Polynomial and contour interfaces for stable boundary models

Stable modes of a boundary model require polynomial projections and contour moments at every nilpotent pole order. This lesson establishes those tools; Stable modes and the algebra of boundary data applies them.

Section 13 of Metric and topological foundations proves the full original scalar and finite-coordinate calculus. Its Section 13.12 gives the exact Gaussian, Rayleigh, logarithmic-contour, matrix and half-line receiving maps, retaining every original factor, endpoint and norm.

The complete original real-number, compactness, extrema and finite-norm proofs are in Section 12 of Metric and topological foundations. Its Section 12.10 gives the exact receiving minimum and spectral-contour maps; the original coordinates, norms and constants are retained.

The route has three steps: scalar polynomial division and Bézout give finite algebraic projections; triangularization gives Cayley–Hamilton without using those projections; and directly integrated power series give the contour moments needed for every nilpotent pole order. This last step concerns globally convergent power-series numerators. It does not assert a Cauchy theorem for arbitrary holomorphic functions on arbitrary domains.

1. Facts used from earlier lessons

We use the entry facts from Fourier transforms, finite spectra and convex separation, including real multivariable calculus, Taylor’s formula, smooth compact cutoffs, absolutely integrable change of variables and Fubini, dominated convergence and differentiation, Cauchy–Schwarz, real completeness, finite-dimensional compactness, and the stated Schwartz-space and continuous-dual conventions. This companion needs only the following finite algebra and calculus part of that base, with elementary conventions made explicit.

Sections 10.1–10.9 prove the following finite algebra facts at the stated real/complex scalar base: field arithmetic over $\mathbb C$; finite polynomial arithmetic and induction on degree; basis extension, rank-nullity and direct sums in finite dimension; matrix multiplication, inverses, adjoints and operator norms; determinant multiplicativity, triangular and block-triangular determinants, elementary row operations, the characteristic-polynomial eigenvalue criterion and invariance under change of basis; continuity of determinants and matrix inversion; and trace and rank for projections. Sections 9.1–9.4 prove complex-root existence and full finite factorization with the original leading coefficient and all multiplicities. Sections 9.6–9.7 prove determinant identities, the actual root-to-kernel map and both invariant quotient constructions used below. Section 10.10 proves their exact receiving maps for the earlier spectral and metric arguments. No Jordan decomposition is assumed.

The calculus requirements are the real chain rule and fundamental theorem of calculus on compact intervals, the elementary derivatives of the real logarithm and arctangent, the scalar exponential series, locally uniform power-series calculus, and differentiation or integration of series when the differentiated series converge uniformly on the compact set in question. Ordinary piecewise $C^1$ path integration is defined below by real parametrization. Sections 11.1–11.4 prove matrix-entry dominated differentiation, its original operator-norm comparison and its exact piecewise path integral; Section 11.5 gives the full ordered resolvent and time-derivative receivers. These are the elementary facts used in the arguments below. In particular, the rule allowing termwise differentiation does not include an unstated theorem that every holomorphic function has a global power series.

2. Scalar polynomial division and Bézout

For a field $F$, $f,g\in F[z]$ and $g\ne0$, the division interface supplies unique polynomials $q,r$ such that $$f=gq+r,\qquad r=0\quad\hbox{or}\quad\deg r<\deg g.$$ For $p,q\in F[z]$ not both zero, the Bézout interface supplies a unique monic greatest common divisor $d$ and polynomials $a,b$ with $$d=ap+bq.$$ Here $d$ divides both $p,q$, and every common divisor divides $d$. In particular, relatively prime polynomials have a combination equal to $1$. We use these assertions with $F=\mathbb C$.

Proof of division. Let $e=\deg g$ and let $b\ne0$ be its leading coefficient. If $f=0$ or $\deg f<e$, take $q=0$ and $r=f$. Otherwise let $d=\deg f$, with leading coefficient $a\ne0$, and subtract $(a/b)z^{d-e}g$ from $f$. The leading terms cancel, so the difference is zero or has degree strictly below $d$. Repeating this step terminates because the degree of a nonzero remainder decreases at every step. The sum of the subtracted monomials is $q$, and the final difference is the required $r$. This includes a constant divisor and the zero dividend. If $f=gq+r=gq'+r'$ with both remainders zero or of degree below $e$, then $g(q-q')=r'-r$. A nonzero left side has degree at least $e$, whereas the right side is zero or has degree below $e$. Therefore $q=q'$ and $r=r'$.

Proof of Bézout and the gcd claim. The set of combinations $ap+bq$, with $a,b\in F[z]$, contains a nonzero polynomial because $p,q$ are not both zero. Keep a least-degree nonzero combination $\delta=a^{(0)}p+b^{(0)}q$ with its original leading coefficient $\kappa\ne0$. Divide each original polynomial by $\delta$, using the full pivot coefficient $\kappa$ in the division step proved above. The remainder is still a combination and has smaller degree; it must therefore be zero. Thus $\delta$ divides both polynomials and every common divisor divides $\delta$. Define the comparison name $d=\kappa^{-1}\delta$, retaining the identity $\delta=\kappa d$ and the full Bézout coefficients $\kappa^{-1}a^{(0)},\kappa^{-1}b^{(0)}$. Formula (CR13) proves this exact scalar-unit map and its inverse. Consequently $d$ has the same ideal and divisibility properties. For example dividing $p$ by $d$ has zero remainder: both $p,q$ belong to the original combination ideal and the divisor $d$ is its displayed nonzero unit comparison. The same map proves that $d$ divides $q$. Every common divisor of $p,q$ divides the original combination and hence divides $d$. The chosen representation of $d$ is the Bézout identity. If another monic polynomial has the same gcd property, each divides the other; both quotients have degree zero and monicity makes them $1$. This proves uniqueness, including when exactly one of $p,q$ is zero. The excluded pair $(0,0)$ has no monic greatest common divisor under this convention.

Thomas W. Judson, Abstract Algebra: Theory and Applications, polynomial division, proves scalar division; the same chapter’s gcd proposition proves Bézout. The two details below specify the cases used here.

Two details in the division and gcd proofs matter here. The original nonzero element of the linear-combination ideal remains the working divisor. Its monic comparison retains the full leading coefficient and inverse in (CR14). Uniqueness of the monic gcd follows from mutual divisibility: the quotient is constant by degree, and monicity makes it $1$. Monicity and equality of degrees alone would not prove equality. The pair $(0,0)$ is excluded from this monic-gcd contract. The division assertion includes the zero dividend and a nonzero constant divisor, and does not assign a degree to the zero remainder.

This is scalar division. The typed division $B=Qp+R$ for a monic $\operatorname{End}(E)$-valued polynomial $p$ and a $\operatorname{Hom}(E,G)$-valued polynomial $B$ is proved separately in Stable modes and the algebra of boundary data, Problem 3. That argument keeps $Q$ on the left and preserves every coefficient order; it is not an application of the scalar reading to a noncommutative coefficient ring.

3. From coprime factors to projections

Let $h_1,\ldots,h_r\in\mathbb C[z]$ be pairwise relatively prime nonconstant polynomials, with $r\geq1$, and put $$H=\prod_{j=1}^r h_j,\qquad H_j=\prod_{k\ne j}h_k.$$ There are polynomials $e_j$ such that $$e_j\equiv1\pmod {h_j},\qquad e_j\equiv0\pmod {h_k}\quad(k\ne j),$$ and $$\sum_j e_j\equiv1\pmod H,\qquad e_j^2\equiv e_j\pmod H,\qquad e_je_k\equiv0\pmod H\quad(j\ne k).$$ If $A\in\operatorname{End}(V)$ satisfies $H(A)=0$, the maps $E_j=e_j(A)$ are commuting projections whose pairwise products are zero and whose sum is $I_V$. They give the direct sum $$V=\bigoplus_{j=1}^r\ker h_j(A),\qquad \operatorname{im}E_j=\ker h_j(A).$$

Proof. First $h_j$ and $H_j$ are relatively prime. For each $k\ne j$, choose $u_kh_j+v_kh_k=1$ by Bézout. Multiply these finitely many identities. The term using every $v_kh_k$ is $(\prod_{k\ne j}v_k)H_j$; every other term has a factor $h_j$. Grouping those terms gives an identity $$a_jh_j+b_jH_j=1.$$ For $r=1$, this is simply $0h_1+1\cdot1=1$. Set $e_j=b_jH_j$. Its stated individual congruences follow immediately.

We also need divisibility by the whole product. If $u,v$ are coprime, $u\mid vw$ implies $u\mid w$: multiply $au+bv=1$ by $w$. Consequently, if both $u$ and $v$ divide $f=uw$, then $v\mid w$ and $uv\mid f$. Multiplying Bézout identities as above shows that any subproduct of the $h_j$ is coprime to the next factor. Induction now proves that divisibility by each $h_j$ implies divisibility by $H$. Apply this to $\sum_j e_j-1$, $e_j^2-e_j$ and $e_je_k$, whose remainders modulo every factor vanish.

Evaluation of scalar polynomials at one operator preserves sums and products: expansion of a product gives the same finite coefficient sum because powers of $A$ commute. Multiples of $H$ therefore evaluate to zero. This proves the projection identities. Further, $h_je_j=b_jH$, so $h_j(A)E_j=0$ and $\operatorname{im}E_j\subseteq\ker h_j(A)$. If $h_j(A)v=0$, evaluating $1-e_j=a_jh_j$ on $v$ gives $v=E_jv$, proving the reverse inclusion. The sum identity represents every vector as a sum from the images. If $\sum_jv_j=0$ with $v_j\in\operatorname{im}E_j$, applying $E_k$ gives $v_k=0$; the sum is direct. This proves every assertion.

The empty list is relevant only with $H=1$. The hypothesis $H(A)=0$ then says $I_V=0$, so $V=0$; the empty direct sum and empty sum of projections give exactly that zero space. Powers of distinct linear factors are pairwise coprime: otherwise a nonconstant common divisor has a complex root, which would have to be both distinct roots. No square-free or simple-root assumption has been made.

4. Cayley–Hamilton from an invariant flag

Every endomorphism of a finite-dimensional complex vector space has an upper-triangular matrix in some basis; see Robert A. Beezer, A First Course in Linear Algebra, Theorem UTMR. The zero-dimensional case is the empty basis. Section 9.7 proves both exact routes from the original eigenvalue: the invariant range of $A-\lambda I$ and the eigenline quotient used in the proof below. Both retain their full coordinate changes and block determinants. The induction uses rank-nullity and basis extension and does not depend on Cayley–Hamilton or a primary decomposition.

Proof of triangularization. Induct on $n=\dim V$. For $n=0$ the empty basis works. For $n>0$, the complete proof in Sections 9.1–9.4 gives a root $\lambda$ of $\det(zI-A)$, so the exact free-coordinate kernel map (CR19) gives a nonzero eigenvector of $A$, which we denote by $v_1$. Its span is invariant. The induced operator on the quotient by this line acts on a space of dimension $n-1$; by induction choose an upper-triangular quotient basis and lift its vectors in order to $v_2,\ldots,v_n$. For each $j>1$, upper triangularity in the quotient says that $Av_j$ is a linear combination of $v_1,\ldots,v_j$. The same is true for $Av_1=\lambda v_1$. Thus the lifted basis makes $A$ upper triangular. The induction uses neither Cayley–Hamilton nor a primary decomposition.

For every finite-dimensional complex vector space $V$ and $A\in\operatorname{End}(V)$, set $$\chi_A(z)=\det(zI_V-A).$$ Then $\chi_A(A)=0$, with no restriction on eigenvalue multiplicities, diagonalizability or invertibility.

Proof. Let $n=\dim V>0$, choose the triangular basis $v_1,\ldots,v_n$, and let its diagonal entries be $\lambda_1,\ldots,\lambda_n$. Put $V_j=\operatorname{span}(v_1,\ldots,v_j)$, including $V_0=0$. Upper triangularity says $AV_j\subseteq V_j$. More precisely, $$(A-\lambda_j I)V_j\subseteq V_{j-1}.$$ Indeed, on $v_j$ subtracting $\lambda_jv_j$ removes the last possible coordinate; on each earlier basis vector both terms already lie in $V_{j-1}$.

For $v\in V_n$, apply the factor $A-\lambda_n I$ first, then $A-\lambda_{n-1}I$, and continue downward. The intermediate vectors lie successively in $V_{n-1},V_{n-2},\ldots,V_0$. Thus $$(A-\lambda_1I)\cdots(A-\lambda_nI)v=0.$$ All factors are polynomials in $A$ and hence commute. The triangular determinant identity gives $$\chi_A(z)=\prod_{j=1}^n(z-\lambda_j),$$ so the vanishing product is exactly $\chi_A(A)$. Both determinant and polynomial evaluation commute with change of basis, and the result holds for the original operator. If $n=0$, the determinant is $1$; evaluation gives $I_V$, which is the zero endomorphism of the zero space. This proves that case as well. The characteristic polynomial is the actual $\det(zI-A)$. Its leading coefficient is computed from every original permutation term in (CR17); the full sign comparison with $\det(A-zI)$ is (CR18).

For completeness, this gives the exact minimal-polynomial interface used next in Stable modes and the algebra of boundary data. There is a nonzero annihilating polynomial, namely $\chi_A$. Keep an original least-degree nonzero annihilator $m$ and its original leading coefficient $c_m\ne0$. Formula (CR15) records its exact comparison with the named monic polynomial $\mu_A$: $m=c_m\mu_A$. For any annihilator $f$, divide by the original $m$, with its full leading coefficient, rather than using $\mu_A$ as a replacement. The remainder also annihilates $A$, so it is zero by minimality. Consequently the original annihilator divides every other one. The exact inverse scalar-unit map in (CR13)–(CR15) proves the corresponding divisibility and uniqueness assertion for $\mu_A$. In the zero space it is $1$. In positive dimension it is nonconstant, since a nonzero constant evaluates to a nonzero scalar identity. Sections 9.1–9.4 factor the original $m=c_mH$ into distinct linear powers with every multiplicity retained. Formula (CR16) applies Section 3 through the actual quotient $M_j=m/h_j=c_mH_j$; each projection keeps both $c_m$ and its inverse in its full evaluation. The order of these steps avoids using primary decomposition to prove the Cayley–Hamilton prerequisite for that same decomposition.

5. Contour moments from entire series

A piecewise $C^1$ path is a continuous map $\gamma:[a,b]\to\mathbb C$ that is $C^1$ on each of finitely many subintervals, with the usual one-sided endpoint derivatives. For a continuous function along it, define $$\int_\gamma f(z)\,dz=\sum_{\text{pieces}}\int f(\gamma(s))\gamma'(s)\,ds.$$ A cycle $\Gamma$ here is a finite integer linear combination of such closed paths. Its integrals are the corresponding finite linear combinations. If $\zeta$ avoids their union, define $$\operatorname{ind}_\Gamma(\zeta) =\frac{1}{2\pi i}\int_\Gamma\frac{dz}{z-\zeta}.$$ No general winding-number or homology theorem is needed to make this definition. The indices for the contours we construct will be computed below.

We first record the primitive rule in the form needed here. If a single-valued function $F$ along a path’s neighborhood has a derivative satisfying $\frac{d}{ds}F(\gamma(s))=f(\gamma(s))\gamma'(s)$ on every piece, real integration of its real and imaginary parts gives $$\int_\gamma f(z)\,dz=F(\gamma(b))-F(\gamma(a)).$$ The intermediate endpoints cancel because both $F$ and $\gamma$ are continuous. Every closed-path integral of such a derivative vanishes. In particular, for any integer $\ell\ne-1$, the function $(z-\zeta)^{\ell+1}/(\ell+1)$ is a single-valued primitive of $(z-\zeta)^\ell$ on $\mathbb C\setminus\{\zeta\}$. This uses ordinary integer powers and the real chain rule, including negative powers.

Suppose explicitly that $$\phi(z)=\sum_{r=0}^\infty c_r(z-\zeta)^r$$ is a power series converging on all of $\mathbb C$, locally uniformly. Then for every integer $k\geq0$ and every cycle avoiding $\zeta$, $$\frac{1}{2\pi i}\int_\Gamma \frac{\phi(z)}{(z-\zeta)^{k+1}}\,dz =\operatorname{ind}_\Gamma(\zeta)\frac{\phi^{(k)}(\zeta)}{k!}.$$

Proof. The paths have compact union and finite total length after weighting by the absolute values of the integer coefficients. Their distance from $\zeta$ is positive. Dividing the locally uniform series by $(z-\zeta)^{k+1}$ therefore preserves uniform convergence on that union. For any continuous $g$, the absolute value of its integral is at most this weighted length times $\sup|g|$, so the series may be integrated term by term. The primitive rule kills every term except $r=k$. That remaining term is $c_k\operatorname{ind}_\Gamma(\zeta)$.

Here is the coefficient-derivative justification, independent of a holomorphic Cauchy theorem. Choose $0<R<R'$. Convergence at $z-\zeta=R'$ bounds the sequence $|c_r|(R')^r$ by a finite constant. For every fixed derivative order $j$, the terms of the formally differentiated series on $|z-\zeta|\leq R$ are consequently bounded by a constant times $r^j(R/R')^r$. This numerical series converges: the ratio of successive terms tends to $R/R'<1$. Hence the derivative series converges uniformly there, and the entry-base termwise differentiation rule applies, repeatedly. Evaluation at $z=\zeta$ gives $\phi^{(k)}(\zeta)=k!c_k$, completing the proof.

For any $a\in\mathbb C$, the exponential series gives $$\frac{1}{2\pi i}\int_\Gamma \frac{e^{az}}{(z-\zeta)^{k+1}}\,dz =\operatorname{ind}_\Gamma(\zeta)e^{a\zeta}\frac{a^k}{k!}.$$ Indeed $e^{az}=e^{a\zeta}\sum_{r\geq0}a^r(z-\zeta)^r/r!$; the exponential addition identity follows by the absolutely convergent product of its two series and the binomial formula. Constants and finite polynomials times exponentials also have the required series about every $\zeta$. This includes $1$, $e^{itz}$ and $z^j e^{itz}$ for every $j\geq0$ and every real $t$. Matrix-valued finite sums are handled entrywise. In Stable modes and the algebra of boundary data the resolvent is explicitly decomposed into finitely many nilpotent principal parts, so all pole orders are covered by the formula above; a general residue theorem adds nothing to that calculation.

The phrase “encloses a spectral point once” means precisely index $1$ at that point; “excludes” means index $0$. These conditions, rather than an implicit picture of a contour, govern the matrix projection formula. This theorem says nothing about a merely holomorphic numerator lacking the asserted global series. A later need for a more general Cauchy theorem remains a separate prerequisite.

Jiří Lebl, Guide to Cultivating Complex Analysis, version 1.9, gives a related treatment of line integrals and winding numbers. The primitive rule and exact contour moments required here have been proved above; the stronger disk and general-cycle theorems are not used.

6. Indices of rectangles and circles

Let $a<b$, $c<d$, and let $\partial R$ be the positively oriented boundary of the closed rectangle $$R=\{x+iy:a\leq x\leq b,\ c\leq y\leq d\}.$$ For every $\zeta\notin\partial R$, $$\operatorname{ind}_{\partial R}(\zeta)= \begin{cases} 1,&\zeta\in\operatorname{int}R,\\ 0,&\zeta\notin R. \end{cases}$$

Proof for an interior point. Translate $\zeta$ to zero. Write the positive distances to the left, right, bottom and top sides as $A,B,C,D$, respectively. The translated rectangle is $[-A,B]\times[-C,D]$. For each angle $\theta$, take the applicable positive numbers in the following list: $$\frac{B}{\cos\theta}\ (\cos\theta>0),\qquad \frac{A}{-\cos\theta}\ (\cos\theta<0),\qquad \frac{D}{\sin\theta}\ (\sin\theta>0),\qquad \frac{C}{-\sin\theta}\ (\sin\theta<0).$$ Let $r(\theta)$ be their minimum. At least one number is applicable. The inequalities defining the rectangle say exactly that its intersection with this ray is the segment of radii $0\leq\rho\leq r(\theta)$. Thus $r(\theta)e^{i\theta}$ is its unique boundary point on the ray.

The active side changes only at the four corner directions. On an interval between corner directions, the relevant displayed quotient is smooth with a nonzero denominator. Near an axis, a quotient with denominator tending to zero cannot be the minimum because the other applicable quotient stays bounded. At a corner the two active quotients have the same positive value. These facts prove that $r$ is positive, continuous, $2\pi$-periodic and piecewise $C^1$. They also give a finite partition on which its parametrization is smooth.

As $\theta$ increases by $2\pi$, these boundary points traverse each of the four sides once in positive order. To check the orientation directly, on the right side the height is $B\tan\theta$ and increases; on the top side the horizontal coordinate is $D\cot\theta$ and decreases; on the left side the height is $-A\tan\theta$ and decreases; on the bottom side the horizontal coordinate is $-C\cot\theta$ and increases. On each relevant angular interval these are monotone parametrizations of the corresponding full side or its part split at the starting ray. The real change-of-variable rule therefore identifies this radial integral with the usual positively oriented rectangle integral.

On every smooth piece, with $\gamma(\theta)=r(\theta)e^{i\theta}$, $$\frac{\gamma'(\theta)}{\gamma(\theta)} =\frac{r'(\theta)}{r(\theta)}+i.$$ The integrals of $r'/r$ are real logarithm differences. They telescope across the finitely many endpoints and total zero because $r(2\pi)=r(0)$. The integral of $i$ is $2\pi i$. Division by $2\pi i$ proves the index is $1$.

Proof for an exterior point. If $\zeta\notin R$, at least one real coordinate lies strictly outside the corresponding closed interval. Translate by $-\zeta$, then multiply by some $\alpha\in\{1,-1,i,-i\}$ so that the entire translated rectangle lies in the open right half-plane. For $w=x+iy$ with $x>0$, define the single-valued function $$L(w)=\tfrac12\log(x^2+y^2)+i\arctan(y/x).$$ Real differentiation gives $L_x=1/w$ and $L_y=i/w$. Hence along any piecewise $C^1$ path $w(s)$ in that half-plane, $\frac{d}{ds}L(w(s))=w'(s)/w(s)$. With $w(s)=\alpha(\gamma(s)-\zeta)$, this is $\gamma'(s)/(\gamma(s)-\zeta)$. The primitive rule gives zero around the closed rectangle. This proves the exterior case. Points on its boundary were excluded because the integrand would be singular there.

For clarity, a positively oriented circle of center $c_0$ and radius $R_0>0$ has the same inside/outside index rule. Put $w=\zeta-c_0$ and parametrize by $c_0+R_0e^{i\theta}$. If $|w|<R_0$, the integrand with its parameter differential is $$\frac{i}{1-(w/R_0)e^{-i\theta}}\,d\theta =i\sum_{n\geq0}(w/R_0)^n e^{-in\theta}\,d\theta.$$ Uniform geometric convergence permits integration; only the constant term remains, giving $2\pi i$. If $|w|>R_0$, write it instead as $$-i\sum_{n\geq1}(R_0/w)^n e^{in\theta}\,d\theta;$$ every term integrates to zero. The elementary exponential integrals follow from their explicit primitives. Boundary points are again excluded. Reversing any contour’s orientation negates the index, and taking integer sums adds indices, directly from the definition.

7. Differentiation along a fixed contour

Let $Y\subseteq\mathbb R^d$ be open, let $A:Y\to\operatorname{End}(V)$ be $C^r$, where $r$ is a nonnegative integer or $\infty$, and let $\Gamma$ be a fixed finite cycle with $zI-A(y)$ invertible for all its points and all $y\in Y$. Then $$R(y,z)=(zI-A(y))^{-1}$$ is continuous on $Y$ times the contour and is $C^r$ in $y$. Integration over $\Gamma$ commutes with these parameter derivatives. In particular, $$\partial_{y_j}R=R(\partial_{y_j}A)R,$$ with the order of the three factors as displayed. For a nonzero multi-index $\beta$ of order at most $r$, repeated differentiation gives the useful recursive identity $$\partial^\beta R =\sum_{0<\eta\leq\beta}\binom\beta\eta R(\partial^\eta A)(\partial^{\beta-\eta}R).$$

Proof. Continuity follows from continuity of inversion. On any compact parameter box inside $Y$, its product with the finite contour is compact, so the inverses have a common bound. The inverse is differentiable wherever the matrix is invertible, as follows either by the adjugate divided by the nonzero determinant or by the inverse difference identity. Differentiating $(zI-A)R=I$ once and multiplying on the left by $R$ gives the first formula. Applying the multi-index product rule and isolating the term with no derivative on $zI-A$ gives the recursive formula. Induction shows all derivatives exist, are continuous and are bounded on the compact product. After parametrizing every path piece, these bounds times $|\gamma'|$ are integrable majorants. The complete matrix-entry dominated-differentiation theorem in Section 11.2, applied by the exact path comparison in Section 11.4, proves this interchange coordinate by coordinate and at every derivative order. For $r=0$, uniform continuity on compact products gives continuity of the integral directly. For $r=\infty$, use every finite order in turn.

The same reasoning applies to multiplication by $e^{itz}$, for $t$ in a compact real interval, and to all of its time derivatives $(iz)^j e^{itz}$. On a fixed compact contour they are bounded uniformly for bounded $t$. On an upper contour with $\operatorname{Im}z\geq\delta>0$, their absolute values for $t\geq0$ are at most $|z|^j e^{-\delta t}$. Ordered products of parameter derivatives of the resolvent require this integration estimate, not an additional evaluation theorem for holomorphic matrix functions.

Here is the exact contour-existence boundary used in Stable modes and the algebra of boundary data. Let $K$ be a compact parameter set, let $A$ be continuous on it, and suppose no $A(y)$ has a real eigenvalue. If $K=\varnothing$, any rectangle lying strictly above the real axis satisfies the contour requirements on $K$ vacuously, and the empty neighborhood suffices for the neighborhood assertion below. Assume now $K\ne\varnothing$, and let $M=\max_K\|A(y)\|$. Every eigenvalue satisfies $|\lambda|\leq M$, by applying the norm inequality to an eigenvector. There is also a uniform positive distance $\varepsilon$ from all those eigenvalues to the real axis. Otherwise a sequence of parameters and eigenvalues has, by compactness and the bound $M$, a subsequence tending to $y_*\in K$ and a real $\lambda_*$; continuity of $\det(\lambda I-A(y))$ would make $\lambda_*$ a real eigenvalue of $A(y_*)$, a contradiction.

Choose $R>M+1$ and $0<\delta<\min(\varepsilon,R)$. The positively oriented rectangle with real sides $-R,R$ and imaginary sides $\delta,R$ has index $1$ on every upper eigenvalue and $0$ on every lower eigenvalue, by Section 6. It meets no spectrum and can be kept fixed over $K$. If the family is defined on an open parameter set containing $K$, compactness and determinant continuity preserve invertibility on this contour in some neighborhood of $K$, allowing the preceding differentiation result there. For $V=0$, all invertibility and contour assertions have their unique empty-dimensional meaning and any positive $\varepsilon$ may be used. This is a compact-parameter statement. It does not claim a uniform gap on a noncompact parameter space, choose eigenvalue branches, impose semisimplicity or assert global triviality of a stable bundle.

8. A repeated pole and why the hypothesis matters

Consider the operator $$A=\begin{pmatrix}2i&1&0\\0&2i&0\\0&0&-i\end{pmatrix}$$ and the positive rectangle with real sides $-1,1$ and imaginary sides $1,3$. On the first two coordinates write $A=2iI+N$, where $N^2=0$ and $N\ne0$. Direct multiplication gives $$(zI-A)^{-1}= \begin{pmatrix} (z-2i)^{-1}&(z-2i)^{-2}&0\\ 0&(z-2i)^{-1}&0\\ 0&0&(z+i)^{-1} \end{pmatrix}.$$ The rectangle has indices $1$ at $2i$ and $0$ at $-i$. Its normalized resolvent integral is therefore $$q=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}.$$ The double pole contributes zero to this constant-numerator integral. With numerator $e^{itz}$, however, the $k=1$ moment gives $$\frac{1}{2\pi i}\int_\Gamma e^{itz}(zI-A)^{-1}\,dz =\begin{pmatrix}e^{-2t}&it\,e^{-2t}&0\\0&e^{-2t}&0\\0&0&0\end{pmatrix} =e^{itA}q.$$ The last equality can also be checked from $e^{itN}=I+itN$, obtained by terminating the exponential series. Dropping the double-pole term would accidentally compute the projection correctly but would give the wrong evolution of the second coordinate vector. Thus a projection calculation alone does not check the treatment of Jordan multiplicities in an evolution formula.

Exercise. Let $\Gamma$ be the positive rectangle with real sides $-1,4$ and imaginary sides $-1,1$. The function $\phi(z)=1/(z-3)$ is holomorphic on neighborhoods of the contour and of $0$. Compute $$J=\frac{1}{2\pi i}\int_\Gamma\frac{\phi(z)}z\,dz.$$ Compare it with $\phi(0)$, identify the failed hypothesis of Section 5, and explain why there is no contradiction with that result. Do the computation using only the rectangle indices proved here.

Solution. Partial fractions give $$\frac{1}{z(z-3)}=-\frac{1}{3z}+\frac{1}{3(z-3)}.$$ Both $0$ and $3$ lie inside this rectangle, so both indices are $1$ and $J=-1/3+1/3=0$. On the other hand $\phi(0)=-1/3$. The moment theorem required a globally convergent power series about $0$, which this function does not have: its geometric expansion there has radius $3$, and it has a pole at $3$ inside the contour. Holomorphicity near the contour and evaluation point does not supply the missing hypothesis. No step of the computation uses a general residue theorem, and the result illustrates why the restricted prerequisite must not be advertised as an unrestricted Cauchy theorem.

9. Complex roots with all original coefficients retained

We now prove the complex-root fact used in Sections 1, 3, 4 and 7. The proof keeps the original polynomial and each coefficient, including its leading factor and every translated remainder. We then prove its exact maps to Bézout combinations, original annihilators, spectral projections and finite matrix kernels.

The scalar entry bases are real completeness, field arithmetic, finite polynomial arithmetic and degree induction, real differentiation and integration on compact intervals, and the finite-dimensional definitions in Section 1. Absolute convergence and the required exponential differentiation are justified directly below. No root theorem, primary decomposition or general Cauchy integral theorem is used to prove its own prerequisite.

9.1. The actual exponential and a direction with any prescribed integer power

Define $E(t)=\sum_{k=0}^{\infty}(it)^k/k!$ for real $t$. On $|t|\leq T$ the absolute series and its first derivative are uniformly convergent: once $k+1>2T$, the ratio of successive absolute majorants is less than $1/2$; finitely many earlier terms and the resulting geometric tail are included. To justify the derivative rule directly, finite partial sums obey their integral derivative formula. Uniform limits of the functions and derivatives let that formula pass to the limit on every compact interval, proving $E'=iE$. Absolute convergence, finite binomial expansion and regrouping of the double series give $$E(s)E(t)=\sum_{m\geq0}i^m \sum_{k=0}^m\frac{s^kt^{m-k}}{k!(m-k)!} =\sum_{m\geq0}\frac{i^m(s+t)^m}{m!}=E(s+t). \tag{CR1}$$ Conjugation of the actual series gives $\overline{E(t)}=E(-t)$; hence $E(t)\overline{E(t)}=E(0)=1$. Put $C(t)=\operatorname{Re}E(t)$, $S(t)=\operatorname{Im}E(t)$. Thus $$C'=-S,\quad S'=C,\quad C(0)=1,\quad S(0)=0,\quad C(t)^2+S(t)^2=1. \tag{CR2}$$ There is a first positive zero $t_0$ of $C$, and $0<t_0<2$. For existence, its actual series gives $C(2)=1-2+2/3+\sum_{k\geq3}(-1)^k2^{2k}/(2k)!\leq-1/3<0$: the remaining terms are decreasing in absolute value, and each consecutive negative/positive pair has nonpositive sum. Continuity and the intermediate value property supply a zero in $(0,2)$. The intermediate value property follows directly from real completeness. For a continuous real function $f$ with $f(a)>0>f(b)$, let $c=\sup\{x\in[a,b]:f(x)>0\}$. Positivity near $a$ and negativity near $b$ give $a<c<b$. Points of that set approach $c$, so continuity gives $f(c)\geq0$. If $f(c)>0$, continuity supplies a point of the set strictly to its right, a contradiction. Thus $f(c)=0$. Reversing the signs covers the other case, and subtracting a requested intermediate value covers arbitrary endpoint values. Since $C>0$ near zero, the closed nonempty zero set in a compact subinterval has a positive infimum that belongs to it by continuity. This gives $t_0$; another change of sign before it would have supplied an earlier zero. Therefore $C>0$ on $[0,t_0)$.

The integral $S(t)=\int_0^t C(s)\,ds$ is strictly positive for $0<t\leq t_0$. Formula (CR2) gives $S(t_0)=1$. Also $C$ decreases from one to zero there, because $C'=-S<0$ on the interior. For any unit complex number $a+ib$ in the first quadrant, the intermediate value property supplies $t\in[0,t_0]$ with $C(t)=a$, and positivity plus (CR2) gives $S(t)=b$. Any unit complex number can be multiplied by one of $1,-i,-1,i$ to lie in that quadrant. Since $E(t_0)=i$, (CR1) then shows that it is $E(t+j t_0)$ for some $j\in\{0,1,2,3\}$. Thus every prescribed unit number $d$ is $E(\theta)$ for a real $\theta$. For any positive integer $q$, $$c=E(\theta/q),\qquad |c|=1,\qquad c^q=d. \tag{CR3}$$ This proves the needed direction without assuming a root theorem for an arbitrary complex polynomial. The parameter $t_0$ serves only this proof; no original circle, contour or Fourier constant is being changed.

For any positive real $r$, there is a unique positive $q$-th root of $r$. The continuous function $x\mapsto x^q$ increases strictly on $[0,\infty)$, since for $y>x\geq0$ its difference is $(y-x)\sum_{j=0}^{q-1}y^{q-1-j}x^j>0$. It is zero at zero and exceeds $r$ at $r+1$. The same intermediate value argument gives existence and the strict inequality gives uniqueness. These are the real roots used in the bounds below.

9.2. A minimum of the original polynomial modulus

Let the original nonconstant polynomial be $$p(z)=\sum_{k=0}^N a_k z^k,\qquad N\geq1,\quad a_N\ne0,\quad A_0=\sum_{k=0}^{N-1}|a_k| . \tag{CR4}$$ Its actual leading coefficient remains $a_N$. For $r=|z|\geq1$, the full reverse triangle inequality gives $$|p(z)|\geq |a_N|r^N-\sum_{k=0}^{N-1}|a_k|r^k \geq r^{N-1}(|a_N|r-A_0). \tag{CR5}$$ Choose $$R=1+\frac{2A_0}{|a_N|} +\left(\frac{2(|a_0|+1)}{|a_N|}\right)^{1/N}. \tag{CR6}$$ For $r\geq R$, (CR5) is at least $|a_N|r^N/2$ and exceeds $|a_0|+1$. Thus the global infimum is attained inside the original disk $|z|\leq R$.

Here is the existence argument for the attained minimum. The nonempty set of values on that disk is bounded below by zero; real completeness defines its infimum $b$. Choose original points $z_j$ in the disk with $|p(z_j)|<b+1/j$. A bounded sequence in a closed complex square has a convergent subsequence: repeatedly divide the square into four closed subsquares, retain one with infinitely many original points, and choose increasing original indices in the nested squares. Their diameters tend to zero, so completeness in each real coordinate gives a common limit. The disk is closed, hence its subsequence limit $z_0$ belongs to it. Polynomial continuity follows directly from the full identity $$p(z)-p(w)=(z-w)\sum_{k=1}^N a_k \sum_{j=0}^{k-1}z^{k-1-j}w^j . \tag{CR7}$$ On any fixed disk its finite sum is bounded; thus the difference tends to zero with $z-w$. Consequently $|p(z_0)|=b$. By the strict outside bound and the available value $|p(0)|=|a_0|$, this is a global minimum.

9.3. The exact translated coefficients force a zero

Suppose $b_0=p(z_0)\ne0$. The complete finite binomial formula is $$p(z_0+w)=\sum_{r=0}^N b_r w^r,\qquad b_r=\sum_{k=r}^N a_k\binom{k}{r}z_0^{\,k-r},\qquad b_N=a_N. \tag{CR8}$$ Let $q$ be the smallest positive index with $b_q\ne0$. It exists because $b_N\ne0$; every intervening coefficient is exactly zero. Define the unit direction $$d=-\frac{|b_q|b_0}{|b_0|b_q},\qquad c^q=d,\quad |c|=1,\qquad H=\sum_{r=q+1}^N|b_r|. \tag{CR9}$$ Existence of $c$ is (CR3), with all original factors in $d$. Put $w=tc$. Choose $t>0$, $t<1$, with $|b_q|t^q<|b_0|$ and, if $H>0$, $tH<|b_q|/2$. Such a choice follows from the proved positive real roots; for example take half of the minimum of $1$, $(|b_0|/|b_q|)^{1/q}$ and $|b_q|/(2H)$, omitting only the last bound when its actual denominator $H$ is zero. Then $$\begin{split} |p(z_0+tc)| &\leq |b_0+b_qt^qc^q|+\sum_{r=q+1}^N|b_r|t^r\\ &= |b_0|-|b_q|t^q+\sum_{r=q+1}^N|b_r|t^r\\ &\leq |b_0|-|b_q|t^q+Ht^{q+1}<|b_0| . \end{split} \tag{CR10}$$ The middle equality uses $b_qc^q=-|b_q|b_0/|b_0|$ and the strict bound on $t^q$. The last inequality also holds when $H=0$, since the full higher sum is then zero. This contradicts the global minimum. Hence $p(z_0)=0$. Every nonconstant original complex polynomial has a root, with all its original coefficients and full translated remainder retained. $\square$

9.4. Full factorization, leading coefficient and multiplicities

For an arbitrary $\alpha\in\mathbb C$, the full division identity is $$p(z)=p(\alpha)+(z-\alpha)q_\alpha(z),\qquad q_\alpha(z)=\sum_{k=1}^N a_k \sum_{j=0}^{k-1}z^{k-1-j}\alpha^j . \tag{CR11}$$ It follows from (CR7) with $w=\alpha$. When $p(\alpha)=0$, the quotient has degree $N-1$ and exactly the same leading coefficient $a_N$; its original finite coefficient sums are shown in (CR11). Applying the proved root theorem to each successive nonconstant quotient, and retaining every zero remainder, yields $$p(z)=a_N\prod_{\ell=1}^N(z-\alpha_\ell) =a_N\prod_{\lambda\in Z(p)}(z-\lambda)^{m_\lambda}, \qquad \sum_{\lambda\in Z(p)}m_\lambda=N . \tag{CR12}$$ Here $Z(p)$ is the finite set of distinct roots of the original polynomial, and $m_\lambda$ counts its occurrences in the list, including repetitions. These multiplicities are intrinsic: in (CR8) centered at a root, the least index $r$ with nonzero coefficient is exactly the largest power of $z-\lambda$ dividing $p$, since division at that power leaves a quotient with nonzero value at the root. The derivative identity $p^{(r)}(\lambda)=r!b_r$, obtained by differentiating every finite term, gives the same characterization with its full factorial. Thus distinct factorizations have the same root set and multiplicities. A nonzero constant has no roots and is its original leading constant times the empty product. The zero polynomial has every complex number as a root and no finite multiplicity at any point; it is excluded from the degree-$N$, $a_N\ne0$ theorem. $\square$

9.5. Scalar units and the exact Bézout and annihilator receiving maps

For every original nonzero scalar $c$, multiplication $$U_c:\mathbb C[z]\longrightarrow\mathbb C[z],\quad f\longmapsto cf, \qquad U_{c^{-1}}U_c=U_cU_{c^{-1}}=I \tag{CR13}$$ is a complex-linear and $\mathbb C[z]$-module automorphism. It is not asserted to be a unital algebra homomorphism unless $c=1$. Both identities are pointwise scalar multiplication. The full evaluation relation is $(U_cf)(A)=c f(A)$. Thus its exact inverse preserves roots, multiplicities, divisibility, principal ideals and annihilation; these assertions follow respectively from multiplication by the nonzero number at every scalar point, from (CR12), from $cf=cgq$ and its inverse, from the two ideal inclusions, and from $cf(A)=0$ precisely when $f(A)=0$.

Keep the original least-degree nonzero Bézout combination $\delta=a^{(0)}p+b^{(0)}q$ from Section 2, with original leading coefficient $\kappa\ne0$. Division by $\delta$, using its actual $\kappa$, shows directly that $\delta$ divides both $p,q$; otherwise a nonzero lower-degree remainder would still be an original combination. Every common divisor divides $\delta$. The monic gcd named there is compared by $$\delta=\kappa d,\qquad d=\kappa^{-1}\delta =(\kappa^{-1}a^{(0)})p+(\kappa^{-1}b^{(0)})q . \tag{CR14}$$ Every multiplier and inverse factor is retained. The original $\delta$ and its combination remain in the calculation; (CR13) proves the exact comparison. If $p,q$ are relatively prime, $\delta$ is an actual nonzero constant and its inverse gives the full identity one needed for the projection. The pair $(p,q)=(0,0)$ has no such nonzero combination.

Existence of the original annihilator is supplied by the invariant-flag Cayley–Hamilton proof in Section 4, after the root and triangularization interfaces are proved in Sections 9.3 and 9.6–9.7. Its nonempty set of nonzero annihilators contains the original characteristic polynomial, so the set of degrees has a least integer. For an original least-degree nonzero annihilator $m$ of $A$, with leading coefficient $c_m$, the comparison with the source’s named monic minimal polynomial is $$m=c_m\mu_A,\qquad m(A)=c_m\mu_A(A)=0. \tag{CR15}$$ Divide every other annihilator by the original $m$, retaining its leading coefficient in the division steps. Its remainder still annihilates $A$ and has smaller degree, so is zero. Hence the original $m$ divides every annihilator. This also proves uniqueness up to its exact nonzero scalar unit, and (CR13) proves the comparison with the named $\mu_A$.

In positive dimension (CR12) applied to the original $m$ gives $m=c_m H$, $H=\prod_j h_j$, with $h_j=(z-\lambda_j)^{\nu_j}$, every multiplicity retained. Put $H_j=\prod_{k\ne j}h_k$ and $M_j=m/h_j=c_m H_j$. Distinct linear powers are coprime: a nonconstant common divisor would, by Section 9.3, have a root equal to both distinct $\lambda_j$’s. Section 3 of this lesson proves $a_jh_j+b_jH_j=1$. In the original $M_j$ calculation this is exactly $$a_jh_j+(c_m^{-1}b_j)M_j=1,\qquad E_j=(c_m^{-1}b_j)(A)M_j(A) =c_m^{-1}b_j(A)c_m H_j(A). \tag{CR16}$$ Evaluation retains every unit factor. The full original annihilation $m(A)=c_mH(A)=0$ implies $H(A)=0$ by that displayed inverse, so the existing proof of all projection identities and their kernel/image maps applies. Thus the projection is the same original operator with every coefficient recovered; no leading unit is lost from the calculation. On the zero vector space every scalar identity evaluates to the zero endomorphism; a least-degree annihilator can be any original nonzero constant $c_m$, and the comparison polynomial is $1$, with $m=c_m$ and the empty root list. $\square$

9.6. Original characteristic polynomial and explicit root-to-kernel map

In any fixed original coordinate basis for $V$, with $n=\dim V$, retain the original matrix entries $a_{jk}$. Its characteristic polynomial is the full determinant $$\chi_A(z)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma) \prod_{j=1}^n(z\delta_{j,\sigma(j)}-a_{j,\sigma(j)}) =\sum_{\sigma\in S_n}\sum_{S\subseteq\{1,\ldots,n\}} \operatorname{sgn}(\sigma)z^{|S|} \prod_{j\in S}\delta_{j,\sigma(j)} \prod_{j\notin S}(-a_{j,\sigma(j)}). \tag{CR17}$$ All zero terms, signs and products are accounted for. When $n>0$, the leading term comes from the full $S$ and identity permutation and is $z^n$; the lower coefficients are the remaining full displayed sums. This leading coefficient one is an exact determinant calculation. When $n=0$, the single empty permutation and empty subset give the empty product one, so there are no characteristic roots. The other sign convention has the exact relation $$\det(A-zI)=(-1)^n\chi_A(z). \tag{CR18}$$ It follows by taking a minus sign from each of the original $n$ rows; the zero-dimensional sign is one. Every root and multiplicity agrees through this nonzero constant map.

The determinant identities used below follow from the full coordinate formula rather than an assumed eigenvalue theorem. For arbitrary original matrices $X,Y$, expanding every product gives $$\det(XY)=\sum_{k_1,\ldots,k_n=1}^n \left(\prod_i x_{i,k_i}\right) \sum_{\sigma\in S_n}\operatorname{sgn}(\sigma) \prod_i y_{k_i,\sigma(i)} . \tag{CR18a}$$ If two $k_i$’s agree, exchanging their assigned columns is a fixed-point-free pairing of the inner permutations with equal products and opposite signs, so the entire inner sum is exactly zero. Every remaining tuple is $k_i=\tau(i)$ for a permutation $\tau$. Relabeling its original rows by $j=\tau(i)$ makes the inner sum $\operatorname{sgn}(\tau)\det Y$, since the new column permutation is $\sigma\circ\tau^{-1}$. Summing these surviving tuples is exactly $\det X\det Y$. This proves determinant multiplicativity with every original summand accounted for. It also proves $\det S\,\det S^{-1}=1$ for an invertible $S$, because their product is the actual identity matrix, whose permutation sum is one. Row multilinearity follows by expanding the changed row in the same sum; interchange of two rows changes its sign by relabeling. A matrix with repeated rows has zero determinant by that same sign pairing. Consequently an elementary row addition has its unchanged original determinant, while a row exchange has the full factor minus one.

For the original block matrix of sizes $r,d$, $$\det\begin{pmatrix}X&C\\0&Y\end{pmatrix} =\sum_{\sigma\in S_{r+d}}\operatorname{sgn}(\sigma) \prod_{j=1}^{r+d}b_{j,\sigma(j)} =\det X\,\det Y . \tag{CR18b}$$ Every permutation sending a lower row to an upper column has an actual zero factor. All remaining permutations send the $d$ lower rows bijectively onto the $d$ lower columns, hence send the upper rows onto upper columns. Their signs and products factor into those of the two original blocks, yielding the displayed product. Thus every term containing an upper-right entry of $C$ has a corresponding lower-left zero factor; $C$ remains in the operator and every such determinant term has been explicitly accounted for. The upper triangular case follows by the same argument one block at a time. Each formula includes zero block sizes by its empty-product convention.

For $n>0$, Sections 9.2–9.4 give a root $\lambda$ of the actual $\chi_A$. To construct its entire eigenspace, apply elimination to $B=\lambda I-A$, retaining every actual pivot. Scan columns in their original order, exchange rows when needed, and eliminate below a pivot $d_j\ne0$ by subtracting the full multiple $b_{ik}/d_j$ of the pivot row. Do not replace the pivot by one. Each row exchange is an invertible permutation matrix. Each row addition is $I-tE_{ij}$, with inverse $I+tE_{ij}$, since $i\ne j$ and $E_{ij}^2=0$. Their ordered product $T$ is invertible; its inverse is the reversed product of these exact inverses. Write $E=TB$. Its nonzero rows have pivot columns $c_1<\cdots<c_r$, with entries $d_1,\ldots,d_r$; the remaining rows are zero.

The determinant changes sign at each row exchange and is unchanged by a row addition, by the multilinear alternating determinant formula in (CR17). If $r=n$, the resulting triangular determinant is the full nonzero product of its pivots, so $\det B=(-1)^s\prod_{j=1}^n d_j\ne0$, where $s$ is the number of exchanges. Since $\chi_A(\lambda)=0$, this is impossible; hence $r<n$.

Let $J$ be the original nonpivot column set. For arbitrary $t=(t_k)_{k\in J}\in\mathbb C^J$, prescribe $v_k=t_k$ on those columns and define the pivot coordinates in reverse order by $$v_{c_j} =-d_j^{-1}\sum_{k>c_j}E_{jk}v_k, \qquad j=r,r-1,\ldots,1. \tag{CR19}$$ Every full row is then zero, and $Ev=0$ implies $Bv=0$ through the actual $T^{-1}$. Conversely any kernel vector has precisely these original free coordinates and is recovered by the same formula. Thus (CR19) is a linear bijection $\mathbb C^J\to\ker(\lambda I-A)$, inverse to the actual free-coordinate projection. Since $J\ne\varnothing$, any nonzero free vector gives a nonzero eigenvector $Av=\lambda v$. This proves the exact root-to-kernel map and all its fibre dimensions; it is not just a determinant nonidentity. Invertibility away from the roots follows by the same full-pivot elimination and its actual inverse matrices. $\square$