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For each fixed $x$ in the nonnegative orthant, the map $A\mapsto x^{T}Ax$ is a continuous linear function on the vector space of real symmetric matrices. Thusis an intersection of closed half-spaces, using the Frobenius inner product $⟨A,B{⟩}_F=\mathrm{tr}(AB)$ on this space. Arbitrary intersections of closed sets are closed, so $K$ is closed.

If $A,B\in K$ and $s,t\ge 0$, then $x^T(sA+tB)x=s{x}^{T}Ax+t{x}^{T}Bx\ge 0$ for every $x\ge 0$. Therefore $sA+tB\in K$, including the zero matrix when $s=t=0$. HenceThe copositive cone is defined in the real space of symmetric matrices; symmetry is needed later to recover every matrix entry from these quadratic forms.

A pointed cone has no nonzero line through the origin: equivalently $K\cap (-K)=\{0\}$. If $A\in K\cap (-K)$, its quadratic form vanishes on the nonnegative orthant. Testing the coordinate vectors gives $A_{ii}=0$. Testing $e_i+e_j$ then givesSince $A$ is a symmetric matrix, every entry is zero. Thus $\boxed{K\cap (-K)=\{0\}}$.

The identity matrix gives an interior point. For a symmetric perturbation $E$ with operator norm $\Vert E{\Vert }_2<1$, the Cauchy-Schwarz inequality givesThis open norm ball around $I$ lies even in the positive semidefinite cone, hence in $K$. Consequently $\boxed{I\in \mathrm{int}K}$ and the interior is nonempty.

More generally, the interior of the copositive cone consists exactly of strictly copositive matrices. Positivity on the compact nonnegative unit sphere has a positive minimum and persists under small perturbations; a zero there is destroyed by an arbitrarily small negative multiple of $I$.

For a convex cone $D$ in an inner product space, use the nonnegative-pairing conventionHere the pairing is the Frobenius inner product on real symmetric matrices. Let $C_0$ be the conic hull of the nonnegative rank-one matrices $x{x}^T$, and let $C=\overline{C_0}$. For $A\in K$, $⟨A,x{x}^T{⟩}_F=x^{T}Ax\ge 0$. This extends to conic combinations, and by continuity to their limits. Hence $C\subseteq K^{\ast }$.

For the converse, if $B\notin C$, separation from a closed convex cone supplies a symmetric $H$ withIn particular $x^{T}Hx\ge 0$ for every $x\ge 0$, so $H\in K$. The negative pairing then excludes $B$ from $K^{\ast }$. ThereforeThe same generator test gives $C^{\ast }=K$. This is the duality of copositive and completely positive cones; the next argument removes the closure.

The real vector space of symmetric matrices has dimension $N=n(n+1)/2$. By the conic Carathéodory theorem, every member of $C_0$ is a conic combination of at most $N$ generators. Absorb each nonnegative coefficient into its vector through $\lambda x{x}^T=(\sqrt{\lambda }x)(\sqrt{\lambda }x{)}^T$.

If $B_{\ell }\in C_0$ converges to $B$, write, padding with zero vectors if necessary,The matrix trace satisfiesThe left side is bounded because $B_{\ell }$ converges. Thus the finite tuple $(x_{\ell 1},\dots ,x_{\ell N})$ is bounded. The Bolzano-Weierstrass theorem gives a subsequence on which every vector converges, say $x_{\ell j}\to x_j\ge 0$. Continuity of the outer product now gives $B={\sum }_{j=1}^N{x}_j{x}_j^T\in C_0$. HenceThis proves closedness of the completely positive cone. The uniform bound on the number of factors and the matrix trace bound are both essential: an arbitrary conic hull of a closed generating set need not be closed.

Suppose $A\mathbf{v}=\lambda \mathbf{v}$ for a nonzero complex vector. Because a real symmetric matrix is Hermitian, ${\mathbf{v}}^{\ast }A\mathbf{v}$ is real. But ${\mathbf{v}}^{\ast }A\mathbf{v}=\lambda {\mathbf{v}}^{\ast }\mathbf{v}$, so $\lambda$ is real. The real matrix $A-\lambda I$ is singular; the real or imaginary part of a nonzero complex null vector supplies a nonzero real eigenvector. Thus every eigenvalue is real and has a real eigenvector.

Let $q$ denote the quotient map and write $[X]$ for a point of Real projective space. For $i=1,\dots ,n+1$, take the open setIt is open in the quotient topology because its inverse image under $q$ is the open set $\{X:X_i\ne 0\}$. These sets cover, since a nonzero vector has at least one nonzero coordinate. Define the manifold chartThe hat means that the $i$th coordinate is omitted. Ratios are unchanged by multiplying $X$ by any nonzero real number, so the chart is well defined. Its inverse inserts $1$ into position $i$ and takes the resulting equivalence class. The ratios and this inverse are continuous, giving a homeomorphism $U_i\to {\mathbb{R}}^n$.

For a compact description of the smooth transition maps, write $a_i=1$ and let $a_k=X_k/X_i$ for $k\ne i$. On $U_i\cap U_j$ one has $a_j\ne 0$, andEach coordinate is a rational smooth function on the open domain $a_j\ne 0$; reversing $i,j$ gives a smooth inverse. Thus these $n+1$ charts form the standard affine atlas of real projective space.

For completeness, the underlying space is Hausdorff: normalizing $X$ identifies it with the antipodal quotient of the unit sphere, and the continuous injective map $[X]\mapsto X{X}^T/(X^{T}X)$ into the space of symmetric matrices separates any two distinct classes by disjoint open neighborhoods. It is also a second-countable space: pull back the countable bases of rational balls in these finitely many charts, and take their union. Therefore

Under an orthogonal change of coordinates represented by $Q$, the component matrix changes byThus $T^{\prime }$ is similar to $T$, so its characteristic polynomial, eigenvalues, and their algebraic multiplicities are unchanged.

If $T$ is symmetric, the spectral theorem for real symmetric matrices makes it diagonalizable, so algebraic and geometric multiplicities agree. Consequently the dimensions of its eigenspaces are also independent of the coordinate frame.

For the stated tensor, choose a unit eigenvector $n$ belonging to the simple eigenvalue $\mu$. Its perpendicular plane is the eigenspace of $\lambda$, soHence

Locally write the real analytic hypersurface as $\Sigma =\{F=0\}$ with $dF\ne 0$. The conormal bundle is spanned by $dF$. The surface is a characteristic hypersurface at $p$ precisely when the principal symbol vanishes on that conormal:Only the matrix $(a_{ij}+a_{ji})/2$, which is a symmetric matrix, contributes to this expression.

The continuous coefficients are uniformly continuous on a compact neighborhood of $\overline{W}$. For every $x_0\in \overline{W}$, choose a ball $B_r(x_0)⋐U$ small enough thatfor all $i,j$. Part 4(b), with the frozen symmetric matrix $A=(a^{ij}(x_0))$, then applies on this ball.

Choose a finite collection of these balls and a smooth partition of unity $({\eta }_k)$ that sums to one near $\overline{W}$, with each ${\eta }_k$ supported in its corresponding ball. Applying part 4(b) to ${\eta }_{k}u$ givesSince $(a^{ij})$ is symmetric, the Leibniz rule gives the commutator formulaThe coefficients and the finitely many derivatives of the cutoff functions are bounded, soFinally $u={\sum }_k{\eta }_{k}u$ near $\overline{W}$. Summing the finite set of local estimates provesThis is the coefficient-freezing interior second-derivative estimate.

At each fixed time, the maximum over nonzero initial conditions is the optimal energy amplification of a linear system:Write $a=e^{{\lambda }_{1}t}$, $d=e^{{\lambda }_{2}t}$ and $b=(a-d)/({\lambda }_1-{\lambda }_2)$. The symmetric matrix $A^{T}A$ has trace $\tau =a^2+b^2+d^2$ and determinant $a^2{d}^2$. Its largest eigenvalue isAn optimal initial condition is a right singular vector of $A(t)$ associated with its largest singular value.

If ${\lambda }_2>{\lambda }_1$, let $\alpha =1/({\lambda }_2-{\lambda }_1)$. Then $A(t)/e^{{\lambda }_{2}t}\to \left(\begin{array}{cc}0& 0\\ \alpha & 1\end{array}\right)$, so the precise long-time asymptotic statement isThe Cauchy-Schwarz inequality shows that the unit initial conditions achieving this leading factor areThis is an adjoint eigenvector, since $L^T(\alpha ,1{)}^T={\lambda }_2(\alpha ,1{)}^T$. It differs from the right eigenvector $v_2=(0,1{)}^T$: the initial condition maximizes the projection onto the slow mode, whereas the eventual state aligns with that right eigenvector. For ${\lambda }_2<0$, even the optimal energy ultimately decays to zero; the prefactor describes enhanced excitation of the slow mode.

For algebraic stability, the weights must be nonnegative and the real symmetric matrix $M=BA+A^{T}B-b{b}^T$, where $B=\mathrm{diag}(b)$, must be a positive semidefinite matrix.

The weights are positive, but $a_{11}=0$ and $b_1=1/6$, soA positive semidefinite matrix cannot have a negative diagonal entry, as its quadratic form at the first standard basis vector would be negative. Thus this Lobatto IIIA method is not algebraically stable, despite its A-stability.

Consider a bounded Lipschitz domain $\Omega$ and homogeneous Dirichlet boundary conditions forAssume $A$ is a bounded real symmetric matrix field with ${\xi }^{T}A(x)\xi \ge a_0|\xi {|}^2$ uniformly, $c\ge 0$ is bounded, and $f$ defines a bounded linear functional on $V=H_0^1(\Omega )$, the zero-boundary Sobolev space. The weak formulation isThe Cauchy-Schwarz inequality gives continuity $|a(u,v)|\le M\Vert u{\Vert }_V\Vert v{\Vert }_V$. Uniform ellipticity and the Poincare inequality give a coercive bilinear form, $a(v,v)\ge \alpha \Vert v{\Vert }_V^2$. The Lax-Milgram theorem therefore supplies a unique weak solution, with $\Vert u{\Vert }_V\le \Vert \ell {\Vert }_{V^{\prime }}/\alpha$. Its proof represents $a$ by a bounded linear operator $T$ using the Riesz representation theorem; coercivity gives an injective operator with closed image, and a zero orthogonal complement makes that image all of $V$. This yields existence, uniqueness and the displayed estimate.

For a conforming finite element space $V_h\subseteq V$, the Galerkin method finds $u_h\in V_h$ such that $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$. With a basis ${\varphi }_1,\dots ,{\varphi }_N$, this is the linear system $KU=F$, where the stiffness matrix is $K_{ij}=a({\varphi }_j,{\varphi }_i)$ and $F_i=\ell ({\varphi }_i)$. Its positive-definite matrix property proves unique discrete solvability. Local support gives a sparse matrix through elementwise assembly.

For the symmetric problem, the Ritz method minimizes $J(v)=a(v,v)/2-\ell (v)$ over $V_h$. Differentiating in every direction $v_h$ gives precisely the Galerkin method equations. Conversely, if $u_h$ solves them, $J(u_h+w_h)-J(u_h)=a(w_h,w_h)/2$, proving it is the unique minimum. The same argument over $V$ gives $J(v)-J(u)=\Vert v-u{\Vert }_a^2/2$, where $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ is the energy norm. Thus the Ritz method chooses the best trial function in that norm.

Subtracting the continuous and discrete equations gives Galerkin orthogonality, $a(u-u_h,v_h)=0$. For arbitrary $v_h\in V_h$ and $e=u-u_h$, it follows that $a(e,e)=a(e,u-v_h)$. Coercivity and continuity prove the Céa lemma:In the symmetric energy norm, the Cauchy-Schwarz inequality improves the constant to one. Equivalently, Galerkin orthogonality gives the Pythagorean identity $\Vert u-v_h{\Vert }_a^2=\Vert u-u_h{\Vert }_a^2+\Vert u_h-v_h{\Vert }_a^2$. These estimates establish stability and show that approximation properties of the trial space determine convergence.

For continuous piecewise-linear elements on a shape-regular mesh in dimensions at most three, a finite element interpolation estimate gives $\Vert u-I_{h}u{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ when $u\in H^2(\Omega )$. Combining it with the Céa lemma yields $O(h)$ energy norm error. More generally, degree-$p$ elements give $O(h^p)$ error in $H^1$ under the corresponding regularity and approximation assumptions. Mere mesh refinement cannot supply an order whose required solution regularity is absent.

An additional order in the L2 norm follows under a dual elliptic regularity assumption. Solve $a(v,z)=(e,v{)}_{L^2}$ and assume $\Vert z{\Vert }_{H^2}\le C\Vert e{\Vert }_{L^2}$. The Aubin–Nitsche duality argument uses Galerkin orthogonality to obtainTherefore $\Vert e{\Vert }_{L^2}\le Ch\Vert e{\Vert }_{H^1}$, giving $O(h^2)$ for the piecewise-linear case with $u\in H^2$. This improvement explicitly requires regularity of the dual problem.

As a concrete example, solve $-u^{\prime \prime }=1$ on $(0,1)$ with zero endpoint values. On a uniform mesh of spacing $h$, the piecewise-linear hat functions give $K_{jj}=2/h$, $K_{j,j\pm 1}=-1/h$ and $F_j=h$. The discrete equations are $2U_j-U_{j-1}-U_{j+1}=h^2$, whose solution is $U_j=x_j(1-x_j)/2$. The exact solution is $u(x)=x(1-x)/2$, so $u_h$ is its piecewise-linear nodal interpolant. On each interval $[x_j,x_{j+1}]$,Integrating the squared error and squared derivative error gives the explicit rateswhere the $H_0^1$ norm here is $\Vert v^{\prime }{\Vert }_{L^2}$.

For nonsymmetric coercive problems, the Galerkin method, Galerkin orthogonality and the Céa lemma still apply, but the symmetric quadratic minimization interpretation of the Ritz method generally does not. Nonconforming trial spaces, inexact integration and noncoercive equations need additional arguments beyond the conforming coercive theory proved here.

Let $A$ be the adjacency matrix of a graph. Since $G$ is a regular graph of degree $d$, every row of $A$ sums to $d$. Therefore, for $\mathbf{e}=(1,\dots ,1{)}^T$,so $d$ is a graph eigenvalue.

If $Av=dv$, then the quadratic form of the Graph Laplacian $L=dI-A$ givesEvery summand is nonnegative, so $v_i=v_j$ along every edge. The graph is connected, hence all coordinates of $v$ are equal and $v$ is a scalar multiple of $\mathbf{e}$. Thus the $d$-eigenspace is one-dimensional. Since $A$ is a symmetric matrix, the spectral theorem for real symmetric matrices makes it diagonalizable, so the algebraic multiplicity is also one. This proves the constant eigenvector of a regular graph result.

The continued-fraction matrix for the original finite continued fraction isEach $M(a_j)$ is a symmetric matrix, so reversing the order gives the transpose:Reading off the columns proves the reversal identity for a finite continued fraction:These are respectively the $n$th and $(n-1)$th convergents of the reversed continued fraction.

Using the rules for the matrix transpose together with $L^T=L$ and $B^T=-B$,Thus the commutator $[B,L]$ is a symmetric matrix.

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.

For a nonzero vector $v\in {\mathbb{R}}^n$, a Householder reflection isIt is immediately symmetric. If $P=v{v}^T/(v^{T}v)$, then $P^2=P$, soHence $H$ is also an orthogonal matrix, with $H^{-1}=H$.

The similarity transformation can be expanded asComputing $v^{T}A$, $Av$, and the scalar $v^{T}Av$ costs $O(n^2)$ arithmetic operations, after which the remaining updates are outer products and scalar multiples, also costing $O(n^2)$. Thus $HA{H}^{-1}$ can be formed in $O(n^2)$ operations rather than by two general $O(n^3)$ matrix multiplications.

The symmetric matrix of the quadratic form in the standard basis isbecause the off-diagonal entries contribute twice to $x^{T}Ax$.

Apply Gram-Schmidt orthogonalization for a symmetric bilinear form toThese vectors are pairwise orthogonal for $B(u,v)=u^{T}Av$, andThus the form is diagonal in the basis $(u_1,u_2,u_3)$, with matrixIt follows that its rank is $3$ and its signature is

By the real spectral theorem, the real symmetric matrix $A$ has an orthonormal basis of eigenvectors. In that basis the associated quadratic form isRescaling the coordinates belonging to nonzero eigenvalues changes every positive coefficient to $+1$ and every negative coefficient to $-1$. Hence the diagonal normal form has one positive square for each positive eigenvalue and one negative square for each negative eigenvalue. By Sylvester's law of inertia, these counts do not depend on the diagonalizing basis, so

The numerical eigenvalues are not invariant under a general change of basis, because the matrix changes by congruence rather than similarity. For example, on a one-dimensional space let $q(e)=1$. Its matrix in the basis $(e)$ is $(1)$, whereas in the basis $(2e)$ it is $(4)$. The eigenvalue changes from $1$ to $4$, although its sign, and therefore the signature, is unchanged.

False under the hypotheses as printed. Nonzero coefficients do not ensure the two-column dominant-subspace condition.

For a counterexample, take $n=3$, eigenvaluesand an orthonormal eigenbasis $w_1,w_2,w_3$ for whichThe two displayed coefficient vectors are orthonormal and can be completed to an orthogonal matrix, so such a real symmetric matrix $A$ exists. Every $b_i$ and $c_i$ is nonzero, butIndeed, with $t=1/\sqrt{2}$,is a nonzero multiple of $w_1$, while the remaining direction in
$\mathrm{span}(A^k{e}_1,A^k{e}_2)$ isThusBy the simultaneous iteration interpretation of the QR algorithm, the leading two columns of the accumulated $Q$ factor span this same space. The leading block $B_k$ is therefore an orthogonal-coordinate representation of the compression of $A$ to this space. Sincewe havefor every relevant $k$, rather than $\{-2,2\}$. Hence the requested Hausdorff convergence does not hold.

For completeness, the intended statement becomes true if one addsLet $Z_k$ consist of the first two columns of ${\overline{Q}}_{k-1}$. The Accumulated QR factorization identity givesAfter division by $|{\lambda }_n{|}^k$, every component along
$w_1,\dots ,w_{n-2}$ tends to zero because of the strict spectral gap, while $\Delta \ne 0$ makes the two surviving dominant components independent. Therefore these two-dimensional subspaces converge to the dominant invariant subspaceThere are two-by-two orthogonal matrices $U_k$ such thatBecauseup to the harmless index convention at $k=0$, we obtainin the matrix 2-norm. Orthogonal similarity preserves the spectrum, and the stated spectral perturbation bound now gives