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For a linear map $T:V\to W$ between finite-dimensional inner product spaces,The first equality follows directly from the defining identity for the adjoint operator; the second follows by taking orthogonal complements. In an infinite-dimensional Hilbert space, the second equality generally requires closure of the image.

An eigenvalue of a matrix $A$ is a scalar $\lambda$ for which $Av=\lambda v$ for some nonzero eigenvector $v$. Its corresponding eigenspace is

Write $v=(a,b,c,d{)}^T$. The displayed matrix is the outer product $A=v{v}^T$, soIts column space is contained in $\mathrm{span}\{v\}$ and is nonzero because $Av=\Vert v{\Vert }^{2}v\ne 0$. Hence $A$ is a rank-one matrix.

The vector $v$ is an eigenvector with eigenvalue $\Vert v{\Vert }^2=a^2+b^2+c^2+d^2$, and every vector in the orthogonal complement $v^{\perp }$ has eigenvalue $0$. ThusSince ${\mathbb{R}}^4=\mathrm{span}\{v\}\oplus v^{\perp }$ is a direct sum of these eigenspaces, $A$ has an eigenbasis and is therefore a diagonalizable.

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.

Away from the rim's $O(h)$ edge region, the geometry and uniform injection have no radial scale other than the factor required by continuity. It is therefore consistent to take $w=w(z)$. Regularity at $r=0$ then lets the continuity equation integrate toThe no-slip boundary condition and prescribed normal velocities areSince $p$ is independent of $z$, the radial equation implies that $w^{\prime \prime \prime }$ is constant. The four boundary conditions giveand henceUsing $w^{\prime \prime \prime }=12V/h^3$ in $p_r=\mu u_{zz}$ givesTaking the pressure at the rim to be atmospheric, $p(R)=p_0$, produces the porous-plate lubrication cushion pressureThe upward pressure force balances the disc's weight:Since $\epsilon =h/R$,

Expanding the two covariant derivatives of a vector, the second partial derivatives cancel in the commutator. The remaining derivatives and products of Christoffel symbols combine into the coordinate definition of the Riemann curvature tensor, giving the Ricci identityFor a type-$(2,0)$ tensor, the connection acts on both upper indices, so the curvature commutator on a contravariant tensor is

Keep $\mathbf{A}=(0,Bx,0)$ and choose the scalar potential $\varphi =-Ex$. ThenFor $p_z=0$, completing the square givesHence the spectrum on ${\mathbb{R}}^3$ is

For a translationally invariant dispersion, the group velocity is $v_y={\hslash }^{-1}\partial E_n/\partial k$. Thereforein agreement with the electric-cross-magnetic-field drift $\mathbf{E}\times \mathbf{B}/B^2$.

In the rectangle, $k=2\pi r/L_y$ as before. The energy now changes with $r$, so the guiding-centre degeneracy is lifted, while the number of states in each tilted band remains approximately $|qB|A/(2\pi \hslash )$. Adjacent states within a formerly degenerate level have the electric-field splitting of a Landau level in a rectangleThus this is the ground-to-first-excited gap when the lowest Landau band contains at least two allowed guiding centres and $\Delta E_{\mathrm{electric}}<\hslash {\omega }_c$. Without that implicit weak-field or large-sample condition, the exact full spectral gap is

An odd composite number $N$ with $\gcd (b,N)=1$ is a Fermat pseudoprime to base $b$ whenFor $N=35$, the Chinese remainder theorem for unit groups reduces this condition to congruences modulo $5$ and $7$. Modulo $5$, Fermat's little theorem givesso $b^{34}\equiv 1(\mathrm{mod}5)$ exactly when $b\equiv \pm 1(\mathrm{mod}5)$. Modulo $7$, the unit group has order $6$, soThe equation $b^4=1$ in the cyclic group $(\mathbb{Z}/7\mathbb{Z}{)}^{\times }$ has $\gcd (4,6)=2$ solutions, namely $b\equiv \pm 1(\mathrm{mod}7)$.

Combining the two independent sign choices by the Chinese remainder theorem givesand hence

Write $A=T-\lambda I$. We first prove that $\mathrm{im}A$ is dense. Otherwise some nonzero $y$ lies in its orthogonal complement, soBy part (b), choose finite-dimensional spaces $E_j$ containing $y$ such that $S_j=P_{E_j}T\to T$ in operator norm. Sincethe operator $S_j^{\ast }-\overline{\lambda }I$ is not injective. Therefore $S_j-\lambda I$ is not surjective; the contrapositive of part (c) makes $\lambda$ an eigenvalue of $S_j$. Choose unit vectors $x_j$ with $S_j{x}_j=\lambda x_j$. ThenCompactness gives a subsequence for which $T{x}_j$ converges. Since $\lambda \ne 0$, the displayed relation makes $x_j$ converge to a unit vector $x$, and continuity gives $Tx=\lambda x$, contradicting the hypothesis. Thus $\mathrm{im}A$ is dense.

Next suppose $A$ were not bounded below, or equivalently that its lower norm were zero. There would be unit vectors $x_j$ with $A{x}_j\to 0$, henceAgain a convergent subsequence of $T{x}_j$ forces the corresponding $x_j$ to converge to a unit vector $x$ satisfying $Ax=0$, another contradiction. Therefore some $c>0$ satisfies

This lower bound makes $\mathrm{im}A$ closed: if $A{x}_j$ converges, then $x_j$ is Cauchy and its limit maps to the same limit. Since the image is both dense and closed, it is all of $H$. We have proved the Fredholm alternative for a compact operator: every nonzero spectral value of a compact operator is an eigenvalue.