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Write $u=u_x$, $w=u_y$ and $B=B_y$. The isothermal equation of state replaces the adiabatic thermal equation by $p=c_s^2\rho$. Since the fields depend only on $x,t$, the solenoidal constraint gives ${\partial }_x{B}_x=0$ and the $x$ component of the ideal magnetohydrodynamic induction equation gives ${\partial }_t{B}_x=0$. Hence $B_x$ is constant in both space and time.

The continuity, two momentum and transverse induction equations areHere $B$ is the transverse magnetic component and the constant $B_x$ is the longitudinal component. The longitudinal magnetic force is the gradient of transverse magnetic pressure; the transverse force is magnetic tension. In primitive variables the isothermal coplanar magnetohydrodynamic characteristic matrix isThe imposed $z$ components stay zero because their corresponding equations have zero right-hand sides under this reduction.

Let $c=v-u$ be a characteristic speed relative to the fluid and define the Alfvén velocity ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$. To derive the characteristic polynomial, an eigenvector with components $(\delta \rho ,\delta u,\delta w,\delta B)$ obeysEliminating $\delta u$ and $\delta w$ on a nondegenerate branch givesTheir consistency yields $(c^2-c_s^2)(c^2-v_{ax}^2)=v_{ay}^2{c}^2$. Equivalently, direct expansion of $\det (\mathbf{A}-vI)$ gives the same polynomial without dividing by $c$ or a possibly vanishing field component:ThusThe discriminant is $(c_s^2-v_a^2{)}^2+4c_s^2{v}_{ay}^2\ge 0$, and both squared speeds are nonnegative. These are the fast magnetosonic wave and slow magnetosonic wave families, transported with the mean flow. The equations are hyperbolic; on a generic state the four distinct real eigenvalues give four characteristic directions. Special field alignments can make speeds coincide or the slow speed vanish. There is no separate entropy characteristic because the thermal model is isothermal, and the independent out-of-plane Alfvén wave polarization is excluded by $u_z=B_z=0$.

To construct a simple wave in magnetohydrodynamics, take $\mathbf{U}=\mathbf{U}(a)$ along a state-space curve whose tangent is a right eigenvector of one selected branch:Substitution into the partial differential equations then gives the scalar transport equationSuch curves exist locally by integrating the eigenvector field away from degeneracies. For example, parameterizing by $a=\rho$ where its eigenvector density component is nonzero gives the explicit state-curve equationswith signed $c$ equal to one of the four relative wave speeds. These formulas are used only where their denominators are nonzero.

A particularly transparent nonlinear example is an isothermal perpendicular magnetosonic simple wave. Set the allowed constant $B_x=0$, choose $B=K\rho$ and keep $w$ constant. For the fast branch,These relations satisfy the eigenvector equations and therefore all four original equations wheneverThe amplitude-dependent characteristic speed has derivativeThis already proves the existence of genuinely nonlinear waves in the stated system.

More generally, the method of characteristics applied to initial data $a(x,0)=a_0(x)$ givesIf the initial speed decreases towards increasing $x$ somewhere, the denominator reaches zero in finite time. When the negative minimum is attained, the first steepening time isFaster trailing states catch slower leading states, giving shock formation by characteristic intersection. For the right-moving perpendicular fast example, a negative density gradient provides such a compression. Beyond $t_{\ast }$ a smooth simple wave is no longer sufficient and an appropriate shock solution is required. Rarefactive profiles instead spread. A linearly degenerate characteristic field has $v^{\prime }=0$ along its state curve and avoids this mechanism; the conclusion is that compressive nonlinear magnetosonic simple waves steepen, not that every possible MHD waveform must do so.

With $y=\dot{x}$, the feedback system isFor $p=0$, $(x,y,z)=(0,0,0)$ is an equilibrium. Its stability matrix and characteristic polynomial areThus there is always a zero eigenvalue, with eigenvector $(1,0,0)$, so it is a nonhyperbolic equilibrium. If $q\ne -1$, the other two eigenvalues are off the imaginary axis: for $q>-1$ their real parts are negative, while for $q<-1$ their product is negative. At $q=-1$, all three eigenvalues are zero, explaining the exclusion in the next part.

The Euclidean matrix norm is the induced operator norm $\Vert C{\Vert }_2={\sup }_{\Vert x{\Vert }_2=1}\Vert Cx{\Vert }_2$. By the spectral theorem, a real symmetric matrix has $A=Q\Lambda Q^T$ with $Q$ orthogonal and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$. HenceOrthogonal transformations preserve the Euclidean norm, and the induced norm of a real diagonal matrix is the maximum absolute diagonal entry. Therefore, for $t\ge 0$,This gives the requested inequality with equality. If an exponent $\beta <\mu (A)$ were to work with prefactor one for every $t$, choose a unit eigenvector for $\mu (A)$. Its image has length $e^{t\mu (A)}>e^{t\beta }$ for every $t>0$, a contradiction. Thus $\mu (A)$ is the smallest possible exponent.

For symmetric matrices the spectral abscissa equals the Euclidean logarithmic norm. Symmetry is essential to this particular exact norm formula; a general non-normal matrix can have transient growth not controlled with prefactor one by its spectral abscissa.

For the state vector $q=(h,u,g^{\prime }{)}^T$, the primitive equations are $q_t+\mathsf{A}{q}_x=S$, withThe characteristic polynomial isConsequently the three characteristic speeds areFor $h>0$ and $g^{\prime }>0$, the hyperbolic system has three distinct real eigenvalues. The long-wave speed relative to the current is $c$, while $u$ transports a buoyancy/contact disturbance.

A left eigenvector for the material characteristic is $(0,0,1)$, givingFor $u\pm c$, left eigenvectors can be chosen asMultiplying $q_t+\mathsf{A}{q}_x=S$ by these rows gives the characteristic compatibility for an entraining triangular-channel current. With all differentials evaluated along the indicated characteristic,The source follows directly from ${\ell }_{\pm }S=-2u{w}_e/h\pm 2c{w}_e/h\mp 2g^{\prime }{w}_e/(3c)$ and $g^{\prime }/c=2c/h$.

Since $4dc=(2c/h)dh+(h/c)d{g}^{\prime }$, an equivalent form isCombining it with the material buoyancy equation also givesOnly when entrainment vanishes and $g^{\prime }$ is spatially constant do $u\pm 4c$ become constant Riemann invariants on these characteristic curves. Dropping the buoyancy differential in the general problem would lose one of the required transport couplings. The characteristic formulas with $1/c$ apply to the nondegenerate region $g^{\prime }h>0$; dry or neutrally buoyant limits require their conservative form instead.

For this alternating arrival and service queue, set $\theta =\lambda (\mu +\beta )/[\mu (\lambda +\alpha )]$. Direct multiplication gives $(\theta ,\lambda /\mu )B=\theta (\theta ,\lambda /\mu )$, so the initial row is a left eigenvector and $({\pi }_{iC},{\pi }_{iW})={\theta }^{i-1}({\pi }_{1C},{\pi }_{1W})$. ConsequentlyThese nonnegative probabilities can be normalized exactly when $\theta <1$, equivalently $\mu \alpha >\lambda \beta$. Summing the geometric series givesWhen $\theta \ge 1$ no positive multiple of the invariant sequence is summable, so no stationary distribution exists. This rules out positive recurrence; absence of a stationary probability alone should not be used to distinguish null recurrence from transience.

Since $A$ is diagonalizable over the complex numbers, write $x^{(0)}={\sum }_i{a}_i{v}_i$. Provided iteration does not encounter zero, its direction is that ofWith one strictly dominant eigenvalue and $a_n\ne 0$, all other contributions are smaller by factors $(|{\lambda }_i|/|{\lambda }_n|{)}^k$. The dominant eigenvalue is real, since nonreal eigenvalues of a real matrix occur in conjugate pairs of equal modulus. The normalized power method therefore converges to a signed unit eigenvector if ${\lambda }_n>0$, and alternates between opposite unit eigenvectors if ${\lambda }_n<0$. A quotient such as $(x^{(k)}{)}^{T}A{x}^{(k)}$ approaches ${\lambda }_n$ in either case. If $a_n=0$, replace it by the largest active modulus; if $A{x}^{(k)}=0$, the prescribed iteration is undefined.

With two dominant eigenvalues of equal modulus and both components active, the iterates approach their real invariant plane rather than one direction. Distinct real equal-modulus eigenvalues are $\rho ,-\rho$ and produce a limiting two-cycle. A conjugate pair $\rho e^{\pm i\theta }$ produces a rotation in suitable coordinates on that plane: the normalized directions are periodic when the angle is rational relative to $2\pi$ and otherwise generally keep rotating without convergence. If the two eigenvalues are equal and real, the initial projection is itself one eigenvector in their repeated eigenspace, and converges or alternates as in the single-value case; one orbit cannot recover every independent eigenvector of that eigenspace.

For distinct dominant values, retain unnormalized consecutive vectors $y,Ay,A^{2}y$ in the nearly invariant plane and solveThe roots of ${\lambda }^2-s\lambda +p$ estimate the two eigenvalues. Corresponding eigenvectors are $Ay-{\lambda }_{2}y$ and $Ay-{\lambda }_{1}y$, since applying $A$ and using the recurrence multiplies them by ${\lambda }_1$ and ${\lambda }_2$. This is dominant eigenpair extraction from a two-step Krylov recurrence. Equivalent extraction uses the $2\times 2$ matrix of $A$ on the computed plane. It needs two independent iterates and nonzero projections onto both distinct eigenvectors.

The given vectors satisfyand $y_k,y_{k+1}$ are independent, so their span is invariant under $A$. On that plane the characteristic polynomial is ${\lambda }^2-2\lambda +2$. ThereforeIndeed $A{v}_{\pm }=y_{k+2}-(1\mp i)y_{k+1}=(1\pm i)v_{\pm }$. No third eigenvalue is determined by these data.

The displayed $x_k=(1,1,1)$ has norm $\sqrt{3}$, although a large-$k$ normalized iterate should have norm one. This harmless source normalization is resolved by dividing all three displayed vectors by $\sqrt{3}$; the recurrence, eigenvalues and eigenvector directions are unchanged.

The shifted matrix $N=D-\lambda I$ is still a normal matrix, sinceand $N{N}^{†}$ has the same expansion except that its first term is $D{D}^{†}$, which equals $D^{†}D$. For the given eigenvector, $Ne=0$. Part (b), applied to $N$, therefore gives $N^{†}e=0$. Since $N^{†}=D^{†}-\overline{\lambda }I$,The corresponding eigenvalue of the Hermitian conjugate is the complex conjugate of the original eigenvalue.

Take eigenvectors $v_i,v_j$ for distinct eigenvalues ${\lambda }_i,{\lambda }_j$. Since the eigenvalues are real and $A$ is a Hermitian matrix,Hence $({\lambda }_j-{\lambda }_i)v_i^{†}{v}_j=0$, so eigenvectors for distinct eigenvalues are orthogonal vectors. They are therefore linearly independent; after normalization the given $n$ eigenvectors form an orthonormal basis of ${\mathbb{C}}^n$.

Use the conventionfor the characteristic polynomial. Choosing $\det (A-tI)$ instead multiplies it by $(-1{)}^n$ and leaves the requested zero identity unchanged. For each eigenvector $v_j$, induction gives $A^r{v}_j={\lambda }_j^r{v}_j$, including $r=0$. Thusbecause ${\lambda }_{j}I-A$ has the nonzero kernel vector $v_j$ and therefore has zero determinant. Part (a) supplies a basis consisting of these $n$ eigenvectors. A linear map vanishing on every basis vector vanishes everywhere. ConsequentlyThis proves the Cayley-Hamilton theorem identity directly in the distinct-eigenvalue case assumed here.

Expand $x={\sum }_j{c}_j{v}_j$ in the orthonormal basis from part (a). ThenFor $x\ne 0$, the weights $|c_j{|}^2/{\sum }_k|c_k{|}^2$ are nonnegative and sum to one. The Rayleigh quotient therefore lies between the smallest and largest eigenvalues. Conversely, normalized extreme eigenvectors $v_{\min },v_{\max }$ give, for $0\le t\le 1$,Thus the exact Hermitian Rayleigh quotient range isWhen $n=1$, the range is the singleton containing the sole eigenvalue.