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For a decision $x$ in the unit ball, conditional expectation under the posterior givesThe symmetric positive semidefinite matrix $C+m{m}^T$ has a Rayleigh quotient maximized over $\Vert x\Vert \le 1$ by any unit eigenvector corresponding to its largest eigenvalue. Bayes decision rule therefore chooses any such eigenvector for the stated utility.

Write $z_i(x)=⟨a_i,x⟩+b_i$. The log-sum-exp function is convex and composition with the affine functions $z_i$ preserves convexity, so $f_{\beta }$ is convex. Directly, its Hessian matrix will also be shown positive semidefinite in part d.

Let $M=f(x)={\max }_i{z}_i(x)$. Factoring $e^{\beta M}$ out of the sum givesAt least one term in the sum is $1$, while every term is at most $1$. Thereforeand henceThus $f_{\beta }$ is a uniform smooth maximum of the affine pieces of $f$.

Define the softmax weightsThey obey $p_i\ge 0$ and ${\sum }_i{p}_i=1$. The gradient is their weighted mean,Differentiating once more gives the covariance-form Hessian matrixFor every unit vector $u$,where $G={\max }_i\Vert a_i{\Vert }_2$. The Hessian is a covariance matrix, so it is positive semidefinite; the displayed upper bound also gives ${\nabla }^2{f}_{\beta }⪯\beta G^{2}I$ in the Loewner order. Consequently $f_{\beta }$ has a Lipschitz gradient with

At $p$, the surface $\tilde{S}$ cannot cross the boundary $S=\partial \mathcal{R}$, because it lies inside $\mathcal{R}$. Consequently the two surfaces are tangent at $p$. Apply a proper Euclidean motion of Euclidean three-space so that $p=0$, their common tangent plane is $z=0$, and the prescribed inward unit normal is $N=e_3$. Locally writewith the region $\mathcal{R}$ on the side $z\ge f(x,y)$. ThenThus $g-f$ has a local minimum at zero and its Hessian matrix is positive semidefinite. The fundamental forms of a graph surface at a horizontal tangent plane giveTaking the trace of the positive-semidefinite difference proves the mean-curvature comparison at tangential contact:

There is no analogous conclusion for $K$. In the same local coordinates takeThen $g\ge f$, so the second graph lies on the inward side of the first, while the Gaussian curvature of a graph surface givesHence the mean-curvature inequality holds butUsing smooth bump functions, these local graph patches can be completed away from $p$ to a compact region with connected smooth boundary and a closed interior surface without changing their germs at $p$. This realizes the counterexample under the global hypotheses.

A symmetric function $k:\mathcal{X}\times \mathcal{X}\to \mathbb{R}$ is a positive-semidefinite kernel when, for every $m$, every $x_1,\dots ,x_m\in \mathcal{X}$, and every $c\in {\mathbb{R}}^m$,Equivalently, every associated kernel matrix is a positive semidefinite matrix.

A Reproducing-kernel Hilbert space $\mathcal{H}$ on $\mathcal{X}$ is a Hilbert space of real-valued functions such that every point-evaluation map $f\mapsto f(x)$ is continuous. The Riesz representation theorem then gives a function $k(x,\cdot )\in \mathcal{H}$ satisfying the reproducing propertyIts reproducing kernel is $k(x,y)=⟨k(x,\cdot ),k(y,\cdot ){⟩}_{\mathcal{H}}$.

Let $\Phi :M_d(\mathbb{C})\to M_D(\mathbb{C})$ be a linear map. It is positive when $X\ge 0$ implies $\Phi (X)\ge 0$, and completely positive whenis positive for every ancillary dimension $r$. In finite dimensions it is enough to check $r=d$.

A finite family of Kraus operators $A^{\alpha }:{\mathbb{C}}^d\to {\mathbb{C}}^D$ defines the Kraus representationThis map is completely positive because, for every positive semidefinite matrix $Y$ on the enlarged space,

Conversely, use the unnormalized maximally entangled vector $|\Omega ⟩={\sum }_{j=1}^d|j⟩|j⟩$. Complete positivity makes the Choi matrixpositive semidefinite. By the spectral theorem for normal operators, $C_{\Phi }={\sum }_{\alpha =1}^R|v_{\alpha }⟩⟨v_{\alpha }|$, where $R=\mathrm{rank}{C}_{\Phi }\le dD$. Reshape each $v_{\alpha }\in {\mathbb{C}}^D\otimes {\mathbb{C}}^d$ into a matrix $A^{\alpha }$ by $|v_{\alpha }⟩={\sum }_{\mu j}{A}_{\mu j}^{\alpha }|\mu ⟩|j⟩$. The Choi matrix inversion formulathen gives $\Phi (X)={\sum }_{\alpha }{A}^{\alpha }X(A^{\alpha }{)}^{†}$. Thus finite Kraus representations characterize finite-dimensional completely positive maps.

The additional normalizationmakes $\Phi$ trace preserving and hence a quantum channel; ${\sum }_{\alpha }{A}^{\alpha }(A^{\alpha }{)}^{†}=I_D$ instead makes it unital.

For $\lambda \ge 0$, a full-column-rank design matrix $B$, and a positive semidefinite matrix $\Omega$, minimize $\Vert y-B\beta {\Vert }^2+\lambda {\beta }^T\Omega \beta$. Differentiation gives $(B^{T}B+\lambda \Omega )\hat{\beta }=B^{T}y$. The fitted-value matrix is $B(B^{T}B+\lambda \Omega {)}^{-1}{B}^T$, whose trace measures the effective degrees of freedom.

A positive linear map sends every positive semidefinite matrix to a positive semidefinite matrix. Positivity need not persist after tensoring with the identity map on an ancillary system; requiring that stronger property gives a completely positive map.

The sums $P+N$ of a real positive semidefinite matrix $P$ and a symmetric nonnegative matrix $N$ form a convex cone inside the copositive cone. Both terms have nonnegative quadratic forms on the nonnegative orthant. The Horn copositive matrix shows that the inclusion is strict in dimension five; the sum of squares criterion for a biquadratic form explains this cone's relation to semidefinite programming.

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 coefficient column $q=(q_0,\dots ,q_d{)}^T$, $M_{ij}=q_i\overline{q_j}$ is a Hermitian matrix and a positive semidefinite matrix because $x^{\ast }Mx=|q^{\ast }x{|}^2$. Expanding $|q(z){|}^2$ on the unit circle gives coefficient $p_k={\sum }_{i-j=k}{M}_{ij}$. Thus $M$ has matrix rank one when $q\ne 0$ and zero otherwise. General feasible Gram matrices need not have matrix rank one.

For differences $D_i$ between corresponding stages and differences $F_i$ between their vector fields, a Runge--Kutta step satisfiesTo derive the identity, expand the squared norm of $d_{n+1}=d_n+h{\sum }_i{b}_i{F}_i$ and substitute $d_n=D_i-h{\sum }_j{a}_{ij}{F}_j$ in the linear term. Symmetrizing the double sum gives the displayed coefficients $m_{ij}$. For a dissipative vector field, the first sum is nonpositive when $b_i\ge 0$, and the second sum is nonnegative when the matrix $(m_{ij})$ is a positive semidefinite matrix. Thus algebraic stability implies B-stability.

A semidefinite program optimizes a linear function subject to equalities between affine functions and positive semidefinite matrix inequalities. It generalizes linear programming: a diagonal matrix is a positive semidefinite matrix exactly when its diagonal entries are nonnegative. Matrix trace expresses the objective as $⟨C,X⟩=\mathrm{tr}(CX)$ for real symmetric matrices.

Replacing $x{x}^T$ by a general positive semidefinite matrix relaxes maximization of $\Vert x{\Vert }_2^2$ subject to $|a_i^{T}x|\le 1$. The relaxed value is an upper bound because $x{x}^T$ has trace $\Vert x{\Vert }_2^2$ and satisfies the same quadratic constraints. Both problems are unbounded if the $a_i$ fail to span the ambient space. If they span it, $H={\sum }_i{a}_i{a}_i^T$ is positive definite and ${\lambda }_{\min }(H)\mathrm{tr}X\le \mathrm{tr}(HX)\le m$, proving boundedness and attainment of the relaxation.

Let $b_i=N{e}_i$ be the interpolating cardinal basis and set ${\Gamma }_{ij}={\int }_a^b{b}_i^{\prime \prime }{b}_j^{\prime \prime }dx$. Linearity gives $J(N\mathbf{v})={\mathbf{v}}^T\Gamma \mathbf{v}$, and this Gram matrix is a positive semidefinite matrix. Its null space consists precisely of vectors $(\alpha +\beta x_i{)}_i$, by the null-space statement for the second derivative roughness penalty. Thus its rank is $n-2$, including the zero matrix when $n=2$.

If $P$ is a real positive semidefinite matrix, then $x^{T}Px=0$ if and only if $Px=0$. Indeed $x^{T}Px=\Vert \sqrt{P}x{\Vert }^2$ by the principal square root of a positive semidefinite matrix, so a zero quadratic form puts $x$ in the kernel of $\sqrt{P}$ and hence of $P$.