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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.

There is no such universal constant. For a real parameter $M$, consider the real matrixIts characteristic polynomial is $(t-1{)}^2$, so both eigenvalues are real and ${\lambda }_{\max }(B_M)=1$. ButFor any proposed finite $C$, choosing $M>2(C-1)$ violates the proposed bound. This is an explicit illustration that nonnormal eigenvalues do not bound a quadratic quotient.

Let $v\ne 0$ be an eigenvector of the Hermitian matrix $A$, with eigenvalue $\lambda$. Multiplying $Av=\lambda v$ by $v^{†}$ givesThe denominator is the positive real number $\Vert v{\Vert }^2$, whileThus the numerator is real, and every eigenvalue of a Hermitian matrix is real. This argument does not require distinct eigenvalues.

The parity operator acts by $(P\psi )(x)=\psi (-x)$. For an even potential, differentiating twice and using $V(-x)=V(x)$ proves $HP=PH$ on the parity-invariant domain of the Hamiltonian operator. Therefore each energy eigenspace is invariant under $P$. The projectors $(I+P)/2$ and $(I-P)/2$ split it into even and odd subspaces, giving a basis of simultaneous eigenstates. An odd state obeys $\psi (0)=-\psi (0)$, hence ${\psi }_{\mathrm{odd}}(0)=0$.

The infinite walls impose $\psi (-a)=\psi (a)=0$, with the wavefunction zero outside. It is continuous at the central delta potential. Integrate the stationary Schrödinger equation over $[-\epsilon ,\epsilon ]$; the energy integral tends to zero and the Dirac delta function contributes $\kappa \psi (0)$. This givesFor the given piecewise positive-energy form, $E={\hslash }^2{\lambda }^2/(2m)$. Replacing $x$ by $-x$ swaps the two pieces and leaves their values unchanged, proving $P\psi =\psi$. The value at zero is $A$, and the one-sided derivatives are $-B\lambda$ and $B\lambda$. The jump condition gives $B=-m\kappa A/({\hslash }^2\lambda )$. The wall at $a$ gives $A\cos (\lambda a)-B\sin (\lambda a)=0$, henceFor $\kappa =0$ the undivided wall equation instead gives the usual free even roots. For $\kappa >0$, set $c=m\kappa /{\hslash }^2>0$ and $k_n=(n+1/2)\pi /a$. In every interval $k_n<\lambda <(n+1)\pi /a$ there is exactly one root: writing $\delta =a(\lambda -k_n)\in (0,\pi /2)$ gives $\cot \delta =\lambda /c$, whose two sides are respectively strictly decreasing and increasing. These are all positive even roots; the quadratic form ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)+\kappa |\psi (0){|}^2$ is nonnegative, so there are no negative eigenvalues. Thus ${\lambda }_n>k_n$ and ${\eta }_n={\hslash }^2({\lambda }_n^2-k_n^2)/(2m)>0$.

The printed limit of this absolute energy difference is false. The exact quantization equation gives ${\delta }_n=\arctan (c/{\lambda }_n)$, so ${\delta }_n\to 0$ but ${\lambda }_n{\delta }_n\to c$. ConsequentlyThus the high-energy shift from a central delta barrier is a finite positive constant, not zero. The corrected vanishing statement is the relative shift ${\eta }_n/E_n^{(0)}\to 0$, or the wave-number difference ${\lambda }_n-k_n\to 0$. Physically, highly energetic particles are weakly affected in relative terms by a fixed point barrier. The free even wavefunction normalized in $[-a,a]$ has density $1/a$ at the origin, so the barrier's first-order expectation is $\kappa /a$, matching the exact limit.

Odd states vanish at zero, so the delta term and the derivative jump both vanish. Their normalized wavefunctions and energies arewith zero wavefunction outside. They do not depend on $\kappa$. These even and odd sectors describe the central delta barrier in a symmetric infinite well.

Subtract the exact solution from the Jacobi method to get $e_n=H{e}_{n-1}$, hence $e_n=H^n{e}_0$. Convergence from every initial vector is equivalent to $H^n\to 0$. If every eigenvalue has modulus below one, a Jordan block $\lambda I+N$ has powers that are finite sums $\left(\genfrac{}{}{0ex}{}{n}{k}\right){\lambda }^{n-k}{N}^k$. The polynomial factor in $n$ is dominated by geometric decay; for $\lambda =0$, the block is nilpotent. Consequently $H^n\to 0$.

Conversely, for a complex eigenvector $v$ with eigenvalue $\lambda$, $H^{n}v={\lambda }^{n}v$. Convergence for real errors also implies convergence for their complexification, so this tends to zero only if $|\lambda |<1$. Therefore the iteration converges from every starting vector if and only if $\rho (H)<1$, where $\rho$ is the spectral radius.

For the specified matrix,Its eigenvalues are $0,\mu ,-\mu$, so $\rho (H)=|\mu |$. Hence $-1<\mu <1$ is the convergence range. The original coefficient matrix has determinant $12(1-{\mu }^2)$, so the endpoints also violate its assumed nonsingularity.

For stereographic projection, choose the north pole $N=(0,0,1)$ of the unit sphere and identify the plane $Z=0$ with the complex plane by $z=X+iY$. Project a point $P=(X,Y,Z)\ne N$ along the line joining it to $N$; adjoin $z=\infty$ for the north pole to identify the whole sphere with the Riemann sphere. A point of that line is $N+t(P-N)$. Its last coordinate vanishes when $t=1/(1-Z)$, givingUsing $X^2+Y^2+Z^2=1$, or solving this line's second intersection with the sphere, gives the inverseA great circle lies in a plane $h_{1}X+h_{2}Y+h_{3}Z=0$ through the origin. Its projected equation isIt is a straight line only if $h_3=0$, meaning the plane contains the north pole and the south pole. Thus if all three sides of a nondegenerate spherical triangle projected to straight sides of a Euclidean triangle, all three distinct side planes would contain both poles. Each pair of side planes could then meet on the sphere only at those same two poles, which cannot supply three distinct triangle vertices. A nondegenerate spherical triangle cannot project to a Euclidean triangle. If a side passes through the projection pole, its image is unbounded and also cannot be a finite triangle side.

A nonidentity rotation in three dimensions has an axis through the origin: an orthogonal three-by-three matrix of determinant one has an eigenvalue one, since nonreal eigenvalues pair as conjugates, real eigenvalues are $\pm 1$, and their product is one. Its axis meets the unit sphere in an antipodal pair, which is fixed. In the identity case every point is fixed, so an antipodal pair can again be selected. Projecting $P$ and $-P$ giveswhenever both points are finite. Thus the associated Möbius transformation fixes an antipodal pair satisfying the claimed relation. The universally valid relation is $q=-1/\overline{p}$ on the Riemann sphere, with $0$ and $\infty$ paired. A rotation about the projection axis fixes exactly that pair, so the PDF's literal finite product needs this extended-coordinate convention. This is the antipodal stereographic coordinate relation.

The monomials $x^i{y}^j$ with $i,j\ge 0$ and $i+j\le n$ form a basis of the given vector space. There are $k+1$ of total degree $k$, soOn that basis the linear operator is diagonal:Both summands of the eigenvalue are nonnegative integers. Their sum is zero precisely when $i,j\in \{0,1\}$. Thus the kernel is the span of those members of $1,x,y,xy$ whose total degree is at most $n$. In particular it is $\mathrm{span}\{1,x,y\}$ for $n=1$ and $\mathrm{span}\{1,x,y,xy\}$ for $n\ge 2$; for $n=0$ it is the constants. The image of a linear map is the span of all the other permitted monomials, since each has a nonzero diagonal eigenvalue and hence is the image of its reciprocal multiple. No excluded monomial can occur in an image because its diagonal coefficient is zero.

The trace is the sum of these eigenvalues. At degrees zero and one all contributions vanish. At degree two the contributions are $2,0,2$, summing to four; at degree three they are $6,2,2,6$, summing to sixteen; at degree four they are $12,6,4,6,12$, summing to forty. ThereforeThis is the diagonal monomial action of a second-order Euler operator.