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Write $B=b+ic$ with $b>0$. Normalization of the Gaussian wave packet gives $|A{|}^2\int e^{-2b{x}^2}dx=1$, hence $|A{|}^2=\sqrt{2b/\pi }$. The supplied Gaussian integral ratio then givesThe momentum operator obeys $p\Psi =2i\hslash Bx\Psi$, and direct second differentiation gives $p^2\Psi =(2{\hslash }^{2}B-4{\hslash }^2{B}^2{x}^2)\Psi$. Taking the expectation value and substituting the first result givessince $2b-B=\bar{B}$. These are the second moments of a complex Gaussian wave packet.

For the solution in part (b), the trigonometric identitygives, writing $D=\cosh (2t)+\cos (2\varphi )$,Thus normalizability requires $\sin (2\varphi )>0$, and the exact expectation values areSince $\cosh (2t)\sim e^{2t}/2$,Their difference is the constant $-\hslash \cot (2\varphi )$, so the mean energy $⟨H⟩=-\hslash \cot (2\varphi )/2$ stays constant despite exponential spreading in both observables.

Set $z=\varphi -it$ with real constant $\varphi$. ThenThus $B=\tan (\varphi -it)/(2\hslash )$ solves the Riccati equation wherever it is finite. Integrating the amplitude equation also giveswith the square root chosen continuously from $t=0$. For the normalizable solutions used in part (c), $\sin (2\varphi )>0$; equivalently one can choose $0<\varphi <\pi /2$ modulo $\pi$. The unrestricted real parameter solves the differential equation locally, but not every value defines a normalizable wavefunction.

For the proposed Gaussian wave packet, differentiation givesInserting these in the Time-dependent Schrödinger equation with Hamiltonian $(p^2-x^2)/2$, its right-hand side becomesEquating the constant and quadratic coefficients with $i\hslash {\partial }_t\Psi$ yieldsConversely these two equations make the substitution an identity, so they are sufficient for the Gaussian evolution in an inverted harmonic oscillator.

Proceed by induction on $\dim V$. The zero-dimensional case has the empty orthonormal basis. In positive dimension, the fundamental theorem of algebra gives a root of the characteristic polynomial, hence a unit eigenvector $e$ of $\alpha$, say with eigenvalue $\lambda$. Part (ii) makes it an eigenvector of ${\alpha }^{\ast }$ as well. Thus the line spanned by $e$ and its orthogonal complement are invariant under both operators by the introductory adjoint argument.

On $\{e{\}}^{\perp }$, the restriction of ${\alpha }^{\ast }$ is the adjoint of the restriction of $\alpha$, and the restrictions still commute. The restricted linear operator is therefore normal. By induction it has an orthonormal basis of eigenvectors. Adjoining $e$ proves an orthonormal eigenbasis exists for $V$, establishing the finite-dimensional spectral theorem by construction.

For the final deduction, use this orthonormal basis $e_j$ and write $\alpha e_j={\lambda }_j{e}_j$. DefineThe real nonnegative diagonal entries make $\beta$ a Hermitian operator, and the modulus-one diagonal entries make $\gamma$ a unitary operator. Both are diagonal in the same orthonormal basis, so they commute and $\beta \gamma e_j={\lambda }_j{e}_j$ even when ${\lambda }_j=0$. This proves the required factorization without assuming invertibility.

Conversely, if $\alpha =\beta \gamma$ with ${\beta }^{\ast }=\beta$, ${\gamma }^{\ast }\gamma =\gamma {\gamma }^{\ast }=I$, and $\beta \gamma =\gamma \beta$, then ${\alpha }^{\ast }={\gamma }^{\ast }\beta$ andThus $\alpha$ is normal. Normality is equivalent to a commuting Hermitian–unitary factorization, with the Hermitian factor even selectable nonnegative. This is the polar decomposition with a unitary extension on the kernel.

Part (ii) shows that ${\alpha }^{\ast }v=\bar{\lambda }v$ for every $v\in E_{\lambda }$, so $E_{\lambda }$ is invariant under ${\alpha }^{\ast }$. Apply the adjoint criterion for an invariant orthogonal complement with ${\alpha }^{\ast }$ in place of $\alpha$, using $({\alpha }^{\ast }{)}^{\ast }=\alpha$. It follows thatThe same argument interchanging $\alpha$ and ${\alpha }^{\ast }$ shows that $E_{\lambda }^{\perp }$ is invariant under both operators, a fact needed for the induction in part (iv).

For any complex $\lambda$, the shifted linear operator $T=\alpha -\lambda I$ is normal: expansion of $T^{\ast }T-T{T}^{\ast }$ leaves just ${\alpha }^{\ast }\alpha -\alpha {\alpha }^{\ast }=0$. Apply part (i) to this shifted normal operator:Hence a nonzero vector is an eigenvector of $\alpha$ with eigenvalue $\lambda$ if and only if it is an eigenvector of ${\alpha }^{\ast }$ with eigenvalue $\bar{\lambda }$. The eigenspaces coincide, with conjugate eigenvalues.

A normal operator satisfies ${\alpha }^{\ast }\alpha =\alpha {\alpha }^{\ast }$. Moving an operator across the inner product givesOne of these norms is zero exactly when the other is zero. Therefore

Integrate $F(z)=(1+z^2{)}^{-(k+1)}$ along the interval $[-R,R]$ closed by the positively oriented upper semicircle, with $R>1$. There is just one enclosed pole, at $i$, of order $k+1$. There,Differentiating $k$ times givesBy part (b),On the semicircle $|1+z^2|\ge R^2-1$, so its integral has absolute value at most $\pi R/(R^2-1{)}^{k+1}$, tending to zero. The real integral converges absolutely, so the residue theorem and the limit $R\to \infty$ give

If the pole has order $k+1$, multiplying the Laurent series by $(z-a{)}^{k+1}$ gives a function $h$ with a holomorphic extension across $a$:Its coefficient of $(z-a{)}^k$ is exactly $c_{-1}$. The Taylor series coefficient formula gives $h^{(k)}(a)=k!c_{-1}$. Continuity of this derivative at $a$ therefore proves

The residue is the coefficient of $(z-a{)}^{-1}$ in the Laurent series:Equivalently it is $\frac{1}{2\pi i}{\int }_{|z-a|=r}f(z)dz$ for a positively oriented circle small enough to enclose no other singularities.

The definition gives $T=DU$. The matrix $D=P^{-1}{D}_{0}P$ is invertible and diagonalizable. In the same basis $U$ is $I+D_0^{-1}N$, which commutes with $D_0$ because $N$ does. Conjugating back givesThis is the multiplicative Jordan decomposition in an affine algebraic group, here applied to the general linear group. The explicit construction also handles repeated eigenvalues and nontrivial Jordan blocks.

For $D=P^{-1}{D}_{0}P$, conjugating $U=D^{-1}T$ givesBecause $D_0^{-1}$ commutes with $N$, and $N^n=0$,Thus $U-I$ is similar to a nilpotent matrix, and is itself nilpotent. Therefore $U$ is a unipotent matrix.

Choose an invertible change-of-basis matrix $P$ so that $J=PT{P}^{-1}$ is the Jordan normal form of $T$. Write each block aswhere $N_{m_j}$ has ones on its first superdiagonal and all other entries zero. DefineEvery ${\lambda }_j\ne 0$ because $T$ is invertible, so $D_0$ is an invertible diagonal matrix. The matrix $N$ is strictly upper triangular and hence nilpotent. On each block $D_0$ is a scalar multiple of the identity, so it commutes with $N$ on that block and therefore on the whole space. This gives

If $U$ is a unipotent matrix, $N=U-I$ is a nilpotent matrix. If $Uv=\lambda v$ for a nonzero eigenvector, then $N^{m}v=(\lambda -1{)}^{m}v=0$ for some $m$, forcing $\lambda =1$.

Conversely, suppose the only eigenvalue is $1$. Over $\mathbb{C}$, the characteristic polynomial is then $(t-1{)}^n$. The Cayley-Hamilton theorem gives $(U-I{)}^n=0$, so $U-I$ is nilpotent. Thus $U$ is unipotent if and only if its only eigenvalue is one.

Continue with $D,D_i$ from part (c). Comparing omission of $x_i$ and $x_n$ givesComparing omission of $x_0$ and $x_n$ gives $D_0=D_n+(x_n-x_0)D$. Eliminating $D$ provesHere $D_n=f[x_0,\dots ,x_{n-1}]$ and $D_0=f[x_1,\dots ,x_n]$, so this is exactly the required omitted-node identities for divided differences. The original nodes need not be ordered, and the formula remains valid even if $\gamma$ lies outside $[0,1]$.

Write $D=f[x_0,\dots ,x_n]$ and let $D_i$ be the divided difference with $x_i$ omitted. By symmetry from part (a), reorder the full list so that its first node is $x_i$ and its last is $x_j$. The recurrence from part (b) then has first omitted-node term $D_i$ and second omitted-node term $D_j$, and givesThis proves the requested identity for every pair $i\ne j$ and fixes its sign explicitly; it is one of the omitted-node identities for divided differences.

For $n\ge 1$, let $p_L,p_R$ be the interpolation polynomials of degree at most $n-1$ on nodes $x_0,\dots ,x_{n-1}$ and $x_1,\dots ,x_n$, respectively. The polynomialhas degree at most $n$. At $x_0$ it equals $p_L(x_0)$; at $x_n$ it equals $p_R(x_n)$. At each common node both interpolants equal $f$, and the displayed weights sum to one, so $p$ also equals $f$ there. Thus it is the full interpolant by uniqueness.

Its coefficient of $x^n$ is the difference of the leading coefficients of $p_R,p_L$ divided by $x_n-x_0$. By the definition of a divided difference,

For distinct interpolation nodes $x_0,\dots ,x_n$, let $p_n$ be the unique polynomial of degree at most $n$ satisfying $p_n(x_i)=f(x_i)$. Existence follows from the Lagrange polynomial expressionuniqueness follows because the difference of two interpolants has $n+1$ distinct zeros but degree at most $n$.

Define the divided difference $f[x_0,\dots ,x_n]$ to be the coefficient of $x^n$ in $p_n$, even when that coefficient is zero. For $n=0$ this gives $f[x_0]=f(x_0)$. Permuting the nodes does not change the interpolation conditions, so it does not change their unique interpolant or this coefficient. The divided difference is symmetric in all the nodes. Equivalently its symmetric expression is

For every $a\in \mathbb{R}$, the nonempty order intervalis open in the lexicographic order topology on the real plane. The sets $U_a$ are pairwise disjoint.

Suppose $\{B_n:n\in \mathbb{N}\}$ were a countable basis of a topology. For each $a$, the basis of a topology property gives a member containing $(a,0)$ and contained in $U_a$. Let $n(a)$ be the least index of such a member. The same nonempty $B_n$ cannot lie in two disjoint $U_a$, so $a\mapsto n(a)$ is an injection from the uncountable set $\mathbb{R}$ into $\mathbb{N}$, a contradiction. This topology is not second countable.

The lexicographic order is a total order. Each point $p=(a,b)$ lies in the order interval between $(a,b-1)$ and $(a,b+1)$, so the order intervals cover ${\mathbb{R}}^2$.

If $p$ lies in both $⟨x_1,y_1⟩$ and $⟨x_2,y_2⟩$, let $x=\max (x_1,x_2)$ and $y=\min (y_1,y_2)$ in that total order. Then $x<p<y$ andIn particular an order interval contains $p$ and is contained in the intersection. Both conditions of part (a) hold, proving that these intervals form a basis of a topology. The resulting topology is the lexicographic order topology on the real plane.