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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 Poincaré 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.

For an unconstrained differentiable function of two real variables, an interior local minimum must solve the simultaneous stationary point equationsApply Newton's method in optimization to this gradient equation. The Jacobian matrix of the gradient is the Hessian matrixAt $p_n=(x_n,y_n{)}^T$, solve the linear system $H(p_n)s_n=-\nabla f(p_n)$, then set $p_{n+1}=p_n+s_n$. When the Hessian matrix is invertible, this isThis follows by setting the first order Taylor expansion $\nabla f(p_n+s_n)\simeq \nabla f(p_n)+H(p_n)s_n$ equal to zero. Solving the linear system avoids explicitly computing a matrix inverse. A stationary point need not be a local minimum: for example, a positive-definite matrix $H(p)$ suffices to classify a stationary point as a strict local minimum, whereas an indefinite Hessian matrix identifies a saddle point of a scalar function.

The gradient and constant Hessian matrix areAt ${\mathbf{x}}_0=(1,0{)}^T$, $\nabla F=(10,-8{)}^T$ and $H_F^{-1}\nabla F=(1,-1{)}^T$. Hence one multivariable Newton iteration givesAt this point the gradient vanishes, so every subsequent iterate is $(0,1)$. The one-step result follows because the gradient of a quadratic polynomial is affine and its linearization is exact. Since the leading principal minors of $H_F$ are $6>0$ and $8>0$, it is a positive-definite matrix. Alternatively,directly proves that $(0,1)$ is the unique global minimum, with value $-1$.

The partial derivatives are $g_x=\sin x$ and $g_y=y$, so all stationary points are $(n\pi ,0)$ for integers $n$. Their Hessian matrix is $\mathrm{diag}(\cos x,1)$. At even multiples of $\pi$ it is a positive-definite matrix, whereas at odd multiples it has one positive and one negative eigenvalue. ConsequentlyThere are no local maxima. Since both $1-\cos x$ and $y^2/2$ are nonnegative, these are also global minima. In the requested open plotting window, the stationary points are $(0,0)$ and $(\pm \pi ,0)$; the local minima at $(\pm 2\pi ,0)$ lie on excluded edges.

For a level set $g=c$, solving for the vertical coordinate givesFor $0<c<2$ these level sets are closed ovals around each local minimum. At $c=0$ the level sets shrink to the local minima. The level set through the saddle points iswhere the two signed branches together describe the same set as $y=\pm 2|\cos (x/2)|$. They cross at $(\pm \pi ,0)$ and reach $(0,\pm 2)$. The crossing slopes at a saddle point are $\pm 1$, as also follows from the local quadratic form $(-\delta x^2+\delta y^2)/2$. For $c>2$, the upper and lower level sets remain separate and extend across the plotting window without meeting the horizontal axis.

The first partial derivatives factor asIf either coordinate vanishes, the other equation forces the other coordinate to vanish as well. If both are nonzero, subtracting the two bracket equations gives $x^2=y^2$, and then $x^2=y^2=1$. Thus the complete list of stationary points is $(0,0)$ and the four points $(\pm 1,\pm 1)$.

The Hessian matrix isAt the origin it is $8I$, which is a positive-definite matrix. At the four other points its eigenvalues are $32$ and $-16$, independently of the signs. The second-derivative test therefore givesThere are no local maxima. The minimum at the origin is not global, since $g(t,t)=8t^2-4t^4\to -\infty$ as $|t|\to \infty$.

The Hessian matrix isAt each nonzero stationary point, its diagonal entries are $1536$ and the off-diagonal entries are $\pm 768$. Its eigenvalues are $768$ and $2304$, both positive. Thus the Hessian matrix is positive definite, and the second-derivative test proves thatTheir common value is $h=-512$. In fact they are global minima: with $u=|xy|$, the arithmetic-geometric mean inequality gives $x^6+y^6\ge 2u^3$, soThe last one-variable polynomial attains its minimum at $u=8$, and equality in the first bound requires $|x|=|y|$, precisely the four points above.

Near a nonzero stationary point $\mathbf{p}$, Taylor expansion of the gradient gives $\nabla h(\mathbf{p}+\mathbf{v})=H(\mathbf{p})\mathbf{v}+O(|\mathbf{v}{|}^2)$. Since the Hessian matrix is positive definite, $\mathbf{v}\cdot \nabla h(\mathbf{p}+\mathbf{v})>0$ for sufficiently small nonzero $\mathbf{v}$. Thus the gradient arrows point away from each of the four local minima, with different strengths in the two eigenvector directions.

At the origin, on the $x$-axis the gradient is $(6x^5,0)$ and on the $y$-axis it is $(0,6y^5)$, so arrows point outward. On $y=x$ the gradient is $6x^3(x^2-8)(1,1)$, and on $y=-x$ it is $6x^3(x^2-8)(1,-1)$, so arrows point toward the origin for small $|x|$. Away from the axes the leading terms are $(-48x{y}^2,-48x^{2}y)$, pointing inward in both coordinate signs. The narrow outward sectors close to the axes are visible in the exact vector field.

Set $L_i={\hat{L}}_i+y_i$. The equilibrium equations cancel the constant gravitational and elastic terms, leaving the small-oscillation mass and stiffness matricesThe off-diagonal mass terms arise because rope-length coordinates are not absolute climber displacements. For any nonzero velocity vector, ${\dot{y}}^{T}M\dot{y}=m_1{\dot{y}}_1^2+m_2({\dot{y}}_1+{\dot{y}}_2{)}^2+m_3({\dot{y}}_1+{\dot{y}}_3{)}^2>0$, so $M$ is a positive-definite matrix.

Within the vertical Hookean model, the displacement equations are exactly linear while all ropes stay taut. Small amplitudes ensure that none reaches the slack regime $L_i\le L_0$.

Complete the square in the Hermitian matrix polynomial:For every nonzero vector $v$,Thus $H$ is a positive-definite matrix, so $\boxed{0\text{ is not an eigenvalue of }H}$. In finite-dimensional spectral theorem terms, each eigenvalue isIf ${\alpha }^2\ge 4\beta$, the scalar polynomial has real roots of a polynomialAn allowed nonzero root can be an eigenvalue of $A$, a Hermitian matrix which is also an invertible matrix, in which case $H$ has a zero eigenvalue. For example, $\alpha =2$, $\beta =1$, $A=-I$ gives $H=0$. The weaker inequality does not force failure for every $A$; it removes the general positive lower bound. There is also the harmless corner $\alpha =\beta =0$, for which $H=A^2$ stays positive definite because $A$ is invertible, even though the strict discriminant hypothesis is absent.

On $X^{+}$, $z^2>x^2+y^2$, so $z\ne 0$ and therefore $a=-2z\ne 0$. Yet $a(I)=-2$ and $a(-I)=2$. The same continuous image of a connected space argument shows that $X^{+}$ is not connected.

Put $r=\sqrt{x^2+y^2}$. The required three disjoint nonempty open sets areTheir union is precisely $X$, the complement of the double cone $z^2=x^2+y^2$. The eigenvalues are $z\pm r$, so $X_1$ consists of positive-definite matrices above the upper sheet, $X_2$ of negative-definite matrices below the lower sheet, and $X_3$ of indefinite matrices outside the double cone. Examples are $I,-I,\mathrm{diag}(1,-1)$ respectively.

The two definite cones are convex sets. The indefinite region is path-connected: shrink $z$ continuously to zero while keeping $r>0$, then change the nonzero vector $(x,y)$ within the punctured plane. Thus these sets are also exactly the three connected components. This is the determinant cone for real symmetric two-by-two matrices.