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The equation of a quadric surface after orthogonal diagonalization of its symmetric matrix and appropriate translation. The spectral theorem for real symmetric matrices removes mixed quadratic terms, and completing squares removes linear terms in directions with nonzero eigenvalues. Linear terms in zero-eigenvalue directions can remain and generate elliptic paraboloids or hyperbolic paraboloids. The signs, zero coefficients and constant distinguish ordinary surfaces, degenerate conics in plane sections, and empty real loci.

For a second-degree equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$, the number ${\Delta }_2=B^2-4AC$ is the discriminant of a binary quadratic form associated with its quadratic part. For a nonempty, nondegenerate real conic section, a negative value gives an ellipse, zero a parabola, and a positive value a hyperbola. It is invariant under a rotation of orthonormal coordinates.^[ref]

A short explanation uses the symmetric matrixAn orthogonal matrix $R$ changes this matrix to $R^{\mathsf{T}}HR$, whose determinant equals $\det H$. The spectral theorem for real symmetric matrices then removes the mixed term. Equal-sign, opposite-sign and one-zero eigenvalues give the three types after the remaining linear terms and offsets are considered.

The qualification matters. The equations $x^2+y^2=0$, $y^2=0$ and $xy=0$ describe a point, a repeated straight line, and two crossing straight lines, respectively. Their quadratic discriminants have the signs expected of ellipse, parabola and hyperbola types, but all three are degenerate conics. Also, $x^2+y^2=-1$ has an empty set of real solutions despite its negative quadratic discriminant.

The continuous-time diffusion model on a finite undirected graph uses the Graph Laplacian in the ordinary differential equationAcross each edge, the higher value sends an amount proportional to the difference to the lower value. The two contributions cancel in the sum over vertices, giving conservation of mass:If the entries represent masses, start with nonnegative entries; signed entries can instead describe deviations from a common level. For the three-vertex path graph, the initial vector $x(0)=(1,0,-1{)}^T$ is an eigenvector with eigenvalue one, so the exact solution isMore generally, an eigenvector with eigenvalue $\lambda$ decays by $e^{-\lambda t}$. The spectral theorem for real symmetric matrices supplies an orthonormal basis of these eigenvectors, and the kernel of a graph Laplacian identifies the surviving terms. On a nonempty connected graph,The limiting value is the original arithmetic mean. On a disconnected graph, each connected component of a graph converges to its own original arithmetic mean. The same matrix therefore measures neighbour disagreement, detects disconnected regions and governs relaxation toward agreement.

For a real positive-definite matrix $A$ in finite dimension, write $A=Q\Lambda Q^T$ by the spectral theorem for real symmetric matrices. In the coordinates in an orthonormal basis $w=Q^{T}x$, the positive level set $q(x)=k$ becomesThis is an ellipsoid, with semi-major and semi-minor axes in two dimensions forming an ellipse; each semi-axis has length $r_j=\sqrt{k/{\lambda }_j}$ and direction given by the corresponding eigenvector. In the opening example, the shorter semi-axis is $\sqrt{k}/2$ along $e_{+}$, while the longer is $\sqrt{k/2}$ along $e_{-}$. A stiffer direction, meaning a larger eigenvalue, reaches the same energy at a smaller displacement. The ratio of longest to shortest semi-axis is $\sqrt{{\lambda }_{\max }/{\lambda }_{\min }}$, the square root of the spectral condition number of a positive-definite matrix.^[ref]

Strict positivity matters because ${\lambda }_{\min }>0$ givesHence a fixed positive level set cannot extend arbitrarily far. The Cholesky decomposition $A=L{L}^T$ gives another view: $q(x)=\Vert L^{T}x{\Vert }^2$, so the ellipsoid is the inverse image of a sphere under an invertible linear map. A singular positive semidefinite matrix has a nonzero kernel; adding a kernel vector preserves its quadratic form, so any nonempty positive level set is unbounded. These statements concern finite dimension; the positive definite symmetric operator entry explains why strict positivity alone need not give a uniform lower bound on an infinite-dimensional Hilbert space.

For an integer $0\le k\le n$, the fantope is the convex hull of rank-$k$ orthogonal projection matrices in ${\mathbb{R}}^n$. Equivalently, it consists of real symmetric matrices with eigenvalues in $[0,1]$ and matrix trace $k$. To prove the equivalence, apply the spectral theorem for real symmetric matrices and express the eigenvalue vector as a convex combination of the zero-one extreme points of the capped simplex. The sum of the largest eigenvalues is its support function.

Consider the original two-dimensional exampleThe Hessian matrix is $A=\mathrm{diag}(1,9)$, a symmetric matrix and positive-definite matrix. Its eigenvectors are the coordinate directions, its eigenvalues are $1$ and $9$, and its Euclidean norm condition number is $9$. The unique global minimum is $x_{\ast }=0$, so the error $e_k=x_k-x_{\ast }$ equals $x_k$. The linear map $I-\eta A$, with $I$ the identity matrix, givesFor $\eta =1/5=0.2$, the two coordinate error multipliers are $0.8$ and $-0.8$. Consequently $x_1=(1.6,-0.8{)}^T$, $x_2=(1.28,0.64{)}^T$, and the second coordinate changes sign at every step while both shrink. The Euclidean norm of the error contracts by $0.8$ each step. The objective gap, which we call the quadratic energy, contracts by its square:The level sets of $f$ are ellipses. Oscillation across their narrow direction is compatible with strictly decreasing energy.

For $\eta =1/4=0.25$, the coordinate error multipliers become $0.75$ and $-1.25$, giving $x_1=(1.5,-1.25{)}^T$. The first coordinate still tends to zero, but the second alternates and grows. Direct substitution givesThus a larger positive step size can turn a strictly convex function into a divergent iteration. Energy and error are different quantities: coordinate errors carry signed multipliers, while each coordinate's contribution to energy carries the squared multiplier.

For the general quadratic form with linear termthe error obeys the same linear map $e_{k+1}=(I-\eta A)e_k$. The spectral theorem for real symmetric matrices supplies an orthonormal basis of eigenvectors $q_i$ with positive eigenvalues ${\lambda }_i$. Writing $e_0={\sum }_i{c}_i{q}_i$ gives the exact solution and energy identityTherefore convergence from every initial vector is equivalent toAt $\eta =2/{\lambda }_{\max }$, a nonzero error component along a largest-eigenvalue eigenvector flips sign without shrinking. Above that threshold it grows. A particular start can still converge if it has zero components in every unstable direction; the interval is the condition for all starts. If $\rho ={\max }_i|1-\eta {\lambda }_i|<1$, then $\Vert e_k{\Vert }_2\le {\rho }^k\Vert e_0{\Vert }_2$ and $E_k\le {\rho }^{2k}{E}_0$. These bounds follow by summing the squared components in the orthonormal basis.

For a real matrix $A$, let $B=A^{\mathsf{T}}A$. It is positive semidefinite, and the spectral theorem for real symmetric matrices gives $x^{\mathsf{T}}Bx\ge {\lambda }_{\min }\Vert x{\Vert }^2$. If ${\lambda }_{\min }>0$, rescaling any longer vector to unit length decreases its image norm, so minimizers lie on the unit sphere. Lagrange multipliers applied to $x^{\mathsf{T}}Bx$ with $\Vert x\Vert =1$ give $Bx=\lambda x$. Thus the minimizers are exactly the unit vectors in the least eigenspace.

If ${\lambda }_{\min }=0$, the minimum is zero and the full argmin consists of all $x\in \ker A$ with $\Vert x\Vert \ge 1$. Indeed, $\ker (A^{\mathsf{T}}A)=\ker A$ because $x^{\mathsf{T}}{A}^{\mathsf{T}}Ax=\Vert Ax{\Vert }^2$. Restricting the minimizers to the sphere would omit longer kernel vectors in this case. For a ball constraint $\Vert x\Vert \le 1$, the argmin is always $\ker A$ intersected with the ball.

For a positive-definite symmetric matrix $A$, the ellipsoid $x^{T}Ax=1$ has nearest points to the origin exactly at $x={\lambda }_{\max }^{-1/2}u$, where $u$ ranges over unit vectors in the largest-eigenvalue eigenspace. Indeed, the spectral theorem for real symmetric matrices gives $x^{T}Ax\le {\lambda }_{\max }|x{|}^2$, with equality precisely in that eigenspace. A simple largest eigenvalue gives two opposite nearest points; multiplicity $m>1$ gives a scaled $(m-1)$-sphere of nearest points.

The restriction of a positive definite form to each $V_m$ is positive definite. By the spectral theorem for real symmetric matrices, choose an orthogonal matrix $P$ such that $P^T{A}_{m}P=\mathrm{diag}(d_1,\dots ,d_m)$.

Each $d_i=\phi (f_i,f_i)>0$, since a basis vector is nonzero. Taking determinants in part (i) givesbecause an orthogonal matrix has determinant $\pm 1$. ConsequentlyThese determinants are the leading principal minors in Sylvester's criterion.

Work for an arbitrary realization of the sequence, and set $u_t=X_t^T{V}_{t-1}^{-1}{X}_t$. In the Loewner order,By the spectral theorem for real symmetric matrices, every eigenvalue of $V_{t-1}$ is at least $\lambda$. The matrix inverse therefore has eigenvalues at most ${\lambda }^{-1}$, andFor $0\le u\le 1$, put $g(u)=2\log (1+u)-u$. Since $g(0)=0$ and $g^{\prime }(u)=(1-u)/(1+u)\ge 0$, we have $u\le 2\log (1+u)$. The matrix determinant lemma now givesTaking logarithms and using a telescoping sum,This elliptical potential bound holds for every realization and every $n$, without a probabilistic exceptional event. The condition $\lambda \ge 1$ is what ensures $u_t\le 1$.

The unbroken group is the stabilizer subgroup of the vacuum expectation value. For a nonzero vector vacuum expectation value, an orthogonal transformation puts $⟨\varphi ⟩=v{e}_N$. A special orthogonal transformation fixing it has the block form $\mathrm{diag}(A,1)$ with $A\in \mathrm{SO}(N-1)$. ConsequentlyThe zero vacuum expectation value leaves the entire special orthogonal group unbroken.

For a real symmetric tensor write its vacuum expectation value as a real symmetric matrix $Q$. The transformation is $Q\mapsto OQ{O}^T$. The spectral theorem for real symmetric matrices gives an orthonormal basis in whichBecause $OQ{O}^T=Q$ is equivalent to $OQ=QO$, an unbroken transformation preserves each eigenspace. The complete unbroken group is thereforeIts identity component is $H^0={\prod }_i\mathrm{SO}(m_i)$. This symmetric tensor vacuum stabilizer shows why the representation alone does not determine a unique breaking pattern:

For example a rank-one tensor $Q=v{e}_N{e}_N^T$ preserves $O(N-1)$ embedded by $A\mapsto \mathrm{diag}(A,\det A)$. Its continuous unbroken group is $\mathrm{SO}(N-1)$, but it also allows reversal of the distinguished axis. Restricting to a traceless symmetric tensor imposes ${\sum }_i{m}_i{q}_i=0$ and otherwise leaves the same classification.

The command centres the log measurements and performs principal component analysis of their covariance matrix. With cor=F, it does not first standardize the variables to variance one. For a unit principal component loading vector $q$, the score $q^T(z_i-\bar{z})$ has sample variance $q^T\hat{\Sigma }q$. By the spectral theorem for real symmetric matrices, expand $q$ in an orthonormal eigenbasis; that variance is a weighted average of the eigenvalues. Its maximum is the largest eigenvalue, reached at the corresponding eigenvector. Repeating subject to orthogonality to previous directions gives the other principal components.

The software uses the divisor-$31$ sample covariance matrix for its component standard deviations, whereas var used divisor $30$. Diagonalizing the displayed matrix gives approximatelyTaking square roots of the latter gives $(0.5671603,0.07312405,0.03064834)$, agreeing with the output to the rounding of the displayed sample covariance matrix. The normalization changes these numerical standard deviations but leaves the eigenvectors and proportions of explained variance of a principal component unchanged.

The first principal component explains about $98.08\%$ of the total log-variable variance, the second another $1.63\%$, and the last about $0.286\%$. Keeping two preserves about $99.714\%$. An original scree plot displays the sharp drop after the first component:

Consequently the centered cloud is very elongated along one direction: a one-dimensional approximation represents almost all of its squared variation. From the displayed covariance matrix, a unit first eigenvector, with its arbitrary sign chosen positive, is approximatelyIts principal component score is therefore $0.3980(\log G_i-\overline{\log G})+0.09514(\log H_i-\overline{\log H})+0.9124(\log V_i-\overline{\log V})$. All coefficients have the same sign, so this direction describes overall tree size, with log volume dominating the unstandardized variation. The summary alone supplies proportions, not the loading directions; the displayed sample covariance matrix supplies the latter. This is a geometric approximation, not evidence of an exact deterministic relation or independent original variables.

One further use is principal component analysis on a correlation matrix, obtained with cor=T, which equalizes marginal variances and answers a different scaling-sensitive question. Another is principal component regression: fit a response to selected principal component scores rather than highly correlated original predictors, choosing the number using prediction assessment. The fitted transformation also projects new measurements into the same low-dimensional coordinates, provided the training centering and scaling are retained.

For a real matrix $A$ with positive determinant, the polar decomposition of an invertible real matrix gives unique factorizationswhere $U$ and $V$ are the right stretch tensor and left stretch tensor, respectively. Positivity here means positive definiteness, not merely nonnegative entries.

To prove existence, $C=A^{T}A$ is symmetric and $v^{T}Cv=|Av{|}^2>0$ for every nonzero $v$, because $A$ is invertible. The spectral theorem for real symmetric matrices supplies an orthogonal matrix $Q$ and positive eigenvalues $c_j$ such that $C=Q\mathrm{diag}(c_1,c_2,c_3)Q^T$. Define $U$ using the principal square root of a positive semidefinite matrix construction:Then $R^{T}R=U^{-1}{A}^{T}A{U}^{-1}=I$, so $R$ is orthogonal. Also $\det R=\det A/\det U>0$; an orthogonal matrix has determinant plus or minus one, hence $\det R=1$. Set $V=RU{R}^T$. It is symmetric positive definite, $V^2=A{A}^T$, and $VR=RU=A$.

For uniqueness, any second factorization $A=R_1{U}_1$ with the same properties gives $U_1^2=A^{T}A=C$. Since $U_1$ commutes with its square, it preserves every eigenspace of $C$. On a $c$-eigenspace its own positive eigenvalues must square to $c$, so it equals $\sqrt{c}$ times the identity there. Thus $U_1=U$, then $R_1=A{U}^{-1}=R$, and finally $V$ is fixed too. This proves both uniqueness and the polar decomposition in continuum mechanics, with the same proper rotation in its left and right forms.

Use the curvature sign convention implicit in part (c): the Jacobi field equation is $D_t^{2}J+K_{\dot{\gamma }}J=0$, so the operator in a positively curved normal direction has positive eigenvalue. With the alternative convention $R_{\mathrm{std}}(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$ used for the catalog's Jacobi curvature operator, the paper's tensor is $R=-R_{\mathrm{std}}$, and its operator is $K_v(x)=R_{\mathrm{std}}(x,v)v$. This fixes the sign throughout Q1.

Let $r(A,B,C,D)=⟨R(A,B)C,D⟩$. Pair interchange for the Riemann curvature tensor givesThus $K_v$ is self-adjoint with respect to the Riemannian metric. The curvature sign does not affect this conclusion. The spectral theorem for real symmetric matrices therefore supplies the real orthonormal eigenbasis used in part (b). Also $K_{v}v=0$, so the tangent direction is a zero-eigenvalue direction.

Put $r=k/h^2$. The update isUnder $k\le h^2/(2\beta )$, all three weights are nonnegative and sum to one. With the imposed zero Dirichlet boundary conditions, this is a convex combination including any boundary values. Thus the monotone half-grid diffusion update givesDifferences of two computed solutions obey the same estimate, proving numerical stability in the discrete maximum norm, uniformly in $h,k,n$.

There is also an L2 norm proof. The diffusion matrix $L_h$ is symmetric and satisfiesSince $(v_{m+1}-v_m{)}^2\le 2(v_{m+1}^2+v_m^2)$, every eigenvalue lies in $[-4\beta /h^2,0]$. Hence the eigenvalues of $I+k{L}_h$ lie in $[-1,1]$ under the same step restriction, and the spectral theorem for real symmetric matrices gives an L2 norm contraction. These stability estimates are valid for the printed recurrence, although they cannot repair its inconsistency with the printed advection equation.

At a fixed material point, freeze the incremental elastic moduli at their local values and fix a unit propagation direction $\nu$. For an arbitrary nonzero polarization $m$, the rank-one matrix $a_{i\alpha }=m_i{\nu }_{\alpha }$ annihilates the added quadratic minor null Lagrangian, sinceThe assumed positivity therefore givesThus the acoustic tensor is positive-definite; this is strong ellipticity in elasticity. Major symmetry also makes it a real symmetric matrix. By the spectral theorem for real symmetric matrices, it has three real positive eigenvalues $q_1,q_2,q_3$ and an orthonormal basis of real polarizations $m^{(1)},m^{(2)},m^{(3)}$. ThereforeEach polarization gives a real plane wave profile through the equation of part (a), in either sign of the propagation speed. Repeated eigenvalues still admit three independent polarizations; the claim is not that the three speeds must be distinct. For an inhomogeneous base deformation these are local, frozen-coefficient wave modes, rather than exact plane waves throughout the entire body.

By the spectral theorem for real symmetric matrices, choose an orthonormal eigenbasis $u_1,u_2,u_3$ with $A{u}_i={\lambda }_i{u}_i$. If $x={\sum }_i{y}_i{u}_i$, thenThus every point of the ellipsoid has distance at least ${\lambda }_1^{-1/2}$ from the origin. Equality holds precisely whenBoth coefficients are strictly positive, so $y_2=y_3=0$. The surface equation then gives $y_1=\pm {\lambda }_1^{-1/2}$. ThereforeExactly two points attain the minimum. The strict ordering makes the largest-eigenvalue eigenspace one-dimensional; this is the equality case in nearest points to the origin on a positive-definite ellipsoid.

For the polar decomposition of an invertible real matrix, start with $C=F^{T}F$. For nonzero $y$, $y^{T}Cy=|Fy{|}^2>0$, since $\det F>0$ makes $F$ invertible. The spectral theorem for real symmetric matrices gives an orthonormal basis $n_a$ with $C{n}_a={\lambda }_a^2{n}_a$ and ${\lambda }_a>0$. DefineThen $U$ is a positive-definite symmetric matrix, $U^2=C$, andbecause $\det U=\sqrt{\det (F^{T}F)}=\det F$. Thus $R$ is an orthogonal matrix with positive determinant. Set $V=RU{R}^T$; it too is a positive-definite symmetric matrix, and $VR=RU=F$. Moreover $V^2=F{F}^T$. This provesThe positive square root is unique: any positive symmetric square root of $C$ commutes with $C$, preserves its eigenspaces and has the positive scalar square root on each of them. Hence $U$, then $R$ and $V$, are unique. These are the right stretch tensor, rotation and left stretch tensor of the polar decomposition in continuum mechanics.

A standard spectral strain measure isTake $f\in C^1(0,\infty )$, $f(1)=0$ and $f^{\prime }(1)=1$, with $f$ strictly increasing; the customary stronger condition $f^{\prime }(\lambda )>0$ gives a regular one-to-one parameterization of all positive principal stretches. The zero condition makes a rigid rotation strain free, monotonicity distinguishes extension from contraction and prevents nonunit stretch from being strain free, and the derivative normalization gives the ordinary infinitesimal strain. Indeed, for $F=I+H$,Under a superposed spatial rotation $F\mapsto QF$, $F^{T}F$ and hence $E^f$ remain unchanged, giving material frame indifference. Choosing $f(\lambda )=\lambda -1$ satisfies all these conditions and gives the Biot strain tensor

The meaning of work-conjugate stress and strain here is that a symmetric material tensor $T^f$ satisfiesfor every admissible rate, with both powers measured per reference volume. To keep the nominal-stress index convention explicit, write $M_{iI}=P_{Ii}$, so $M=P^T$ and the power is $M:\dot{F}$. Differentiating $F=RU$ gives $\dot{F}=R(\Omega U+\dot{U})$, where $\Omega =R^T\dot{R}$ is skew. Conservation of angular momentum gives symmetry of $M{F}^T$, or equivalently of $R^{T}MU$. ThereforeThe first term vanishes because a symmetric tensor has zero contraction with a skew tensor. Since ${\dot{E}}^{(1)}=\dot{U}$, the symmetric Biot stress isIf the nominal stress is instead stored spatial-index first as $M$, the same answer is $\mathrm{sym}(R^{T}M)$. The symmetrization is essential: $PR$ need not itself be symmetric.

Principal component analysis replaces the original coordinates by orthogonal linear combinations, ordered so that the first few retain as much variance as possible. It gives a lower-dimensional summary for visualization or compression and reveals the main directions of variation. A direction of large variance need not be the direction most useful for classification.

Write $\mu =\mathbb{E}X$. For a unit vector $u$, the centered score $Y=u^T(X-\mu )$ has varianceMaximizing this Rayleigh quotient with a Lagrange multiplier givesThus a maximizing direction must be an eigenvector of the covariance matrix. To identify the maximum, use the spectral theorem for real symmetric matrices to choose an orthonormal basis of eigenvectors $v_1,\dots ,v_p$, with eigenvalues ${\lambda }_1\ge \cdots \ge {\lambda }_p\ge 0$. Writing $u={\sum }_j{c}_j{v}_j$ givesEquality is attained at $u=v_1$. Consequently the first population principal component isFor each subsequent principal component, maximize the same variance over unit coefficient vectors orthogonal to all the previously chosen ones. Restricting the expansion above to the remaining eigenvectors gives $v_2$, then $v_3$, and so on. ThusThe principal components are uncorrelated; they need not be independent unless additional distributional assumptions, such as joint Gaussianity, hold. Their total variance is $\mathrm{tr}\Sigma ={\sum }_j{\lambda }_j$. Signs of eigenvectors are arbitrary, and a repeated eigenvalue permits any orthonormal basis in its eigenspace. In sample principal component calculations, replace $\mu$ by the sample mean and $\Sigma$ by the sample covariance matrix.

Two real symmetric bilinear forms are congruent if an invertible linear map takes one to the other; in matrices, $B=S^{\mathsf{T}}AS$ for some invertible $S$. The spectral theorem for real symmetric matrices gives an orthogonal eigenbasis. Rescaling each basis vector with nonzero eigenvalue then reduces any such form toSylvester's law of inertia asserts that $(p,q,r)$ is independent of the congruence reduction. To prove it, $p$ is the greatest dimension of a subspace on which the form is positive definite: the positive coordinate subspace attains it, whereas any subspace of dimension greater than $p$ meets the nonpositive coordinate subspace of dimension $n-p$ and therefore cannot be positive definite. Applying the same argument to the negative of the form determines $q$. Then $r=n-p-q$ is determined too.

The rank of a quadratic form is $p+q$, and its signature of a quadratic form is $p-q$ under the signed-difference convention. They determine $p,q,r$ and hence the congruence class. Counting all pairs $p,q\ge 0$ with $p+q\le n$ givesFor $n\ge 1$, exactly two classes have every level set bounded: the positive-definite and negative-definite classes. Their nonempty nonzero levels are ellipsoids, and their zero level is the origin. A degenerate form has an entire null line in its zero level. An indefinite nondegenerate form also has a nonzero isotropic vector, for example the sum of one positive and one negative normalized basis vector, whose scalar multiples make the zero level unbounded.

For the displayed three-dimensional form,Its leading principal minors are $3,14,32$, all positive. Sylvester's criterion makes it positive definite, so the surface is a bounded ellipsoid. Directly, its quadratic expression iswhich also proves positivity and boundedness of the unit level.