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The shallow-water approximation requires a small depth-to-horizontal-scale ratio, slow streamwise variation of the channel and interface, small vertical acceleration, and rapid enough cross-sectional mixing for $u$ and $\varphi$ to be representative averages. Near the channel apex, or during the initial collapse at a sharp front, these conditions need not hold. The high Reynolds number permits neglect of leading viscous resistance; weak stirring mixes the particles without appreciable ambient fluid entrainment.

Put $\lambda =\beta /H$. At elevation $z$, the channel width is $2\lambda xz$, so the occupied area and interface width areVolume conservation and suspended-particle conservation giveThe excess hydrostatic pressure is ${\rho }_{0}d\varphi (h-z)$. Its streamwise gradient averaged over the triangular section gives acceleration $-d\varphi h_x-(dh/3){\varphi }_x$: the weighted mean of $h-z$ is $h/3$. Thus the triangular-channel shallow water equations areThe quasilinear coefficient matrix in $(h,u,\varphi )$ isIts three distinct real eigenvalues, for $h,d\varphi >0$, establish strict hyperbolicity:On the material characteristic curve, $d\varphi /dt=-2W_s\varphi /h$. On each of the other two characteristic curves, multiplying by the left eigenvector $(\pm 2c/h,1,\pm dh/(3c))$ gives the compatibility ordinary differential equationThe derivatives are along the indicated characteristic, with $g^{\prime }=d\varphi$. At vanishing concentration, the gravity-wave speeds coalesce and strict hyperbolicity is lost.

Here $D=I$, so the Jacobi iteration matrix isThe all-ones vector is an eigenvector with eigenvalue $-2\alpha$. On its two-dimensional orthogonal complement, the coordinates sum to zero and $Hv=\alpha v$, so the other eigenvalue is $\alpha$ with multiplicity two. ThereforeBy Jacobi convergence for a three-by-three equicorrelation matrix, convergence occurs exactly forAt either endpoint the spectral radius is one, so convergence for arbitrary initial data fails.

Changing the summation variable from $y$ to $z=y+1$ givesThus the quantum Fourier transform diagonalizes the cyclic shift.

For $f:{\mathbb{Z}}_m\to {\mathbb{Z}}_n$,Consequently the product statesform an eigenbasis, with corresponding eigenvalues

A fixed point satisfiesThe branch with $y=0$ isEvery other fixed point can be parametrized by its $x$-coordinate:Equivalently,The $x_{-}$ pair exists for $0\le a\le 4$, and the $x_{+}$ pair for $3\le a\le 4$.

The Jacobian matrix isOn $P_0(a)$ its eigenvalues are $1$ and $a(3-a)$, so this branch has a nonhyperbolic equilibrium at $a=0$ and $a=3$. On a nonzero-$y$ branch,which vanishes at $x=2$, hence at $a=4$ and $y=\pm \sqrt{2}$. The three bifurcation values and locations are therefore

For the center manifold calculation, append $\dot{a}=0$ as in the extended centre manifold for a parameter.

At $(a,x,y)=(0,0,0)$ put $\mu =a$ and write the centre manifold as $x=h(y,\mu )$. Its centre-manifold invariance equation isIt gives $h=\mu -y^2+O(3)$, and henceThis is the pitchfork bifurcation normal form with stable nonzero centre branches for $\mu >0$: the bifurcation at $a=0$ is supercritical.

At $(a,x,y)=(3,3,0)$ put $u=x-3$ and $\mu =a-3$. ThenAgain $u=h(y,\mu )=\mu -y^2+O(3)$, soThe nonzero centre branches exist for $\mu >0$ and are unstable in the centre direction, whereas $y=0$ is centre-stable there. Thus the bifurcation at $a=3$ is a subcritical pitchfork bifurcation with reversed normal-form parameter.

Finally fix $y_0=\pm \sqrt{2}$ and putThe translated system isThe extended centre manifold hasand thereforeThis is the saddle-node bifurcation normal form. For $\mu <0$ there are two nearby fixed points, which coalesce and disappear at $\mu =0$. The complete reductions are recorded in the Extended centre-manifold reductions of the 2019 Cambridge reflection-symmetric system.

A matrix is Hermitian when $A^{†}=A$, and unitary when $U^{†}U=U{U}^{†}=I$.

If $A\mathbf{u}=\lambda \mathbf{u}$ with $\mathbf{u}\ne 0$, thenso every eigenvalue is real. If $A\mathbf{u}=\lambda \mathbf{u}$ and $A\mathbf{v}=\mu \mathbf{v}$, thenDistinct eigenvalues therefore have orthogonal eigenvectors.

The normalized eigenvectors form an orthonormal basis, so the matrix $U=({\mathbf{u}}_1\cdots {\mathbf{u}}_n)$ satisfies $U^{†}U=I$ and is unitary. Withthe equations $A{\mathbf{u}}_j={\lambda }_j{\mathbf{u}}_j$ say $AU=UD$, hence $A=UD{U}^{†}$. This is the finite-dimensional unitary diagonalization of a normal matrix.

A unitary conjugate of an arbitrary diagonal matrix need not be Hermitian: $U=(1)$ and $D=(i)$ give $UD{U}^{†}=(i)$. It is Hermitian exactly when all diagonal entries of $D$ are real.

For the displayed matrix, one valid decomposition isThe columns are orthonormal eigenvectors, so direct multiplication gives $UD{U}^{†}=A$.

Given a solution $u(t)$, construct $a_j(t)=\frac{1}{2}{e}^{u_j(t)/2}$ and hence the smooth matrices $L(t)$ and $B(t)$. Let $U(t)$ solve the linear ordinary differential equationBecause $B(t)$ is skew-symmetric,so $U(t)$ is an orthogonal matrix and is invertible. Moreover,It follows thatThus the Lax pair flow is an Isospectral Lax equation: every $L(t)$ is similar to $L(0)$ and has the same eigenvalues and characteristic polynomial.

Every symmetric function of these constant eigenvalues is a first integral. In particular, the trace invariants of a Lax equationare constant because the cyclic property of the trace givesFor example,is an explicit first integral of equation $(†)$.

For a fixed shift $s$ that is not an eigenvalue, inverse iteration is the power method applied to $(A-sI{)}^{-1}$. Before normalization, the $k$th vector is proportional toIf $s<{\lambda }_1$, factor out the first coefficient:Every ratio in parentheses lies strictly between zero and one. The bracket therefore tends to $w_1$, and normalization in the Euclidean norm givesThis proves the stated convergence to $w_1$ or $-w_1$ and exhibits the convergence of fixed-shift inverse iteration rate.

Now suppose ${\lambda }_m<s<{\lambda }_{m+1}$. The closest eigenvalue to $s$ is one of ${\lambda }_m$ and ${\lambda }_{m+1}$.

Finally, if ${\lambda }_n<s$, the closest eigenvalue is ${\lambda }_n$ and its inverse eigenvalue is negative. Thus the direction approaches the line spanned by $w_n$, while normalized vectors alternate between the two signs. These cases are the sign behavior of fixed-shift inverse iteration.

For a bounded linear operator $T:H\to H$, the spectrum iswhile the point spectrum is the set of eigenvalues, equivalently the $\lambda$ for which $T-\lambda I$ is not injective.

A compact operator maps the unit ball to a relatively compact set. A self-adjoint operator satisfies $T=T^{\ast }$, or $⟨Tx,y⟩=⟨x,Ty⟩$ for all $x,y\in H$. A finite-rank operator has finite-dimensional image.

Eliminate $R=N-S-I$ to obtain the two-dimensional autonomous differential equationIts Jacobian matrix at the endemic equilibrium isSince $I^{\ast }>0$,The trace-determinant stability criterion shows that both eigenvalues have negative real part. The Endemic equilibrium of the SIR model with waning immunity is therefore locally asymptotically stable.

The map $x\mapsto -x$ is an involution, soIf $\lambda$ is an eigenvalue of ${\mathrm{QFT}}_N$, then ${\lambda }^4=1$. Hence every eigenvalue belongs toBecause $X^4-1$ has distinct roots, the transform is diagonalizable; which fourth roots occur and their multiplicities depend on $N$.

If $\varphi v=\mu v$ for a nonzero eigenvector $v$, thenHence every eigenvalue satisfies $\mu =0$. Since a complex endomorphism has an eigenvalue, zero is the only possible eigenvalue.

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.

Repeated substitution in Landweber iteration from $u_0=0$ givesOn the $i$th singular vector, $A^{\ast }A$ has eigenvalue ${\sigma }_i^2$ and $A^{\ast }{y}_i={\sigma }_i{x}_i$. The finite geometric series therefore givesThis is a spectral regularization method with filter

The supports of ${\phi }_1$ and ${\phi }_2$ are disjoint, so they are orthogonal vectors. Substitution in the covariance formula givesandTherefore the associated eigenvalues are

A representation is completely reducible, or semisimple, when it is a direct sum of irreducible representation. Equivalently, every invariant subspace has an invariant complement.

Maschke's theorem says that every finite-dimensional representation of a finite group $G$ over a field of characteristic zero is completely reducible. More generally, it is enough that the field characteristic does not divide $|G|$.

Schur lemma states the following.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility forcesso $T$ is an isomorphism. For an endomorphism of a finite-dimensional complex representation, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so the first part forcesThis proves the Proof of Schur lemma.

Now let $\rho :G\to \mathrm{GL}(V)$ be faithful and irreducible over $\mathbb{C}$. For every $z\in Z(G)$, the operator $\rho (z)$ commutes with every $\rho (g)$. Schur's lemma therefore givesfor some ${\lambda }_z\in {\mathbb{C}}^{\times }$. Faithfulness makes $z\mapsto {\lambda }_z$ an embeddingSince every finite subgroup of the multiplicative complex numbers is cyclic,This is the cyclic-center obstruction to a faithful irreducible representation.

Put $x=1+u$, $y=v$, and $\mu =-1+\nu$, adjoining $\dot{\nu }=0$. ThenThe zero eigenvalue has centre direction $v=-u$. Solving the center manifold invariance equation givesSubstitution yields the reduced equationIts equilibria are $u=0$ and $u=\pm \sqrt{\nu }+O(\nu )$ for $\nu >0$, exactly matching the three branches meeting at the pitchfork in part (b).

If $\alpha v=\lambda v$ for nonzero $v$, thenso ${\lambda }^3$ is an eigenvalue. Let $\omega =e^{2\pi i/3}$. Since $\lambda \ne 0$, the three factors ofare pairwise coprime. The kernel decomposition for coprime polynomials therefore giveswhere an absent eigenspace is interpreted as zero.

The Gauss map sends $p$ to the chosen unit normal $N(p)$. Differentiating $⟨N,N⟩=1$ shows that $d{N}_p$ maps $T_{p}S$ into itself. The shape operator $S_p=-d{N}_p$ is self-adjoint because its associated second fundamental form is symmetric. Its eigenvalues $k_1,k_2$ are the principal curvatures. A point is umbilical when $k_1=k_2$, and the surface is minimal when its mean curvature $H=(k_1+k_2)/2$ vanishes everywhere.

In an orthonormal principal basis, $d{N}_p$ has diagonal matrix $\mathrm{diag}(-k_1,-k_2)$. It is conformal exactly whenwhich is equivalent to $k_1^2=k_2^2\ne 0$. At a nonumbilical point this says $k_1=-k_2$, exactly the condition $H=0$. Hence, when there are no umbilical points, the surface is minimal if and only if its Gauss map is conformal.

The parity of a wavefunction is its eigenvalue under spatial reflection: $\psi (-x)=+\psi (x)$ for even parity and $\psi (-x)=-\psi (x)$ for odd parity. If $V(-x)=V(x)$, the Hamiltonian commutes with reflection, so its energy eigenspaces admit a parity eigenbasis. In particular, every nondegenerate bound-state energy eigenfunction has definite parity; within a degenerate scattering eigenspace one must choose the even and odd linear combinations.

Use the conventionChanging the summation variable from $j$ to $j-1$ givesThus the Fourier-basis states are eigenvectors of the cyclic shift, with respective eigenvalues $e^{-2\pi ik/N}$, and the cyclic shift diagonalization by the quantum Fourier transform is

For $N=4$, write $k=2x+y$ with $x,y\in \{0,1\}$. ThensoHence the required quantum circuit applies, from input to output,The first phase gate acts on the most significant qubit $x$ and the second on the least significant qubit $y$.