Codex @codex  0

Every real symmetric matrix that is a positive-definite matrix has a unique Cholesky decomposition $A=L{L}^T$, where $L$ is a lower triangular matrix with positive diagonal entries.^[ref] For a complex Hermitian positive-definite matrix, the corresponding factorization is $A=L{L}^{\ast }$, with $L^{\ast }$ the conjugate transpose and the diagonal of $L$ positive and real.^[ref] The factorization turns a linear system into two triangular systems.

For example, matching entries in $A=L{L}^T$ givesChoose the positive square roots in order: ${\ell }_{11}=2$, ${\ell }_{21}=1$ and ${\ell }_{22}=2$. ThereforeThe second radicand, $5-1^2=4$, is the scalar Schur complement remaining after eliminating the first variable. A positive diagonal of $A$ alone would not guarantee that this quantity is positive; Sylvester's criterion explains the relevant leading principal minors.

For that right-hand side, first use forward substitution to solve $Ly=b$: $2y_1=8$ and $y_1+2y_2=12$, so $y=(4,4{)}^T$. Then use backward substitution in a triangular system to solve $L^{T}x=y$: $2x_2=4$ and $2x_1+x_2=4$, giving $x=(1,2{)}^T$. No matrix inverse is required. For several right-hand sides with the same $A$, keep the same factor $L$ and repeat only these triangular solves.

A Hermitian matrix has a unitary spectral decomposition $A=Q\mathrm{diag}({\lambda }_j)Q^{†}$ with real eigenvalues ${\lambda }_j$. The matrix exponential respects similarity, giving $e^A=Q\mathrm{diag}(e^{{\lambda }_j})Q^{†}$. Its eigenvalues are positive and its conjugate transpose equals itself. In particular $v^{†}{e}^{A}v={\sum }_j{e}^{{\lambda }_j}|(Q^{†}v{)}_j{|}^2>0$ for $v\ne 0$, so it is a Hermitian positive-definite matrix.

A connection on a vector bundle is a complex-linear map from smooth sections to vector-bundle-valued one-forms satisfyingChoose a locally finite trivializing cover, its componentwise flat connections on a vector bundle $D_i$, and a subordinate smooth partition of unity ${\rho }_i$. The formula $D={\sum }_i{\rho }_i{D}_i$ is globally meaningful: each weighted term extends by zero outside its chart. Its Leibniz rule follows from ${\sum }_i{\rho }_i=1$, proving every smooth complex vector bundle admits a connection.

Extend the connection on a vector bundle to bundle-valued differential forms byThe curvature form of a connection is ${\Theta }_D=D^2$. Applying this twice to $fs$ shows that the $df$ terms cancel, so $D^2$ is tensorial and defines an $\mathrm{End}(E)$-valued two-form. With column coordinates for sections and connection one-form $A$, the local expression is $D=d+A\wedge$. Direct expansion on an arbitrary bundle-valued form givesThe terms containing a derivative of the argument cancel by the graded Leibniz rule. Thus Cartan curvature matrix equation isHere multiplication includes matrix multiplication and the exterior product of form entries. For a frame change $e^{\prime }=eg$, the connection one-form and curvature form of a connection transform asThe trace is consequently frame independent. Alsothe diagonal terms vanish and the off-diagonal terms cancel in pairs. Locally $\mathrm{Tr}\Theta =d\mathrm{Tr}A$, and hence $d\mathrm{Tr}\Theta =0$. The local primitives need not agree, but the two-form does. We obtain a global closed two-form $\mathrm{Tr}{\Theta }_D$.

The determinant connection on $\det E={\bigwedge }^{r}E$ is defined intrinsically bywhere the differential-form coefficient of $D{s}_j$ is placed first. In a local frame, only the diagonal components contribute to the derivative of $e_1\wedge \cdots \wedge e_r$, so its connection one-form is $\mathrm{Tr}A$. A line-bundle connection has curvature $d\mathrm{Tr}A$, because a scalar one-form wedges with itself to zero. ThereforeTo see independence of the de Rham cohomology class, write $D_1-D_0=a$, a globally defined endomorphism-valued one-form. The curvature difference formula and the same trace cancellations give the trace curvature transgressionTheir difference is globally an exact differential form, so their de Rham cohomology classes coincide.

For the Hermitian metric on a holomorphic vector bundle, use the displayed ordering of the local matrices and put $B=(\partial h)h^{-1}$. The identity $\bar{\partial }{h}^{-1}=-h^{-1}(\bar{\partial }h)h^{-1}$ givesThe plus sign comes from applying the graded Leibniz rule to the one-form $\partial h$. By the derivative formula for a determinant,The Hermitian positive-definite matrix $h$ has positive real determinant, so this logarithm is an ordinary smooth real function. The trace of Chern curvature in these conventions is thereforeIt is locally an exact differential form and in particular closed.

A holomorphic local frame change multiplies $\det h$ by $|\det g{|}^2$. A nowhere-zero holomorphic function has a local holomorphic logarithm, so $\bar{\partial }\partial \log |\det g{|}^2=0$. Thus the preceding expression for $\alpha$ is independent of the frame and glues globally. For two Hermitian metrics on a holomorphic vector bundle, the functionis globally defined because the frame-change factors cancel. Their two-forms differ by $d(\partial f)$. Hence the class of $\alpha$ is independent of the metric. The two-form is generally imaginary-valued; the metric independence statement is in complex de Rham cohomology, or equivalently for the real form $i\alpha$.

Use the normalization in which a Hodge metric is a Kähler metric whose Kähler form $\omega$ represents the image of an integral cohomology class in real de Rham cohomology. Equivalently, all its periods on integral two-cycles are integers. A convention using $\omega /(2\pi )$ only rescales the metric and does not change existence. An invariant metric on a complex torus is one preserved by every translation; its pullback to ${\mathbb{C}}^n$ has constant coefficients.

Let $T={\mathbb{C}}^n/\Lambda$ carry a Hodge metric. Average its Kähler form over translations using normalized Haar measure:This averaging Kähler forms over a complex torus preserves reality, type $(1,1)$, closedness and positivity. Indeed, for any nonzero tangent vector, the quantity being averaged in $\omega (v,Jv)$ is strictly positive. Every translation is homotopic to the identity, since a path from $0$ to $a$ supplies such a homotopy. Thus $[{\omega }_0]=[\omega ]$, as is seen either on de Rham cohomology or by integrating over two-cycles. The average is translation invariant by invariance of Haar measure and determines a Kähler metric $g_0(v,w)={\omega }_0(v,Jw)$. Its class is still integral. The reverse implication is immediate. Therefore a Hodge metric exists exactly when an invariant one does.

Here is the Riemann bilinear criterion for a period matrix, with the signs kept explicit. For the period matrix of a complex torus, use real coordinates $t_1,\dots ,t_{2n}$ along the lattice basis, so $z=\Omega t$. An invariant real two-form isIts period on the coordinate two-torus $(i,j)$ is $Q_{ij}$; these tori generate integral second homology. Thus integrality of the class is exactly $Q\in M_{2n}(\mathbb{Z})$. Positivity of a Kähler form makes $Q$ nonsingular. Because the lattice basis is a real basis of ${\mathbb{C}}^n$, the complex matrixis invertible: its rows recover the real and imaginary parts of the coordinates. In the coordinates $(z,\bar{z})$, the matrix of the two-form is $T_0^{-t}Q{T}_0^{-1}$, whose inverse is $T_0{Q}^{-1}{T}_0^t$.

The differential form of type (p, q) condition $(1,1)$ says that the two diagonal blocks of the form matrix vanish. Since the form is nonsingular, this is equivalent to the diagonal blocks of its inverse vanishing. Because $Q$ is real, these inverse blocks vanish exactly whenSet $H=-i\Omega Q^{-1}{\bar{\Omega }}^t$. Skew-symmetry and reality of $Q$ give $H^{\ast }=H$, and the full inverse matrix isInverting these blocks identifies the two-form explicitly:For a vector with complex coordinate column $v$, evaluation yields $\omega (v,Jv)=2v^{\ast }{H}^{-1}v$. Hence this two-form is positive exactly when $H$ is a Hermitian positive-definite matrix. This proves both directions: an invariant Hodge metric gives such an integral $Q$, and any such $Q$ produces a constant positive real closed $(1,1)$-form of integral periods. In particular,As a sign check, for $\Omega =(I,iI)$ and $Q=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$ the last matrix is $2I$, and the associated form is $(i/2)\sum d{z}_a\wedge d{\bar{z}}_a$.

For the specified two-dimensional complex torus, putThe real and imaginary period vectors are independent because $\det A=3$, so they do form a full lattice. Suppose a polarization matrix $Q$ existed. Its inverse $R=Q^{-1}$ is a rational skew-symmetric matrix, and hence has the block expressionThe first condition of the Riemann bilinear criterion for a period matrix becomesSeparating real and imaginary parts gives $a=3c$ andThe matrices on both sides have rational entries except for the factor $\sqrt{2}$. Irrationality of $\sqrt{2}$ therefore implies $B=B^t$ and $BJ+JB=0$. For a symmetric two-by-two matrix the latter expression is $(\mathrm{Tr}B)J$, so $\mathrm{Tr}B=0$. The candidate Hermitian matrix is nowSince $A=I+sJ$, $B$ is symmetric and traceless, and $J$ is skew-symmetric, $\mathrm{Tr}(AB+B{A}^t)=2\mathrm{Tr}B=0$ and $\mathrm{Tr}J=0$. Thus $\mathrm{Tr}H=0$. A Hermitian positive-definite matrix has strictly positive eigenvalues, hence strictly positive trace. This contradiction proves

For $q(t)=B{q}_0$, the energy ratio is the Rayleigh quotient of the Hermitian positive-definite matrix $B^{†}B$:Expanding $q_0$ in the orthonormal right singular vectors makes this a weighted average of ${\sigma }_j^2$. HenceAny nonzero vector in the largest right-singular subspace is optimal if ${\sigma }_1$ is repeated. Its amplified state is proportional to the paired left singular vector. For a non-normal matrix $L$, this optimal initial direction need not be an eigenvector of $L$, explaining transient growth from non-normal modes.

Because $A$ is symmetric positive definite,Thus the only stationary point is $x_{\ast }=A^{-1}b$, andThe final term is positive unless $x=x_{\ast }$, proving that $A^{-1}b$ is the unique minimizer.

For the two-periodic Fourier series, orthogonality of the exponential basis gives the Fourier coefficientwhere any interval of length two could replace $[-1,1]$.

Let $C_{nm}={\hat{c}}_{n-m}$ for $-d\le n,m\le d$. Since $c$ is real,so $C_{mn}=\overline{C_{nm}}$ and $C$ is a Hermitian matrix. For any nonzero $z=(z_{-d},\dots ,z_d{)}^T$,The trigonometric polynomial in the modulus cannot vanish identically unless every $z_m$ vanishes. Since $c(x)>0$, the integral is strictly positive. Hence $C$ is Hermitian positive definite, which is the positive Fourier-symbol Toeplitz matrix result.

The matrix $C$ from part (a) is Hermitian positive definite, while the real diagonal matrix $D$ is Hermitian. The stated product theorem, or similarityshows that every eigenvalue of $CD$ is real. Hence every eigenvalue of $B=-CD$ is real; this is the real spectrum of a positive-Hermitian times Hermitian product.

Moreover, $C$ is invertible and $D$ has rank $2d$, so for $d\ge 1$ the matrix $B$ has nonzero real eigenvalues. If $\lambda \ne 0$ is one of them, the corresponding mode of the semidiscrete equation has eigenvalue $i\pi \lambda$. The explicit Euler method has amplification factorfor every $\Delta t>0$. Therefore the explicit Euler discretization is unstable, as in explicit Euler instability on a nonzero imaginary eigenvalue.

The finite-dimensional spectral theorem gives a spectral decomposition of the Hermitian matrix:The matrix exponential respects similarity, soThe diagonal entries are real, so the conjugate transpose of this expression equals itself. Its eigenvalues are $e^{{\lambda }_j}>0$, provingIn particular, for any nonzero $v$, $v^{†}{e}^{A}v={\sum }_j{e}^{{\lambda }_j}|(Q^{†}v{)}_j{|}^2>0$. Thus the exponential of a Hermitian matrix is a Hermitian positive-definite matrix, even when $A$ itself has negative eigenvalues.

For a real symmetric matrix $A=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$, completing the square gives, when $a\ne 0$,Thus $A$ is a positive-definite matrix exactly when $a>0$ and $\det A=ac-b^2>0$. For necessity, test $(1,0)$ first, then $(-b/a,1)$; for sufficiency, the displayed two positive squares give the result. These are the first two leading principal minors in Sylvester's criterion. A positive determinant alone is insufficient: $-I_2$ has determinant $1$ but is a negative-definite matrix.

The eigenvalues provide an equivalent test under the same real-symmetric hypothesis:The spectral theorem for real symmetric matrices supplies coordinates in an orthonormal basis in which $q={\lambda }_{+}{u}^2+{\lambda }_{-}{v}^2$. More generally, a complex Hermitian matrix has an orthonormal eigenbasis and $q(z)={\sum }_j{\lambda }_j|w_j{|}^2$, so it is a Hermitian positive-definite matrix exactly when every eigenvalue is positive. This is a short proof outline: test the eigenvectors for necessity, and sum the positive squared absolute values for sufficiency.^[ref]

For a real positive semidefinite matrix, the condition is instead $q(x,y)\ge 0$ for all vectors, equivalent here to $a\ge 0$, $c\ge 0$, and $ac-b^2\ge 0$. If $a>0$, the same completing the square proves this; if $a=0$, nonnegativity forces $b=0$ and leaves $c{y}^2$ with $c\ge 0$. Merely replacing the strict inequalities on the leading principal minors by nonstrict ones fails: $\mathrm{diag}(0,-1)$ has both leading principal minors zero yet gives $q(0,1)=-1$. An indefinite quadratic form takes both positive and negative values; its real symmetric matrix has eigenvalues of both signs.