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If one retains the Euler approximation $A=I+J\delta t$, its exact norm matrix isThe additional term can mix the instantaneous symmetric-part eigenvectors, so they no longer generally diagonalize finite-time error amplification. However, when $\delta t^2$ is material, a reliable physical propagator must itself be calculated beyond first-order Euler: a large Euler-step amplification can be a numerical artifact. For constant $J$, the exact map is $\Phi =e^{J\Delta t}$; for a changing reference trajectory it solves $\dot{\Phi }=J(t)\Phi$, $\Phi (0)=I$.

Use its singular value decomposition, $\Phi =U\Sigma V^T$. The initial orthonormal right singular vectors ${\mathbf{v}}_i$ diagonalize ${\Phi }^T\Phi$ and satisfyA singular value $s_i>1$ gives finite-time norm growth, $s_i<1$ decay, and the leading right singular vector is the initial perturbation with greatest amplification in the stated norm. The left singular vector describes the resulting final error pattern. Non-normal dynamics can give transient growth even if every Jacobian matrix eigenvalue has negative real part, so selecting observations from those eigenvalues alone is insufficient. Over a genuinely short interval, $s_i^2=1+{\lambda }_i\delta t+O(\delta t^2)$, recovering the symmetric-part result. Initial observations should target the relevant growing right singular projections, with prior covariance and observational precision included when they are not isotropic.

For the linear three-reservoir carbon model, with concentrations as a column vector, the linear ordinary differential equation isA homogeneous normal mode has $\mathbf{x}=\mathbf{v}{e}^{\lambda t}$, where $J\mathbf{v}=\lambda \mathbf{v}$. For negative real eigenvalues the e-folding relaxation time is $\tau =-1/\lambda$. The row sums vanish, so ${\mathbf{v}}_3=(1,1,1{)}^T$ has ${\lambda }_3=0$ andEqual effective concentrations generate no exchange. Summing the reservoir budgets givesThus injected carbon dioxide is conserved after emissions cease: the model redistributes it but contains no permanent removal. The zero normal mode records this conserved excess mass, rather than a long but finite decay.

To check the other normal modes are damped, set $D=\mathrm{diag}(R_a,R_b,R_d)$ and $L=-DJ$. ThenThe similarity transformation $D^{1/2}J{D}^{-1/2}=-D^{-1/2}L{D}^{-1/2}$ is a symmetric negative-semidefinite matrix with a one-dimensional kernel, so the remaining two eigenvalues are strictly negative.

In the three-dimensional Earth, a network identifies P- and S-wave arrivals and fits source coordinates $(x_0,y_0,z_0)$ and origin time $t_0$ using a travel-time model. Time differences constrain path lengths without requiring a common absolute clock zero; calibrated absolute times then fix the origin time. Real inversions trace refracted rays through a layered or heterogeneous velocity model and use many stations, azimuths, and phases, rather than literal circles at fixed wave speed. Residuals and the sensitivity matrix quantify uncertainty and expose poor station geometry or a poor velocity model.

Depth and origin time are often the weakest pair, especially for a shallow event observed by distant surface stations. In the homogeneous example,If the depth sensitivities are small or nearly alike across the network, a deeper source's longer travel time can be offset by an earlier origin time. In contrast, horizontal displacement often gives opposite-sign residuals on opposite sides of an event and is better constrained by good azimuthal coverage. Nearby stations and depth-sensitive reflected phases can break the earthquake depth–origin-time trade-off; the ranking is not universal for every network or event depth.

A real symmetric $3\times 3$ matrix has three real eigenvalues and a complete orthonormal basis of eigenvectors. For a stable elastic material, strong ellipticity makes ${\mathbf{a}}^{\mathsf{T}}B\mathbf{a}>0$ for every nonzero polarization, so its eigenvalues are positive squared wave speeds. There are therefore three orthogonal plane-wave polarizations in a given direction, with three potentially distinct speeds in an anisotropic medium. Degenerate eigenvalues allow arbitrary orthonormal combinations within their eigenspace; they do not remove polarization degrees of freedom. General anisotropic modes need not be exactly longitudinal or transverse.

For an isotropic medium let $\lambda =\kappa -2\mu /3$. Contract the isotropic stiffness with ${\hat{p}}_k{\hat{p}}_l$:UsinggivesFor $\mathbf{a}\parallel \hat{\mathbf{p}}$ the longitudinal projector has eigenvalue one, so $B\mathbf{a}={\alpha }^2\mathbf{a}$: this is the P wave. For either of two independent $\mathbf{a}\perp \hat{\mathbf{p}}$ the projector vanishes, so $B\mathbf{a}={\beta }^2\mathbf{a}$: these are two degenerate transverse S wave polarizations. An isotropic solid has one P speed and one S speed, but three polarization modes: one longitudinal and two transverse. A fluid limit with zero shear modulus has no propagating S wave, so the positive-speed conclusion requires a solid with positive shear stiffness.

For $n=0$, the identity matrix gives the assertion. Suppose it holds at $n$. Conditioning on the state at time $n$ and using the Markov property givesThe nonnegative countable sum is justified by the law of total probability. Induction proves the $n$-step transition matrix is $P^n$.

The entries in the two slack columns give the inverse basis matrixThus the ordered basic variables are $x_3,x_1$. Multiplying the original coefficient matrix and right-hand side by $B^{-1}$ gives the completed constraint rows. With the bottom row defined as reduced costs $c^T-c_B^T{B}^{-1}A$ and right-hand entry $-c_B^T{B}^{-1}b$, the complete tableau isThis bottom-right entry is the negative of the optimal objective under this row-operation convention. If a tableau records the positive objective value separately in that position, it is $26/3$; the completed constraint rows and reduced costs are the same.

The inverse function theorem says that a continuously differentiable map on an open set, with invertible derivative at a point, restricts to a $C^1$ diffeomorphism between neighborhoods of that point and its image. The inverse derivative is the inverse matrix of the original derivative at the corresponding preimage.

Put $s=1+x+y\ne 0$. Differentiating each coordinate givesThe determinant never vanishes on $\Omega$, so the theorem applies at every point. The image point $(1/3,1/3)$ has preimage $(1,1)$: the equations $x/s=y/s=1/3$ imply $x=y=1$. Therefore

For small oscillations about equilibrium, put $M_{ij}=T_{ij}(0)$ and $K_{ij}={\partial }_i{\partial }_{j}V(0)$. The potential has no linear term at equilibrium; to quadratic order the Lagrangian is $L_2={\dot{x}}^{\mathsf{T}}M\dot{x}/2-x^{\mathsf{T}}Kx/2$. Its equations are $M\ddot{x}+Kx=0$. A normal mode $x=v{e}^{i\omega t}$ satisfies the generalized eigenvalue problem $(K-{\omega }^{2}M)v=0$. For a positive-definite mass matrix, diagonalization $M^{-1/2}K{M}^{-1/2}$ gives the squared frequencies and independent normal coordinates. A strictly positive Hessian gives positive frequencies; zero eigenvalues require separate higher-order stability analysis.

Here the spring extensions are $x_1$, $x_2-x_1$, and $-x_2$. ThusThe stiffness matrix is $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and $M=mI$. Its eigenvectors $(1,1)$ and $(1,-1)$ have stiffness eigenvalues $k$ and $3k$, givingIn the first mode the masses move together with equal displacements and the middle spring keeps its length. In the second they move oppositely with equal amplitudes, stretching and compressing the middle spring twice as much as either end spring.

Condition on the full sampled history through time $n$. The Markov property of $X$ implies that its future after $2n$ depends on that history only through $X_{2n}=Y_n$. Summing over the intermediate state givesThus $Y$ is a Markov chain with transition matrix $P^2$. This is a deterministic-time sampled Markov chain.

Within the class $\{1,2\}$, the two-step matrix has eigenvalues one and $1/9$, with stationary distribution $(1/2,1/2)$. Diagonalizing this symmetric matrix, or solving the recursion $p_{n+1}=4/9+p_n/9$ from $p_0=0$, givesConsequently the limit is $1/2$. Restriction to even times is essential because the original chain has period two.

Eliminate the first column with multipliers ${\ell }_{21}=2$ and ${\ell }_{31}=-1$. The new second and third rows are $(0,2,4)$ and $(0,-1,3)$. The next multiplier is ${\ell }_{32}=-1/2$, giving final row $(0,0,5)$. Recording these multipliers gives the LU decompositionFor an upper-triangular matrix with nonzero diagonal, the back substitution algorithm works from bottom to top:In the separate triangular system printed in the question, $5x_3=10$, $2x_2+4x_3=6$, and $x_1+2x_2+2x_3=3$. Hence

Any nonzero vector in ${\mathbb{F}}_2^3$ extends to a basis. Sending one such basis to another gives an element of the general linear group, so its group action on the seven nonzero vectors is a transitive group action. Counting successive independent columns givesBy the orbit-stabilizer theorem, the stabilizer $H$ of a chosen nonzero vector $v$ has order $168/7=24$.

There are three two-dimensional linear subspaces containing $v$, corresponding to the three one-dimensional linear subspaces of ${\mathbb{F}}_2^3/⟨v⟩$. The stabilizer $H$ acts transitively on them: extend a basis of either plane beginning with $v$ to a basis of the whole vector space. Thus the stabilizer of a flag $⟨v⟩<P$ has order $24/3=8$.

If $Q\le G$ has order eight, each orbit of a group action of $Q$ on the seven nonzero vectors has size a power of two. Since seven is odd, at least one orbit of a group action has size one. Thus $Q$ fixes a nonzero vector $v$. Its group action on the three planes containing $v$ likewise has a fixed plane $P$. Therefore $Q$ is contained in, and by order equals, the stabilizer of that flag. The group action of $G$ is transitive on these flags, so all such subgroups are conjugate subgroups.

Distinct flags give distinct stabilizers. In a basis adapted to a flag, its order-eight stabilizer consists ofIts only common fixed nonzero vector is $e_1$. Its only invariant plane is $⟨e_1,e_2⟩$: an invariant plane is the kernel of a linear map of a nonzero linear functional, and over ${\mathbb{F}}_2$ that linear functional must be fixed; the displayed matrices have only the fixed nonzero linear functional $e_3^{\ast }$. Hence each subgroup specifies its flag uniquely. There are seven choices of $v$ and three choices of $P$, giving

Differentiability at $a$ means that there is a linear map $L:V\to V_1$ withThis map is the total derivative $D{f}_a$. In the Frobenius norm, $|AB|\le |A||B|$, by the Cauchy-Schwarz inequality applied to matrix entries. Expanding with the order of all factors preserved givesThe remainder after the terms linear in $H$ has norm at most $3|A||H{|}^2+|H{|}^3=o(|H|)$. HenceFor $|C|<1$, the Neumann series $I-C+C^2-\cdots$ converges absolutely in the finite-dimensional space of matrices. Multiplying its partial sums by $I+C$ and passing to the limit shows that it is $(I+C{)}^{-1}$. Its remainder after $I-C$ has norm at most $|C{|}^2/(1-|C|)=o(|C|)$, proving the required expansion.

If $A$ is an invertible matrix and $|H|<1/|A^{-1}|$, then $A+H=A(I+A^{-1}H)$ is an invertible matrix. Thus the general linear group is an open subset of the matrix space. The preceding expansion also givesApply the chain rule to $g(X)=(X^{-1}{)}^3$, using the cubic total derivative at $A^{-1}$ and this total derivative of inversion:The three terms generally cannot be combined by commuting $A$ and $H$.

The Gershgorin circle theorem states that every complex eigenvalue of a square matrix $A$ lies in at least one closed disc $D_i=\{z:|z-a_{ii}|\le {\sum }_{j\ne i}|a_{ij}|\}$. To prove it, take a nonzero eigenvector $v$ for $\lambda$ and choose $i$ with $|v_i|$ maximal. The $i$th row of $Av=\lambda v$ givesDivision by $|v_i|>0$ proves the claim. The proof applies to real and complex matrices alike.

Write $q=1/4+ϵ$ and take $0<ϵ<1/4$, as intended by the small-parameter assumption. The row discs have centers $1,\dots ,n$ and radii $ϵ,q,\dots ,q,1/4$. They are pairwise disjoint because the sum of neighboring radii is below one. The second Gershgorin circle theorem places one eigenvalue in each disc. Since $A$ is real, nonreal eigenvalues occur in conjugate pairs; a disc centered on the real axis would have to contain both members, contradicting its count of one. Thus all eigenvalues are real andAlternatively, a positive diagonal change of basis symmetrizes the tridiagonal matrix, with transformed off-diagonal entries $\sqrt{ϵ}/2$, also proving reality.

The claimed endpoint bounds with radius $ϵ$ are false for positive corner perturbations. For example, take $n=3$, $ϵ=1/1000$ and $\delta =9/10<1-ϵ$. The characteristic polynomial isThroughout $|\lambda -1|\le 1/1000$, the absolute value of the sum of its first two terms is below $0.003$, whereas $9/160=0.05625$. Hence this interval contains no root, contradicting the first requested bound. The actual eigenvalues are approximately $1.029144,1.943599,3.027257$, which also violate the last requested bound.

There is a valid real-eigenvalue conclusion with wider endpoint bounds. For $0<ϵ<1/4$ and $\delta <(n-1)/2-ϵ$, the Gershgorin discs from rows have one enlarged component on the left; the Gershgorin discs from columns have one enlarged component on the right. Each has an enlarged radius $r=\delta +ϵ<(n-1)/2$. A small disc that meets the enlarged left disc has index $j$ satisfying $j-1\le r+q<n/2$. A small disc meeting the enlarged right disc satisfies $n-j<n/2$. Thus, when $n=2m$, every disc with index at most $m$ is isolated in the column family and every disc with index at least $m+1$ is isolated in the row family. Their counts give all $n$ eigenvalues, one in each disjoint small disc, so all are real.

When $n=2m+1$, the same argument isolates the first $m$ discs in the column family and the last $m$ in the row family. It accounts for $2m$ real eigenvalues. The one remaining eigenvalue is real by conjugation, and belongs to the central small disc: the enlarged left and right discs are disjoint, and their two components can intersect only within the disc centered at $m+1$. Therefore the valid uniform bounds areThe sharper endpoint radii $ϵ$ hold in part (d) when $\delta =0$, and cannot simply be carried over to the perturbed matrix.

Begin with $u=v=0$ and the printed matching $\{(1,1),(3,2),(5,3)\}$. Grow an alternating tree from unmatched row two. Its tight edge reaches column two, whose matched row is three. Thus the reached row set is $S=\{2,3\}$ and the reached column set is $T=\{2\}$. Among unreached columns the reduced costs are $(6,5,4,4)$ from row two and $(3,4,2,3)$ from row three, so their minimum is two.

Increase $u_2,u_3$ by two and decrease $v_2$ by two. The new tight edge $(3,4)$ reaches an unmatched column. Augment along $2\to 2\to 3\to 4$, replacing $(3,2)$ by $(2,2),(3,4)$. The matching becomes $\{(1,1),(2,2),(3,4),(5,3)\}$.

Now grow from unmatched row four. Edge $(4,1)$ reaches matched row one, and edge $(1,5)$ reaches unmatched column five. Augment along $4\to 1\to 1\to 5$, replacing $(1,1)$ by $(4,1),(1,5)$. This yieldsNo further potential update is needed. The final potentials are $u=(0,2,2,0,0)$ and $v=(0,-2,0,0,0)$. Their reduced-cost matrix isEvery entry is nonnegative and every chosen entry is zero. The dual sum is $\sum u_i+\sum v_j=2$, so part (a) certifies optimality.

Relative to the three successive blocks of sizes $1,2,2$, the stabilizer consists of the block upper triangular matricesThe off-diagonal blocks contain $2+2+4=8$ arbitrary field entries. Thus its order is $(q-1)|\mathrm{GL}(2,q){|}^2{q}^8$. By orbit-stabilizer theorem,Equivalently, choose the line first and then a two-dimensional subspace of the four-dimensional quotient:This also checks the stabilizer count independently.

The first column is five times $(4,0,3,0{)}^{\mathsf{T}}/5$. Its orthogonal companion in the first-and-third-coordinate plane is $(3,0,-4,0{)}^{\mathsf{T}}/5$, and the second column has positive residual seven times that vector. The remaining coordinate directions can therefore be read off immediately rather than performing a general four-dimensional orthogonalization. One obtainsThe columns of $Q$ are visibly orthonormal. Taking their inner products with each column of the given matrix gives the displayed $R$, which also verifies $QR=A$ and positive diagonal entries.

At step $k$, choose a real shift ${\mu }_k$, factor $A_k-{\mu }_{k}I=Q_k{R}_k$, and setThe last equality proves orthogonal similarity at every step. Thus all iterates have the same eigenvalues; if $A_k$ is symmetric, the last expression is symmetric, proving symmetry by induction.

For the specified matrix, let $e=(1,\dots ,1{)}^{\mathsf{T}}$. Then $A=I-2e{e}^{\mathsf{T}}/n$ and $e^{\mathsf{T}}e=n$, so $A^{\mathsf{T}}=A$ and $A^2=I$. It is a Householder reflection, hence orthogonal. With the standard positive-diagonal QR convention, part (a) gives $Q=A$, $R=I$ at the first unshifted step, and therefore $A_1=RQ=A$. The same repeats forever: the unshifted algorithm is stationary.

If arbitrary diagonal signs in QR are allowed, $R=D$, $Q=AD$ for a sign matrix $D$, and the next iterate is $DAD$. Its off-diagonal magnitudes are unchanged, so even then only sign changes occur, not progress toward a diagonal form. This explains the role of the usual convention in the literal stationary-iterate statement.