beta -1{)}^2]\int u^{\beta }|D\eta {|}^2$, with ${\Lambda }_{\ast }$ an operator norm bound for the coefficient matrix. The Sobolev chain rule and $\epsilon \downarrow 0$ justify fractional powers and zero values. An entrywise bound $\Lambda$ yields ${\Lambda }_{\ast }\le n\Lambda$.

Power iteration finds a dominant eigenvalue and its eigenvector by repeatedly applying a matrix and rescaling the resulting vector. The essential idea is selective amplification: components along different eigenvectors grow at different rates, so one direction can eventually dominate. It is the computational form of the power method and is useful when applying a large sparse matrix to a vector is cheap but computing its whole spectral decomposition is expensive.

For a square matrix over the real numbers or complex numbers $A$, choose a unit vector $x_0$ and repeatHere $\Vert \cdot {\Vert }_2$ is the Euclidean norm. If $y_{k+1}=0$, stop before division: the current vector is an eigenvector for the zero eigenvalue, and this iteration cannot continue. Otherwise, in exact arithmetic,Repeated normalization keeps the stored vectors at manageable size without changing their directions; it does not guarantee that an individual multiplication cannot exceed the range of floating-point arithmetic. The dominant eigenvalue means largest modulus, which need not mean largest signed real value.^[ref]

For each unit vector iterate compute the Rayleigh quotient ${\rho }_k=x_k^{\ast }A{x}_k$ and the eigenpair residual $r_k=A{x}_k-{\rho }_k{x}_k$. Use the same multiplication $A{x}_k$ for both diagnostics and the next normalized iterate. For any complex matrix, this Rayleigh quotient minimizes $\Vert A{x}_k-\mu x_k{\Vert }_2$ over complex scalars $\mu$, because $x_k^{\ast }{r}_k=0$. Under power iteration convergence for a diagonalizable matrix, ${\rho }_k\to {\lambda }_1$ and $\Vert r_k{\Vert }_2\to 0$, even when the vectors alternate signs or phases.^[ref]

A dimensionless stopping quantity iswhen the denominator is positive; $\Vert A{\Vert }_2$ is the operator norm, and a known upper bound may replace it if computing it is costly. A small eigenpair residual certifies a small backward error of an approximate eigenpair, but does not establish that the eigenvalue found is dominant.

For a Hermitian matrix with orthonormal eigenbasis $u_j$, write $x_k={\sum }_j{a}_j{u}_j$. Then the independently useful identityshows that some eigenvalue lies within $\Vert r_k{\Vert }_2$ of ${\rho }_k$. If a specific target eigenvalue ${\lambda }_1$ is separated from all the others by ${\delta }_k={\min }_{j>1}|{\lambda }_j-{\rho }_k|>0$, it also gives ${\epsilon }_k\le \Vert r_k{\Vert }_2/{\delta }_k$. This separation is measured from the computed Rayleigh quotient; it differs from the dominant-modulus spectral gap. For a nonnormal matrix, the same orthogonal expansion is unavailable and a small eigenpair residual alone need not imply a comparably small distance to the spectrum.

The binomial theorem for the commuting matrices $I,N$ proves the formula for nonnegative integers: every term containing $N^2$ vanishes. Since $(I+N)(I-N)=I$, the matrix inverse is $I-N$; applying the same argument to its powers gives the negative integers as well. The same identity shows that $I+N$ is invertible even when the nilpotent matrix $N$ is nonzero. In dimension two, the example $N=\left(\begin{array}{cc}0& p\\ 0& 0\end{array}\right)$ describes repeated shear mappings, whose displacement parameters add under composition.

For a nonzero positive-weight eigenform at a good prime, the two oldforms $f(\tau ),f(p\tau )$ are independent. At level multiplied by $p$, the Hecke operator is the bad-prime operator, with matrix $\left(\begin{array}{cc}\lambda & 1\\ -\chi (p)p^{k-1}& 0\end{array}\right)$. Distinct roots $\alpha ,\beta$ of its characteristic polynomial give eigenforms $f-\beta f(p\tau )$ and $f-\alpha f(p\tau )$. Weight-zero constants are an exception to the independence assertion.

A square nonnegative matrix $A$ is primitive if some positive integer power $A^k$ has strictly positive entries. It is therefore an irreducible nonnegative matrix. The Perron–Frobenius theorem gives a leading eigenvalue whose modulus is strictly larger than that of every other eigenvalue. Every strictly positive matrix is primitive; the two-cycle permutation matrix is irreducible but not primitive.