# Spectral algebra and contour projections This companion supplies the finite-dimensional complex algebra used in [Quadratic Hamilton maps and positive complex planes](quadratic-hamilton-maps-and-positive-complex-planes.md). All spaces here are finite dimensional. No diagonalizability assumption is made for a general complex matrix. Original programme proofs and illustration: GPT-6 Astra (OpenAI), 5 October 2026; CC0. Earlier components retain their own notices. ## A0. Exact earlier inputs [U001 Q5](../20261004-free-stationary-phase/quadratic-stationary-phase.md) proves real orthogonal diagonalization, including singular forms. [M0a](../20261004-free-intrinsic-graph/prerequisites/relative-maslov-line.md) proves the invariance and additivity of inertia. [C0](../20261004-free-intrinsic-graph/prerequisites/conic-frequency-coordinates.md) supplies basis extension, rank and annihilator dimensions. The [U001 finite-calculus proofs](../20261004-free-stationary-phase/prerequisite-completions.md) supply matrix inversion and the inverse theorem. Its [finite algebra and compactness proofs](../20261004-free-stationary-phase/elementary-proof-completions.md) and [complex exponential and circle proofs](../20261004-free-stationary-phase/exponential-prerequisite-completions.md) give determinant identities, compact extrema, convergent geometric sums, complex arithmetic, all circle angles and differentiation of exponentials. Exact entries are recorded in the proof map. Gaussian elimination works over the complex field just as over the real field: choose a nonzero pivot, divide its row by that pivot and subtract multiples from the other rows. The remaining rectangle has one fewer row and column. Induction constructs a basis of the kernel by assigning its free coordinates, and pivot columns give a basis of the image. In particular a square matrix has nonzero determinant exactly when it is invertible, by the determinant row operations and the cofactor inverse. The complex versions of rank-nullity and basis extension used below follow from these constructions. ## A1. Complex polynomial roots and factorization Every nonconstant complex polynomial has a root and factors completely into linear factors with their multiplicities. **Proof.** If its degree is \(d\), its leading term and the triangle inequality give \(|p(z)|\to\infty\) as \(|z|\to\infty\): for \(|z|\ge1\), the sum of all lower terms is at most \(C|z|^{d-1}\), whereas the leading term has size \(|a_d||z|^d\). Thus \(|p|\) attains a global minimum at some \(z_0\), by compactness of a sufficiently large closed disk. Suppose \(p(z_0)\ne0\). Expand the finite polynomial at that point: \[ \frac{p(z_0+w)}{p(z_0)} =1+a w^k+\sum_{j=k+1}^d b_jw^j,\qquad a\ne0,\quad k\ge1. \tag{A1} \] Such a first nonzero coefficient exists, since translation and division by a nonzero constant do not make a nonconstant polynomial constant. The circle-angle theorem P16.3 supplies a real \(\theta\) for which \(a e^{ik\theta}=-|a|\). Explicitly, write \(a/|a|=e^{i\phi}\) and take \(\theta=(\pi-\phi)/k\). For \(w=re^{i\theta}\), \(0k}|b_j|\). Choose \(r\) so small that \(|a|r^k<1\) and \(Cr<|a|\). Then \[ \left|\frac{p(z_0+w)}{p(z_0)}\right| \le 1-|a|r^k+Cr^{k+1}<1, \tag{A2} \] contradicting the minimum. Hence \(p(z_0)=0\). Polynomial division by \(z-z_0\) gives a polynomial of degree \(d-1\) and a constant remainder equal to \(p(z_0)\). This division is obtained by subtracting the leading multiple repeatedly, so each step reduces the degree. The remainder is zero. Induction proves complete factorization. The same division proves that a nonzero degree-\(d\) polynomial has at most \(d\) distinct roots. \(\square\) ## A2. The characteristic identity For a complex \(d\)-by-\(d\) matrix \(T\), its characteristic polynomial \(p(z)=\det(zI-T)\) satisfies \(p(T)=0\). **Proof.** The dimension-zero case is immediate. For \(d\ge1\), cofactor expansion gives the polynomial matrix identity \[ (zI-T)\operatorname{adj}(zI-T)=p(z)I. \tag{A3} \] Write the adjugate as \(\sum_{j=0}^{d-1}A_jz^j\) and \(p(z)=z^d+\sum_{j=0}^{d-1}c_jz^j\). Comparing coefficients gives \[ A_{d-1}=I,\qquad A_{j-1}-TA_j=c_jI\ (1\le j\le d-1), \qquad -TA_0=c_0I. \tag{A4} \] Successive substitution expresses \(A_0\) as \(T^{d-1}+c_{d-1}T^{d-2}+\cdots+c_1I\). The last equation is \(p(T)=0\). This coefficient argument avoids substituting a matrix into a polynomial identity whose other coefficients might not commute with it. \(\square\) ## A3. The full generalized spectral splitting Let the distinct roots of \(p\) be \(\lambda_1,\ldots,\lambda_s\), with multiplicities \(m_i\). Then \[ E=\bigoplus_{i=1}^s V_i,\qquad V_i=\ker(T-\lambda_i I)^{m_i} =\ker(T-\lambda_i I)^d. \tag{A5} \] Each summand is \(T\)-invariant, and \(T=\lambda_i I+N_i\) on it, with \(N_i^{m_i}=0\). **Proof.** Put \(f_i(z)=(z-\lambda_i)^{m_i}\) and \(q_i=p/f_i\). There are polynomials \(u_i,v_i\) with \(u_iq_i+v_if_i=1\). Indeed polynomial long division gives the Euclidean algorithm; successive nonzero remainders have decreasing degree. Its last nonzero remainder divides both polynomials. By A1, a nonconstant common divisor would have a common root, which these two factors do not have. The remainder is a nonzero constant; reverse the divisions and divide by that constant to obtain the asserted identity. Set \(P_i=u_i(T)q_i(T)\). A2 gives \(f_i(T)P_i=0\), so its image lies in \(V_i\). On \(V_i\), the polynomial identity gives \(P_i=I\); on \(V_j\), \(j\ne i\), it gives \(P_i=0\) because \(q_i\) contains the factor \(f_j\). Also \(\sum_i u_iq_i-1\) is divisible by every \(f_i\): reduce modulo \(f_i\) and use the same identity. Pairwise coprimality implies divisibility by their product. For two factors, if \(a\mid r\), \(b\mid r\), and \(ua+vb=1\), writing \(r=ak\) gives \(k=uak+vbk\), whose two terms are divisible by \(b\); hence \(ab\mid r\). Induct over the factors. Consequently \(\sum_iP_i=I\) by A2. The previously checked restrictions prove that this is a direct sum and that the \(P_i\) are its projections. If \(\mu\ne\lambda_i\), then on \(V_i\) \[ (T-\mu I)^{-1} =\sum_{j=0}^{m_i-1} \frac{(-N_i)^j}{(\lambda_i-\mu)^{j+1}}. \tag{A6} \] Multiplication telescopes, proving the formula. Thus the \(d\)th kernel in (A5) has no component in any other summand, and it equals \(V_i\). Invariance follows because all the operators used are polynomials in \(T\). If \(T\) is real, conjugation in (A5) sends \(V_\lambda\) to \(V_{\bar\lambda}\). \(\square\) ## A4. Nilpotent chains and their invariants Every nilpotent endomorphism has a basis consisting of finite chains \(v,Nv,\ldots,N^{m-1}v\). The number of chains is \(\dim\ker N\); their lengths are determined by the dimensions of the kernels of all powers. Applying this on the summands in A3 proves the full Jordan decomposition, over \(\mathbb C\). For a real nilpotent map, the entire construction is real. **Proof.** If the space is nonzero, let \(m\) be the smallest integer with \(N^m=0\), and choose \(v\) with \(N^{m-1}v\ne0\). Its \(m\) chain vectors are independent: in a nontrivial linear relation choose the smallest power \(j\) with nonzero coefficient, then apply \(N^{m-1-j}\). Only that coefficient times \(N^{m-1}v\) survives, a contradiction. Extend these vectors to a basis and choose a linear functional \(\ell\) equal to one on \(N^{m-1}v\) and zero on all the earlier chain vectors. Define \[ P x=\sum_{j=0}^{m-1}\ell(N^{m-1-j}x)\,N^jv. \tag{A7} \] On \(N^kv\) the only nonzero coefficient is the one with \(j=k\); powers at least \(m\) vanish. Thus \(P\) is the identity on the chain span \(C\), has image \(C\), and \(P^2=P\). In \(PNx\) its \(j=0\) coefficient vanishes because \(N^m=0\). For \(j\ge1\), its coefficient at \(N^jv\) is \(\ell(N^{m-j}x)\), exactly as in \(NPx\). Hence \(PN=NP\). It follows that \(E=C\oplus\ker P\) is an invariant splitting. The latter space has smaller dimension and its restriction is nilpotent. Induction completes the chain basis, including \(m=1\). A length-\(m\) chain contributes \(\min(j,m)\) to \(\dim\ker N^j\). Therefore \[ \dim\ker N^j-\dim\ker N^{j-1} =\#\{\text{chains of length at least }j\}. \tag{A8} \] These differences determine the number of each exact length. At \(j=1\) they count all chains. In particular, a nonzero nilpotent space with a one-dimensional kernel consists of one chain. The matrix of \(T=\lambda_i I+N_i\) in its chain basis has diagonal entries \(\lambda_i\); hence the characteristic polynomial on that summand is \((z-\lambda_i)^{\dim V_i}\). Comparison of the factors in A3 gives \(\dim V_i=m_i\). Thus all multiplicities and Jordan lengths used in the lesson are actual conjugacy invariants. \(\square\) ## A5. Circle moments and the exact resolvent projection Let \(\gamma(\theta)=c+Re^{i\theta}\), \(0\le\theta\le2\pi\), with \(R>0\), traversed counterclockwise. For \(\lambda\) off the circle, \[ \frac1{2\pi i}\oint_\gamma\frac{dz}{z-\lambda} =\begin{cases}1,&|\lambda-c|R,\end{cases} \qquad \oint_\gamma\frac{dz}{(z-\lambda)^{j+1}}=0\quad(j\ge1). \tag{A9} \] **Proof.** In the inside case put \(a=(\lambda-c)/R\). Substituting \(dz=iRe^{i\theta}d\theta\) gives \(i/(1-ae^{-i\theta})\). Its geometric series converges uniformly because \(|a|<1\). The constant term integrates to \(2\pi i\); every other term integrates to zero, by \(\int_0^{2\pi}e^{ik\theta}d\theta=0\) for a nonzero integer \(k\), which follows from differentiation and the period \(2\pi\). Uniform tails justify integration term by term. In the outside case expand \[ \frac1{c-\lambda+Re^{i\theta}} =\frac1{c-\lambda} \sum_{k\ge0}\left(-\frac R{c-\lambda}\right)^ke^{ik\theta}. \tag{A10} \] After multiplying by \(iRe^{i\theta}\), every frequency is a positive integer, so the integral is zero. Again the series is uniform. For \(j\ge1\), the integrand has the single-valued primitive \(-(z-\lambda)^{-j}/j\) along the circle. The real chain rule and the fundamental theorem give zero because its endpoint values agree. This proves (A9) without a residue theorem. For a matrix \(T\), restrict to a generalized summand of A3. There the finite identity \[ (zI-T)^{-1} =\sum_{j=0}^{m_i-1}\frac{N_i^j}{(z-\lambda_i)^{j+1}} \tag{A11} \] follows by multiplication, using nilpotence. Applying (A9) shows that the integral \((2\pi i)^{-1}\oint_\gamma(zI-T)^{-1}dz\) is the identity on every inside summand and zero on every outside summand. It is exactly the projection onto the full inside generalized spectral sum and commutes with \(T\). In particular this is valid for nontrivial Jordan blocks. \(\square\) ## A6. A common contour for a continuous matrix family Suppose \(T_t\) is a continuous complex matrix family on \(0\le t\le1\), and every \(T_t\) has no real eigenvalue. There are \(M>0,\delta>0\) for which every spectral value satisfies \[ |\lambda|\le M,\qquad |\operatorname{Im}\lambda|\ge\delta. \tag{A12} \] One fixed upper-half-plane circle encloses all upper spectra and no lower spectra. The associated projections are continuous, with locally constant rank and locally continuous bases of their ranges. **Proof.** The Euclidean norm estimate \(\|T\|\le d\max_{jk}|T_{jk}|\) follows from scalar Cauchy–Schwarz. A continuous finite collection of entries is bounded on the parameter interval; choose \(M>0\) bounding these operator norms. Every eigenvalue has an eigenvector by A0, so \(|\lambda|\le\|T_t\|\le M\). If there were no positive \(\delta\), take parameters \(t_j\) and eigenvalues \(\lambda_j\) with imaginary parts tending to zero. Compactness gives a subsequence with \((t_j,\lambda_j)\to(t,\lambda)\), where \(\lambda\) is real. The determinant is a polynomial in its entries, so \(\det(\lambda I-T_t)=\lim_j\det(\lambda_jI-T_{t_j})=0\). This is a forbidden real eigenvalue. Reduce \(\delta\), if necessary, so \(0<\delta\le M\). Put \[ L=M^2/\delta+\delta,\qquad R=L-\delta/2,\qquad \gamma(\theta)=iL+Re^{i\theta}. \tag{A13} \] The circle and its interior lie above the line \(\operatorname{Im}z=\delta/2\). For an upper spectral value, \[ |\lambda-iL|^2 \le M^2+L^2-2L\delta 0\), set \(b=h(x,y)\) and expand \(0\le h(x-(b/c)y,x-(b/c)y)=h(x,x)-|b|^2/c\). If \(h(y,y)=0\), nonnegativity of \(h(y+tx,y+tx)\) for all complex \(t\) forces \(h(x,y)=0\): otherwise choose the phase of a sufficiently small \(t\) to make its real linear term negative, dominating the quadratic term. This proves (A15). A positive definite Hermitian matrix \(G_0\) has a positive minimum \(a\) on the complex Euclidean unit sphere, regarded as a compact real sphere. If \(\|G-G_0\|0\) for \(v\ne0\). Thus strict positivity is open. For a continuous family of projections of fixed rank, apply this to the Gram matrices of the continuous bases in A6. For later use, a nonempty subset of \([0,1]\) that is both relatively open and closed is the whole interval. If not, choose one point in it and one outside. After reversing the interval if needed, the first is left of the second. The supremum of its points between them belongs to it by closedness and is less than the second point. Openness then gives a still larger point in it, a contradiction. \(\square\) ## A8. Positive square roots and orthonormal bases For a positive definite real symmetric matrix \(A\), Q5 gives an orthogonal \(U\) and positive \(a_j\) with \(A=U\operatorname{diag}(a_j)U^T\). Therefore \[ D=U\operatorname{diag}(\sqrt{a_j})U^T \tag{A16} \] is real symmetric positive definite, invertible, and satisfies \(D^2=A\). The scalar positive roots and their signs are among the U001 inputs. This proves the square-root construction used in the graph normalization. It is the unique positive symmetric square root. If \(E\) is another one, \(EA=EE^2=E^2E=AE\), so \(E\) preserves each eigenspace of \(A\). Its symmetric restriction there has an orthonormal eigenbasis by Q5. On the eigenspace with eigenvalue \(a>0\), every eigenvalue of that restriction is positive and has square \(a\), so it equals \(\sqrt a\). Hence \(E=D\). For any positive real bilinear inner product \(g\), a real basis \(v_1,\ldots,v_k\) can be made \(g\)-orthonormal. After \(e_1,\ldots,e_{j-1}\) have been chosen, set \[ w_j=v_j-\sum_{l