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Use the shape-rate convention for the Gamma distribution. DefineThe zero null space assumption makes $B$ a positive-definite matrix, and $u_{\ast }$ is the unique ordinary least squares solution. The Gaussian likelihood contributes ${\gamma }^{k/2}$, which must be retained because $\gamma$ is unknown. Multiplying it by the improper prior and the Gamma distribution gives the normal-gamma posterior with a flat prior:Here the constant density of the improper prior cancels on normalization. PutThe normal equation $A^{T}r=0$ makes the residual orthogonal to the range of $A$. The orthogonal decomposition by a closed subspace into that range and its orthogonal complement givesIntegration over $u$ by the Gaussian integral, followed by integration over $\gamma$ using the Gamma distribution, yieldsThus the posterior distribution is proper, despite the improper prior; here $k\ge d$ follows from the zero null space assumption.

Completing the square in $u$ and collecting the powers of $\gamma$ give the two conditional distributions:Explicitly their probability density functions areFor a joint maximum a posteriori estimate relative to $dud\gamma$, first fix $\gamma >0$. The unique maximizing $u$ is $u_{\ast }$. The remaining log density is $(a-1)\log \gamma -b_0\gamma$ plus a constant. Consequently, if $a>1$,The stated assumptions do not always imply $a>1$. If $a=1$, the supremum occurs as $\gamma \downarrow 0$ but is not attained on $\gamma >0$; if $a<1$, the density is unbounded there. In either case there is no joint maximum a posteriori estimate in the stated parameter space. Calling zero a boundary mode does not make it an admissible positive precision.

If “MAP estimators” instead means separate modes of the marginal distributions, integration givesThus the marginal maximum a posteriori estimate of $u$ is always $u_{\ast }$, and that of $\gamma$ is $(s-1)/b_0$ when $s>1$, with the same nonattainment at zero when $s\le 1$. Joint and marginal maximization are different operations; the usual joint interpretation gives the preceding boxed pair.

Differentiating the ridge regression objective givesFor every nonzero $z$,so $X^{T}X+\lambda I_p$ is positive definite and hence invertible. The strictly convex objective therefore has the unique closed-form ridge regression estimator

Take the Dulac function $B=1$. Fordirect differentiation givesThe quadratic form in parentheses has matrixwhose eigenvalues are $14$ and $40$, so it is positive definite. For $0<k\le 3$ the divergence is nonpositive and vanishes only at the origin when $k=3$. The Bendixson-Dulac criterion therefore gives

Because $A$ is a positive-definite matrix, it has a positive-definite inverse square root. Choose the exact inverse-square-root preconditionerThenwhich has one distinct eigenvalue, so preconditioned CG terminates in one step.

This construction is of little computational use because obtaining and applying $A^{-1/2}$ is generally at least as expensive in storage and arithmetic as directly factoring or solving the original system. A useful preconditioner must only approximate this transformation while remaining cheap to construct and apply.

Here the only nonzero coefficients of $a$ areTherefore, for each $(m,n)\in {\Lambda }_N$, the truncated system iswhere a coefficient is omitted when its index is outside ${\Lambda }_N$ or equals $(0,0)$.

For wavevectors $\mathbf{k}=(m,n)$ and $\mathbf{\ell }=(p,q)$, the matrix entry isBecause ${\hat{a}}_{\mathbf{r}}={\hat{a}}_{-\mathbf{r}}$, $A_{\mathbf{k},\mathbf{\ell }}=A_{\mathbf{\ell },\mathbf{k}}$, so the matrix is symmetric. Its diagonal entry is $5{\pi }^2|\mathbf{k}{|}^2$. The sum of absolute off-diagonal entries is at mostbecause $|\mathbf{k}{|}^2\pm k_j\ge 0$ for nonzero integer $\mathbf{k}$; truncation can only reduce the sum. Hence $A$ is strictly diagonally dominant with positive diagonal. The Gershgorin circle theorem places every eigenvalue in the positive intervalfor at least one row. Since the symmetric matrix has only real eigenvalues, all are positive, and $A$ is a positive-definite matrix. Any one-dimensional ordering merely conjugates $A$ by a permutation matrix, preserving symmetry and positive definiteness.

Apply symmetric Gaussian elimination without row exchanges. At step $k$, let $d_k$ be the leading diagonal entry of the remaining symmetric Schur complement. If $d_k\le 0$, stop and report that $A$ is not positive definite. If $d_k>0$, use it to eliminate the rest of its row and column. If all steps succeed, this constructs an LDL decompositionwhere $L$ is unit lower triangular and $D=\mathrm{diag}(d_1,\dots ,d_n)$ has positive diagonal.

The test is correct from first principles. If all $d_k>0$, then for every nonzero $x$,because $L$ is invertible. Conversely, if $A$ is positive definite, its first pivot is $a_{11}>0$, and completing the square givesChoosing $s=-b^{T}y/a$ shows that the Schur complement is positive definite. Induction forces every pivot to be positive. This is also Sylvester's criterion.

At step $k$, updating the remaining matrix costs $O((n-k{)}^2)$ arithmetic operations. Hence the total iswhich proves the existence of the required algorithm.

The multivariable chain rule gives, at $(x,y)=(a+t\cos \gamma ,b+t\sin \gamma )$,Sufficient conditions for a strict local minimum at $t=0$ are $g^{\prime }(0)=0$ and $g^{\prime \prime }(0)>0$.

At a stationary point, $g^{\prime }(0)=0$ in every direction. Its second derivative is the quadratic form of the Hessian matrix. Ifthen completing the square givesfor every nonzero direction $(u,v)$. The Hessian is therefore positive definite, and the stationary point is a strict local minimum.

The power supplied to overcome drag equals the viscous dissipation:For a nonzero rigid translation the strain cannot vanish throughout a fluid at rest at infinity, so the integral is strictly positive. Since $\mathbf{F}=-\mathbf{RU}$,Thus the symmetric resistance matrix is positive definite.

Fix $j\in A$ and writeThe matrix $R$ is positive definite. Applying the Sherman–Morrison formula to $R+X_j{X}_j^T$ yields

Because $A$ is a real symmetric positive-definite matrix, the finite-dimensional spectral theorem supplies an orthonormal eigenbasis $u_1,\dots ,u_n$ withIts eigendecomposition is also its singular value decomposition. If $e=y^{(\delta )}-y$, thensoThe operator norms satisfy $\Vert A{\Vert }_2={\lambda }_n$ and $\Vert A^{-1}{\Vert }_2=1/{\lambda }_1$. Therefore the worst-case relative perturbation bound isThe ratio ${\kappa }_2(A)$ is the spectral condition number of a positive-definite matrix. A large ratio means that data noise aligned with an eigenvector for the smallest eigenvalue is strongly amplified, so the inverse problem is ill conditioned.

If $T(w)=w$, the defining equation becomes $M(w)+F(w)=M(w)$, so $\boxed{F(w)=0}$.

A map is firmly nonexpansive in the $M$-inner product whenPut $p=T(v)$ and $q=T(w)$. The two implicit equations giveBecause $F$ is a monotone operator,Thereforewhich is precisely firm nonexpansiveness. In particular, the preconditioned proximal point algorithm map $T=(M+F{)}^{-1}M$ is nonexpansive in the norm induced by the positive-definite matrix $M$.

Assume $p\in L^{\infty }(-1,1)$ with $p(x)\ge p_0>0$, $q\in L^{\infty }(-1,1)$ with $q\ge 0$, and $f\in L^2(-1,1)$. The natural energy space is the Sobolev space $V=H_0^1(-1,1)$. Multiplyingby $v\in V$ and using integration by parts gives the weak formulationwhereThe form is bounded, and the Poincare inequality together with $p\ge p_0$ makes it coercive:The functional $\ell$ is bounded by the Cauchy-Schwarz inequality. The Lax-Milgram theorem therefore gives a unique weak solution.

Define the energyIf $y$ is the weak solution, then for every $w\in V$,Thus $y$ is the unique minimizer of $J$. Conversely, differentiating $J(y+tw)$ at $t=0$ recovers $a(y,w)=\ell (w)$, so the minimization and weak problems are equivalent.

The Ritz method chooses a finite-dimensional conforming space $V_h\subset V$ and minimizes $J$ over $V_h$. Its minimizer $y_h$ satisfiesFor a basis $\{{\varphi }_j\}$ this becomes $Kc=F$, withCoercivity makes the stiffness matrix $K$ symmetric positive definite, so the discrete minimizer exists and is unique.

Subtracting the continuous and discrete equations gives Galerkin orthogonality,In the energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$, for any $v_h\in V_h$,After cancellation and minimization over $v_h$, this proves the sharp symmetric form of the Céa lemma:

For a mesh $-1=x_0<x_1<\cdots <x_N=1$, take $V_h$ to be the continuous functions that are affine on each interval and vanish at the endpoints. The interior piecewise-linear hat functions form a basis, and each overlaps only its two neighbours, so $K$ is symmetric tridiagonal. In the illustrative case $p=1$ and $q=0$ on a uniform mesh, its diagonal and neighbouring entries areIf $y\in H^2(-1,1)$, its nodal interpolant $I_{h}y\in V_h$ satisfies $\Vert y-I_{h}y{\Vert }_{H^1}=O(h)$. The best-approximation estimate therefore gives $\Vert y-y_h{\Vert }_{H^1}=O(h)$, and a standard duality argument improves the $L^2$ error to $O(h^2)$ under the corresponding elliptic regularity. This completes the route from the boundary-value problem through its variational principle to a sparse, stable finite-element approximation.

Condition on the information available before time $n$. The current wealth is then fixed, while ${\xi }_n\sim N(b,\Sigma )$ is independent of that information. PutFor any portfolio $\theta$, use the inner product induced by the positive-definite matrix $\Sigma$ to writeThe linear image of a multivariate normal vector shows that the two portfolio returns are jointly normal, andThus independence of uncorrelated jointly normal variables makes the residual return ${\eta }^T{\xi }_n$ independent of the aligned return; it also has expected value zero.

The continuation value $V(n,\cdot )$ is concave by part (c). Conditional Jensen inequality therefore shows that adding the independent centered residual cannot improve the objective:where $m=(1+r)X_{n-1}$. Since the optimal portfolio is unique, its residual must be zero.

It remains to determine the sign. If $\lambda <0$, the portfolios $\lambda {\Sigma }^{-1}b$ and $-\lambda {\Sigma }^{-1}b$ have the same return variance, namely ${\lambda }^{2}B$, while their means are $\lambda B$ and $-\lambda B$. The latter return has the distribution of the former plus the positive constant $-2\lambda B$. Since $V(n,\cdot )$ is increasing, replacing $\lambda$ by $-\lambda$ cannot reduce the objective, contradicting uniqueness. Hence $\lambda \ge 0$.

Applying this conditional argument at every time gives nonnegative, past-measurable random variables ${\lambda }_n$ such thatThis is the Gaussian one-fund theorem.

Forthe gradient is $\nabla F(x)=Ax-b=-r$. The steepest descent method with exact line search for a positive-definite quadratic therefore uses

For $A=\mathrm{diag}(1,\gamma )$ and $b=0$, the minimizer is $x^{\ast }=0$. PutStarting from $x^{(0)}=(\gamma ,1{)}^T$, direct substitution gives ${\alpha }_k=2/(\gamma +1)$ and, inductively,The Euclidean norm is unchanged by the alternating sign, soFor a positive-definite matrix, the spectral condition number of a positive-definite matrix isHere ${\kappa }_2(A)=\gamma$, and hence

The Conjugate gradient method starts with $p_0=r_0$ and updatesBy finite termination of the conjugate gradient method, it reaches the exact solution in at most the number of distinct eigenvalues. This matrix has two, so at mostare required in exact arithmetic.

Suppose a feature map $\varphi :{\mathbb{R}}^d\to {\mathbb{R}}^p$ represented the Gaussian kernel. Choose $n>p$, a unit vector $e$, and points $x_i=iae$ for $i=0,\dots ,n-1$. Their kernel matrix isFor sufficiently large $a$, hence sufficiently small $r$, every row satisfiesThus $K$ is a symmetric strictly diagonally dominant matrix with positive diagonal and is therefore a positive-definite matrix, so $\mathrm{rank}K=n$.

On the other hand, if $\Phi$ is the $n\times p$ matrix whose $i$th row is $\varphi (x_i{)}^T$, then $K=\Phi {\Phi }^T$ and $\mathrm{rank}K\le p<n$, a contradiction. Hence every feature-space realization of the Gaussian kernel requires an infinite-dimensional vector space.

The differential of the log-determinant is $d\log \det \Omega =\mathrm{Tr}({\Omega }^{-1}d\Omega )$. The subdifferential of the entrywise ${\ell }^1$ norm consists of symmetric matrices $Z$ withThe Karush-Kuhn-Tucker conditions for the Graphical Lasso are thereforeBecause $-\log \det \Omega$ is strictly convex on the positive-definite matrices, these conditions characterize the unique minimizer whenever it exists.

At $t=1$, the diagonal entries of $U-L$ are $2$, while every off-diagonal $(i,j)$ entry isHence $U-L$ is exactly the Coxeter Gram matrix $G(W)$, and part c givesFor a Finite Coxeter group the Gram matrix is positive definite, and for a Hyperbolic Coxeter group it is nondegenerate with Lorentzian signature. In either case $\det (I-C)\ne 0$, so $1$ is not an eigenvalue of $C$ and the Coxeter element fixes no nonzero vector.

For an Affine Coxeter group, the Gram form has a nonzero radical. If $0\ne v\in \mathrm{rad}G(W)$, then $⟨v,e_i⟩=0$ for every $i$, and every generating reflection satisfiesTheir product $\sigma (c)$ therefore fixes $v$. Thus every affine Coxeter element has a nonzero fixed vector.

Given a positive-definite matrix $M$, the preconditioned proximal point map isIt is firmly nonexpansive in the weighted inner product $⟨v,w{⟩}_M=⟨v,Mw⟩$ whenever $F$ is monotone.

For $-u^{\prime \prime }+u=f$ with homogeneous Dirichlet boundary conditions, the conforming piecewise-linear hat functions on a uniform grid give a diffusion stiffness matrix $S$ and a mass matrix $M$. Their nonzero entries are $S_{ii}=2/h$, $S_{i,i+1}=-1/h$, $M_{ii}=2h/3$, and $M_{i,i+1}=h/6$, with symmetric counterparts. The sum is a positive-definite matrix because its coefficient quadratic form is $\int ((u_h^{\prime }{)}^2+u_h^2)$. For forcing $f(x)=x$, the load at $x_i$ is $h{x}_i$.

The stiffness matrix represents a variational bilinear form $a$ in a finite element basis. A symmetric coercive bilinear form gives a positive-definite matrix. Local support of the piecewise-linear hat functions produces a sparse matrix, allowing efficient numerical solution.