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Use the determinant-normalized slash operator and let ${\mathcal{M}}_n$ be all integral matrices of positive determinant $n$. Their left $S{L}_2(\mathbb{Z})$ cosets have the unique determinant-n matrix representatives for Hecke operatorsInteger row operations reduce the first column to $(a,0{)}^T$, and a row shear then reduces $b$ modulo $d$. DefineChanging representatives does not affect the sum, while right multiplication by the modular group permutes the cosets. Hence $T_{n}f$ has the same modular transformation law. Rational positive-determinant slash transforms are holomorphic on $\mathbb{H}$ and preserve modular cusp vanishing: choose an integral matrix carrying infinity to their rational image modular cusp, after which the remaining matrix is upper triangular. Its positive dilation sends the original decaying modular cusp expansion to another decaying expansion. Thus these are well-defined Hecke operators on $S_k$.

For the Fourier expansion, insert $f(z)={\sum }_{m\ge 1}a(m)e^{2\pi imz}$. The sum over $b$ vanishes unless $d\mid m$, in which case it is $d$. Writing $m=d\ell$, the resulting exponential is $q^{a\ell }$, and its factor is $n^{k-1}{d}^{1-k}=a^{k-1}$. ThereforeIn particular, the first coefficient of $T_{n}f$ is $a(n;f)$.

Assume $f\ne 0$ is a simultaneous Hecke eigenform. The first-coefficient identity gives $a(n)=\lambda (n)a(1)$. If $a(1)=0$, all coefficients vanish, contradicting $f\ne 0$. After dividing by $a(1)$, we may therefore suppose $a(1)=1$, and then $\lambda (n)=a(n)$.

We derive the relations needed for the Euler product. For coprime $m,n$, applying the coefficient formula twice splits the divisors into their disjoint prime supports and yields $T_m{T}_n=T_{mn}$. For one prime define formal Fourier operators $U_p(\sum a(r)q^r)=\sum a(pr)q^r$ and $V_{p}f(z)=f(pz)$. The coefficient formula givesMultiplying this finite sum by $T_p=U_p+p^{k-1}{V}_p$ shows thatThese auxiliary operators need not separately preserve level one; the identities concern their action on Fourier series. Applied to the eigenform, they giveThe prime-power generating function consequently satisfiesMultiplicativity and unique prime factorization now prove the Euler product of a Hecke eigenform:For the original unnormalized form, the right side is multiplied by $a(1)$ and $a(p)$ in the factors is replaced by $\lambda (p)=a(p)/a(1)$. The zero form has $L=0$ and is excluded from the nonzero-eigenform product assertion.

These manipulations initially hold in an absolutely convergent right half-plane. For example, $y^{k/2}|f(z)|$ is bounded by modular invariance and modular cusp decay. The coefficient integral at $y=1/n$ gives $|a(n)|\le C{n}^{k/2}$, so $\mathrm{Re}s>k/2+1$ is sufficient. No analytic continuation is needed to justify the initial product.

Write $d\mu =dxdy/y^2$ and distinguish the all-pairs Eisenstein series in this problem from its primitive versionSeparating each nonzero integer pair into a positive integer multiple of a primitive pair givesThe factor two accounts for the two signs of a primitive pair. It must not be omitted in the Rankin–Selberg unfolding.

Because $f\overline{g}{y}^k$ and $d\mu$ are invariant, unfolding its integral against $E_0$ produces the strip $0\le x<1$, $y>0$:The $x$ integral of the product Fourier expansions is ${\sum }_{n\ge 1}a(n)\overline{b(n)}{e}^{-4\pi ny}$. The Gamma function integral then givesThese steps first follow by absolute convergence sufficiently far right. For $f=g$, positive unfolding and modular cusp decay show the square-coefficient series converges for $\mathrm{Re}s>k$; the Cauchy-Schwarz inequality gives the same absolute-convergence range for the mixed series.

Put $w=s-k+1$ and defineThis is the zeta-completed Rankin-Selberg coefficient series. Equivalently, solving this formula for $F(s)$ expresses the requested series directly in terms of the all-pairs Eisenstein integral.

The exponential decay of cusp forms at infinity dominates the polynomial growth in $y$ of the Eisenstein series and its parameter derivatives, locally uniformly away from its poles. Thus its stated continuation can be passed through the integral. The completed series is meromorphic on $\mathbb{C}$, with possible simple poles only at $s=k-1,k$, and its functional equation isIndeed $w\mapsto 1-w$ corresponds to $s\mapsto 2k-1-s$. If $⟨f,g⟩=0$, both pole residues vanish and the completion is entire.

The normalization of the poles can also be made explicit. The constant term of the given Eisenstein series isThe second term has residue one at $w=1$, since the pole of $\zeta (2w-1)$ has residue $1/2$; the functional equation gives residue minus one at $w=0$. Hence the residues of $\mathcal{R}$ at $k$ and $k-1$ are respectively $⟨f,g⟩$ and $-⟨f,g⟩$, the Petersson inner product.

It is important to separate this completed statement from the raw $F$. Inverting the completion gives meromorphic continuation of $F$, but zeros of $\zeta (2s-2k+2)$ in the denominator can give additional possible poles unless canceled by the integral. Nor does $F$ itself satisfy the unweighted symmetry $s\leftrightarrow 2k-1-s$. At the two displayed completion poles, for a nontrivial modular cusp modular weight,The latter is finite because the pole of $\Gamma (w)$ cancels that of the integral at $w=0$, using $\zeta (0)=-1/2$. In particular, taking $f=g\ne 0$ gives a raw-series pole at $k$ and regularity at $k-1$. The direct analogue of the stipulated two-pole Eisenstein properties is therefore the completed function $\mathcal{R}$, not the raw series with all completion factors suppressed.

Finally suppose the two $L$-series have Euler products. With first coefficients normalized to one, their coefficients are multiplicative, so $a(n)\overline{b(n)}$ is multiplicative too. Thereforeinitially in its absolute-convergence half-plane. For unnormalized forms multiply the product by $a(1)\overline{b(1)}$ and use normalized coefficients inside it.

For the usual degree-two Hecke products of Question 3, take roots ${\alpha }_p,{\beta }_p$ with sum $a(p)$ and product $p^{k-1}$, and ${\gamma }_p,{\delta }_p$ with sum $b(p)$ and the same product. The prime-power recurrence gives $a(p^r)=({\alpha }_p^{r+1}-{\beta }_p^{r+1})/({\alpha }_p-{\beta }_p)$ and the analogous expression for $b$. Multiplying and summing four geometric series proves the Euler factor of a coefficientwise product of Hecke eigenforms:Repeated roots are handled by continuity, or directly by the recurrence. Multiplication by $\zeta (2s-2k+2)$ cancels the local numerators and yields the degree-four convolution product. This also explains why the zeta factor belongs in the natural analytic completion.

Write $a=|A|$ and $b=|B|$, so $p\ge a>b\ge 1$. To cover both the unsaturated and saturated cases with the same coefficient calculation, setChoose $B^{\prime }\subseteq B$ of size $b^{\prime }$. These parameters satisfy $1\le b^{\prime }<a\le p$ and $1\le r\le p$.

Assume, for a contradiction, that the restricted sumset $C$ has fewer than $r$ elements. Extend it to a subset $D\subseteq {\mathbb{F}}_p$ of size $r-1$, and formOn $A\times B^{\prime }$, this polynomial vanishes: when $x=y$ the first factor vanishes, and when $x\ne y$ the sum belongs to $C\subseteq D$. Its total degree is $r=(a-1)+(b^{\prime }-1)$. Only the leading homogeneous part $(x-y)(x+y{)}^{r-1}$ contributes to the coefficient of $x^{a-1}{y}^{b^{\prime }-1}$. That coefficient isThe equality includes $b^{\prime }=1$, with an out-of-range binomial coefficient interpreted as zero. All factorial arguments are between zero and $p-1$, and $0<a-b^{\prime }<p$, so this coefficient is nonzero in ${\mathbb{F}}_p$. The Combinatorial Nullstellensatz applied to $A\times B^{\prime }$ forces a nonzero value, contradicting the vanishing just proved. Hence the restricted sumset bound for unequal subsets of a prime field isThe strict size inequality is used precisely in the nonzero factor $a-b^{\prime }$. The argument includes $r=p$ without dividing by $p!$, so no separate saturation assumption is hidden.

Choose a fixed prime $p\ge k$, and put $s=2p-2$. We first derive the polynomial step: a graph $H$ on $N$ vertices and $M>(p-1)N$ edges, with maximum degree at most $2p-1$, contains a nonempty $p$-regular subgraph. Use a Boolean variable $x_e$ for each edge and work over ${\mathbb{F}}_p$. DefineThe first product has degree at most $(p-1)N<M$, while the full squarefree monomial ${\prod }_e{x}_e$ has coefficient $(-1{)}^{M+1}\ne 0$ in the second term with its minus sign. The Combinatorial Nullstellensatz, with every grid $\{0,1\}$, gives a nonzero value of $R$. Since $R(0)=0$, this is a nonempty edge selection. The second product is then zero, so the first is nonzero. Fermat's little theorem makes each selected vertex degree zero modulo $p$. The maximum-degree bound implies that every positive selected degree is precisely $p$. This proves the prime-regular subgraph from Boolean polynomial constraints.

We next turn density into an almost-regular bipartite subgraph, keeping the logarithmic loss explicit. Let the original graph have $n\ge 2$ vertices and $m>0$ edges. Some bipartition retains at least $m/2$ edges: choose the sides independently with probability $1/2$ and average the number crossing. Its average degree is at least $m/n$. Repeatedly delete a vertex of degree at most half the current average degree. Such a deletion cannot lower that average, sinceThe process cannot delete the last positive-density graph, because the maintained average is positive. At termination the minimum degree is an integer $\sigma >m/(2n)$.

Choose the larger side $A$ of this remaining bipartite graph, with opposite side $B$. Keep exactly $\sigma$ edges at each vertex of $A$; choices at different vertices of $A$ do not conflict. Remove any now-isolated vertices of $B$. Thus $|A|\ge |B|$, and every vertex of $A$ has degree $\sigma$.

Such a half-regular graph contains a half-regular restriction with a perfect matching. Choose a minimal nonempty $X\subseteq A$ satisfying $|N(X)|\le |X|$, which exists because the whole side $A$ satisfies this inequality. In fact equality holds: if it were strict, deleting a vertex of $X$ would retain the inequality, contradicting minimality; the singleton case has at least one neighbor. Every proper subset $Y\subset X$ satisfies $|N(Y)|\ge |Y|$. Hall's marriage theorem therefore supplies a perfect matching on $X\cup N(X)$. All $\sigma$ edges incident with each vertex of $X$ remain inside that restriction.

Delete this perfect matching and repeat the restriction argument. The left degrees decrease by one at each step. We obtain $\sigma$ edge-disjoint perfect matchings $M_0,\dots ,M_{\sigma -1}$ on nested nonempty vertex sets, with sizesFor any $i$ with $i+s\le \sigma -1$, the union of $s+1=2p-1$ consecutive matchings in a graph has maximum degree at most $2p-1$ and at least $(s+1)N_{i+s}/2$ edges on $N_i$ vertices. Ifit has more than $(p-1)N_i$ edges, so the polynomial step yields a $p$-regular bipartite subgraph. Every regular bipartite graph decomposes into perfect matchings: for a set in one part, counting incident edges gives Hall's inequality, so a perfect matching exists; remove it and repeat. Choosing $k$ of these $p$ matchings in a graph gives a nonempty $k$-regular subgraph. This is the nested matching criterion for a regular subgraph.

If no $k$-regular subgraph exists, every displayed size ratio must therefore satisfy the reverse inequality. Iterating at indices $0,s,2s,\dots$ gives, with $r=\lfloor (\sigma -1)/s\rfloor$,Consequently $r\le \log n/\log ((s+1)/s)$ andCombining with $\sigma >m/(2n)$ proves the required edge bound. For example the finite constantworks for every $n\ge 2$, and the case $n=1$ has no edges. ThusThe logarithm comes from how often a nested matching in a graph vertex set can shrink by a fixed factor before becoming empty; it is not supplied by the polynomial step alone.

The coefficient form of the Combinatorial Nullstellensatz is as follows. Let $F$ be a field, let $P\in F[x_1,\dots ,x_r]$ have total degree at most $t_1+\cdots +t_r$, and suppose the coefficient of $x_1^{t_1}\cdots x_r^{t_r}$ is nonzero. For finite subsets $S_i\subset F$ with $|S_i|>t_i$, there is a point of $S_1\times \cdots \times S_r$ where $P$ is nonzero.

For completeness, restrict to $|S_i|=t_i+1$. The coefficient identity isBy Lagrange interpolation, the one-variable weighted sum on the right kills powers below $t_i$ and equals one on the power $t_i$. Any monomial of total degree at most $\sum t_i$, other than the specified one, has some exponent below $t_i$ and therefore contributes zero; the specified monomial contributes its coefficient. Every denominator is nonzero in the field. Hence a nonzero coefficient forces a nonzero grid value. The two applications below use this coefficient form.

For order six, the support-midpoint sites are $x_i=i+3$, soBy part (b), the complete finite matrix is the symmetric five-diagonal matrixOnly indices from $0$ through $n$ are retained; the first and last rows are truncated, with no exterior coefficients or wraparound terms. For a row with all possible neighbors, the strict diagonal dominance margin is $(66-26-26-1-1)/120=1/10$. Boundary rows have an equal or larger margin.

Here is the appropriate inverse estimate with its proof. If a square matrix $A$ has $|a_{ii}|-{\sum }_{j\ne i}|a_{ij}|\ge d>0$, choose an index $i$ maximizing $|z_i|$. The reverse triangle inequality givesThus $Az=0$ implies $z=0$, and a square finite matrix is invertible. Applying the same estimate with $z=A^{-1}b$ proves $\Vert A^{-1}b{\Vert }_{{\ell }^{\infty }}\le d^{-1}\Vert b{\Vert }_{{\ell }^{\infty }}$. The inverse infinity-norm bound from diagonal dominance consequently yieldsThis also proves existence and uniqueness of the requested interpolating spline for every data vector and every finite $n$. The quintic cardinal spline midpoint collocation matrix includes the cases of one or two coefficients, where the dominance margin is even larger.

For real continuous $f$, the Chebyshev alternation theorem says that $p\in {\mathcal{P}}_n$ is the unique best uniform approximation to $f$ precisely when there are $n+2$ ordered points $0\le x_0<\cdots <x_{n+1}\le 1$ and a sign $\sigma \in \{-1,1\}$ such thatHere ${\mathcal{P}}_n$ means degree at most $n$. Best approximants exist: a minimizing sequence is bounded in the finite-dimensional polynomial space, and finite-dimensional equivalence of norms gives a convergent coefficient subsequence. When $f=p$, the zero error is treated separately; the derivative-sign hypothesis below excludes that case.

Let $p_n$ be a best degree-at-most-$n$ approximant and suppose for contradiction that $E_n(f)=E_{n+1}(f)$. Since $f^{(n+1)}>0$, $f$ is not a degree-at-most-$n$ polynomial and its error is nonzero. The same $p_n$ would be best in ${\mathcal{P}}_{n+1}$, so the Chebyshev alternation theorem at that degree supplies $n+3$ alternating error extrema. The intermediate value theorem gives at least $n+2$ distinct zeros of $e=f-p_n$, one between each consecutive pair of extrema.

Repeated Rolle's theorem then forces a zero of $e^{(n+1)}$. But $p_n^{(n+1)}=0$, whereas $e^{(n+1)}=f^{(n+1)}>0$ everywhere. This contradiction provesThe argument proves strict decrease of best polynomial approximation under a derivative sign; an everywhere negative $(n+1)$st derivative gives the same strict conclusion after replacing $f$ by $-f$.

The ordinary interpolation formulation is unambiguous for strictly increasing knots; repeated knots require specified confluent interpolation or a limiting convention. First take distinct knots and $k\ge 2$, and for fixed $t$ write $F_t(u)=(u-t{)}_{+}^{k-1}$. The two degree-at-most-$k-1$ Lagrange interpolation polynomials ${\ell }_i(\cdot ,t)$ and ${\ell }_{i+1}(\cdot ,t)$ agree at the $k-1$ shared knots $t_{i+1},\dots ,t_{i+k-1}$. Their difference is thereforeThis remains valid if their difference has lower degree: the coefficient is then zero. The leading coefficient of a degree-at-most-$k-1$ interpolant is its order-$(k-1)$ divided difference. Comparing leading coefficients and using the divided-difference recursion givesConsequently the Lee interpolation identity holds for every real $x,t$:No differentiability of the truncated power in its argument was needed for this distinct-node proof.

Sum the identity over $i=1,\dots ,n$. The right side telescopes, giving ${\sum }_i{\omega }_i(x)N_i(t)={\ell }_{n+1}(x,t)-{\ell }_1(x,t)$. When $t_k<t<t_{n+1}$, all the nodes for ${\ell }_1$ lie below $t$, so its interpolation data are zero and ${\ell }_1\equiv 0$. All the nodes for ${\ell }_{n+1}$ lie above $t$, where $F_t(u)=(u-t{)}^{k-1}$ is already a polynomial of the admissible degree. Uniqueness of polynomial interpolation gives ${\ell }_{n+1}(x,t)=(x-t{)}^{k-1}$. Therefore the Marsden identity isFor order one the empty product is one, the interpolants are constants and a consistent half-open interval convention for $(u-t{)}_{+}^0$ gives the same telescoping proof. Nondecreasing repeated-knot sequences can be treated by the appropriate confluent or limiting convention wherever that definition is valid; merely specifying repeated data values does not uniquely determine the interpolation polynomials. The basic open knot interval in the Marsden identity avoids endpoint convention ambiguity.

Write $r=x-u^{\ast }$. If $u^{\ast }$ is a best approximation in the finite-dimensional linear subspace, then for every $v$ in it and every real $t$,A nonzero real linear coefficient would make this negative for sufficiently small $t$ of the appropriate sign. Hence $\mathrm{Re}(r,v)=0$. In a complex inner product space, apply the argument to $iv$ too; the imaginary part also vanishes. Thus $(x-u^{\ast },v)=0$ for every $v$ in the subspace.

Conversely, if that orthogonality holds, every other candidate has the form $u^{\ast }+v$, and the Pythagorean theorem in the inner product norm givesEquality requires $v=0$. ThereforeThe next part constructs this orthogonal projection explicitly. No completeness of the whole inner product space is required for projection onto a finite-dimensional subspace.

Take the inner product linear in its first argument; for a real space no conjugation convention is needed. Insert $u^{\ast }={\sum }_{j=1}^n{a}_j{u}_j$ into the orthogonality condition against each basis vector $u_i$. The normal equations areEquivalently $Ha=b$, where $H_{ij}=(u_j,u_i)$ and $b_i=(x,u_i)$. This orientation of the Gram matrix matches the stated inner-product convention. It is Hermitian and positive definite: for a nonzero coefficient vector $a$,by linear independence of the basis. Hence $H$ is invertible and $a=H^{-1}b$ gives existence and uniqueness of $u^{\ast }$. For an orthonormal basis this simplifies to $a_i=(x,u_i)$; for a general basis the Gram inverse correctly accounts for interactions between basis vectors.

First $N(A)$ is finite: by Cauchy-Schwarz, $\Vert Ax{\Vert }_2\le ({\sum }_{ij}{a}_{ij}^2{)}^{1/2}\Vert x{\Vert }_2$. It is nonnegative and absolutely homogeneous. If $N(A)=0$, every $A{e}_j=0$, so every column is zero and $A=0$; the converse is immediate. The triangle inequality for the Euclidean norm gives $N(A+B)\le N(A)+N(B)$ after taking suprema. Thus $N$ is a norm.

For any vector $x$, rescaling gives $\Vert Ax{\Vert }_2\le N(A)\Vert x{\Vert }_2$. Applying this twice givesFurthermore $|a_{ij}|\le \Vert A{e}_j{\Vert }_2\le N(A)$, proving the requested strict entry bound when $N(A)<\epsilon$.

For a polynomial in matrix entries, expand each monomial at $A+H$. The terms involving exactly one entry of $H$ give a linear map $Df(A)[H]$, with coefficients polynomial in the entries of $A$. Every remaining term has at least two entries of $H$ and is $O(N(H{)}^2)$ for $N(H)\le 1$, because $|h_{ij}|\le N(H)$. This proves differentiability. Its derivative is continuous as an operator-valued function: each coefficient is continuous, and for $N(H)\le 1$ every $h_{ij}$ is bounded by one, so the operator-norm difference is bounded by the sum of the coefficient differences. Coordinate continuity follows from the same entry bound. Hence every such polynomial function is continuously differentiable.

In the Leibniz expansion of $\det (I+H)$, the identity permutation supplies $1+{\sum }_i{h}_{ii}$ plus terms of degree at least two. Any nonidentity permutation moves at least two indices, so each of its terms has at least two $H$ entries. ConsequentlyIf $A$ is invertible, multiplicativity of the determinant and the preceding expansion giveThe adjugate identity gives $\mathrm{adj}A=\det (A)A^{-1}$. Thus $d^{\prime }(A)[H]=\mathrm{tr}((\mathrm{adj}A)H)$ for invertible $A$.

For singular $A$, take invertible $A_r\to A$, using the permitted density result. Both $d^{\prime }$ and the adjugate entries are continuous polynomial expressions, so taking limits for every fixed $H$ proves the derivative of the determinant formula on all matrices:Here the standard adjugate is the transpose of the cofactor array, as required by $(\mathrm{adj}A)A=(\det A)I$. In components this derivative is ${\sum }_{ij}{\mathrm{cof}}_{ij}(A)h_{ij}$, also valid at singular matrices.

For an orthogonal change of Cartesian coordinates $x_i^{\prime }=Q_{ij}{x}_j$, an ordinary rank-$n$ tensor transforms byAn isotropic tensor is invariant under such rotations. Orthogonality gives $Q_{ip}{Q}_{jq}{\delta }_{pq}={\delta }_{ij}$, so transforming any of the three products of two Kronecker deltas leaves it unchanged. Their linear combination with arbitrary scalar coefficients is therefore isotropic.

For completeness, these three products exhaust the rank-four isotropic tensors in three dimensions. Half-turns about coordinate axes force a nonzero component to contain each coordinate label an even number of times. Axis interchanges then leave just the values $X=c_{1111}$, $\alpha =c_{1122}$, $\beta =c_{1212}$ and $\gamma =c_{1221}$. Rotating the first two axes by $\pi /4$ gives $X=\frac{1}{2}X+\frac{1}{2}(\alpha +\beta +\gamma )$, so $X=\alpha +\beta +\gamma$. These component values are exactly those of the three-delta-product expression. Thus isotropy of the elastic medium gives this form for its elasticity tensor.

Contracting with the symmetric strain givesThese are the Lamé parameters. Taking the trace determines the scalar part of the strain:Since ${\sum }_{ij}{e}_{ij}^2=3p^2+{\sum }_{ij}{d}_{ij}^2$, the stored strain energy density isA nonzero traceless test strain proves necessity of $\mu \ge 0$, and a pure scalar strain proves necessity of $3\lambda +2\mu \ge 0$. Conversely these two conditions make both terms nonnegative for every strain. Thus nonnegative isotropic elastic strain energy is equivalent to

Assume for contradiction that $n>m$. Apply part b to the matrix of the given relation coefficients, whose columns are nonzero. For its first dependent prefix choose $b_1,\dots ,b_k$ withMultiply the $j$th given vector relation by $b_j$ and sum. Interchanging the finite sums givesThe $w_i$ are a basis, so all $z_i$ must be zero, contradicting part b. Therefore $n\le m$.

Choose a fixed reference basis and let $V,W$ have columns $v_i,w_i$. Both matrices are invertible. The vectors $v_i+\lambda w_i$ are dependent exactly when $\det (V+\lambda W)=0$. This is a polynomial in $\lambda$ of degree $m$, whose leading coefficient is $\det W\ne 0$. A nonzero degree-$m$ polynomial has at most $m$ distinct real roots. Thus there are at most $m$ exceptional parameters. This is the determinant argument for the matrix pencil $V+\lambda W$.

Set $f=p(x)e^{-x^2/2}$. Differentiating givesThus the polynomial must satisfy $p^{\prime \prime }-2x{p}^{\prime }-(\mu +1)p=0$. If it has degree $n$ and nonzero leading coefficient, comparison of highest powers gives $\mu =-(2n+1)$. For this value the equation becomes the Hermite differential equationWrite $p={\sum }_{j=0}^n{a}_j{x}^j$, with coefficients beyond $n$ zero. Comparing powers yieldsThe equation at $j=n-1$ forces $a_{n-1}=0$, and downward recurrence removes all terms of the opposite parity and determines all same-parity terms from $a_n$. Taking $a_n=1$ produces the explicit polynomialIt satisfies every coefficient equation, so existence as well as uniqueness up to a nonzero scalar is proved. HenceThe monic $p_n$ is $2^{-n}$ times the usual physicists' Hermite polynomial.

Multiplication by $x$ transforms to $id/d\lambda$. Write $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$ and use $\hat{g}=cg$ for the Gaussian $g=e^{-x^2/2}$. ThenSuccessive differentiation of the Gaussian gives a polynomial of degree $j$ with leading term $(-1{)}^j{\lambda }^j$ multiplying that same Gaussian. Thus $q_n$ has leading coefficient $c{a}_n(-i{)}^n\ne 0$. This proves $q_n$ is a polynomial of degree exactly $n$, rather than merely at most $n$.

By part a the transformed function solves the same equation with ${\mu }_n$, and by part c its polynomial factor has degree $n$. The uniqueness proved in part b therefore gives $q_n=c_n{p}_n$. Comparing their leading coefficients givesFor the unitary convention used above, the Gaussian integral gives $c=1$, so $c_n=(-i{)}^n$. With the unnormalized phase-$e^{-i\lambda x}$ convention, $c=\sqrt{2\pi }$ instead. This proves that Hermite functions are Fourier eigenfunctions, with the normalization-dependent scalar fully specified.

The displayed matrices are the three Pauli matrices, in the order ${\sigma }_3,{\sigma }_1,{\sigma }_2$. They satisfy $\mathrm{tr}{A}_i=0$ and $\mathrm{tr}(A_i{A}_j)=2{\delta }_{ij}$. Every Hermitian matrix has a unique expression $aI+{\sum }_i{v}_i{A}_i$ with all four coefficients real. Therefore they form a basis together with $I$.

The preceding trace identities give $b(I,I)=-1$, $b(I,A_i)=0$, and $b(A_i,A_j)={\delta }_{ij}$. The Gram matrix is thusConsequently rank $=4$ and inertia signature $(3,1)$. If signature is defined as the number of positive squares minus the number of negative squares, its value is $2$. This is the Hermitian matrix representation of Minkowski four-vectors, with the negative-determinant sign convention.