Spectral product from traces of powers

If $H=\{0\}$ or $T=0$, then every positive-degree exterior power of $T$ is zero by [A15], so the determinant series gives $D_T\equiv 1$. There are no nonzero eigenvalues, so the product is empty and equals one. We henceforth assume $H\mathrm{\ne }\{0\}$ and $T\mathrm{\ne }0$.

By [A1], AC supplies the Dependent Choice and Countable Choice assumptions used below. By [A2], $T$ is compact and bounded. The algebra result [A3] applies to the nonzero complex Hilbert space $H$, so the Neumann series [A4] and polynomial spectral mapping [A5] apply in $\mathcal{B}(H)$. The operator spectrum and the algebra spectrum agree by [A5].

For each nonzero eigenvalue $\lambda$ of $T$, an eigenvector $x\mathrm{\ne }0$ satisfies $\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }x\mathrm{\parallel }=\mathrm{\parallel }Tx\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$. Hence $\mathrm{\mid }{\lambda }_j(T)\mathrm{\mid }\le \mathrm{\parallel }T\mathrm{\parallel }$ for every $j$. Let $P_N(z):={\prod }_{j=1}^N(1+z{\lambda }_j(T))$, with $P_0=1$. For $\mathrm{\mid }z\mathrm{\mid }\le R$ and $m<N$, $\mid {\prod }_{j=m+1}^N(1+z{\lambda }_j(T))-1\mid \le {\prod }_{j=m+1}^N(1+R\mathrm{\mid }{\lambda }_j(T)\mathrm{\mid })-1\le e^{R{\sum }_{j>m}\mathrm{\mid }{\lambda }_j(T)\mathrm{\mid }}-1.$ Also $\mathrm{\mid }{P}_m(z)\mathrm{\mid }\le e^{RS}$. Since the right-hand tail tends to zero, $(P_N)$ is uniformly Cauchy on every closed disk of finite radius. Its limit $\mathrm{\Phi }(z):={\lim }_{N\to \mathrm{\infty }}{P}_N(z)$ is locally uniform, and [A13] makes $\mathrm{\Phi }$ entire. For a finite eigenvalue list, zero padding makes $P_N$ eventually constant; for an empty list, every $P_N$ is $1$.

Since $T\mathrm{\ne }0$, $\mathrm{\parallel }T\mathrm{\parallel }>0$. Choose $r>0$ with $r\mathrm{\parallel }T\mathrm{\parallel }<1$ and $rS<1\mathrm{/}2$ (the second condition is automatic when $S=0$). For $\mathrm{\mid }z\mathrm{\mid }<r$, every factor is nonzero, and for every finite $N$, $\mathrm{\mid }{P}_N(z)\mathrm{\mid }\ge {\prod }_{j=1}^N(1-\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }{\lambda }_j(T)\mathrm{\mid })\ge 1-\mathrm{\mid }z\mathrm{\mid }{\sum }_{j=1}^N\mathrm{\mid }{\lambda }_j(T)\mathrm{\mid }>1\mathrm{/}2.$ The finite-product inequality follows by induction from ${\prod }_j(1-a_j)\ge 1-{\sum }_j{a}_j$ for $0\le a_j\le 1$. Passing to the limit shows $\mathrm{\mid }\mathrm{\Phi }(z)\mathrm{\mid }\ge 1\mathrm{/}2$, so $\mathrm{\Phi }$ has no zeros on this disk. Also $\mathrm{\parallel }zT\mathrm{\parallel }<1$, so [A4] makes $I+zT$ invertible; [A16] then gives $D_T(z)\mathrm{\ne }0$ there.

For every integer $n\ge 1$, [A2] shows inductively that $T^n$ is trace class and $\mathrm{\parallel }{T}^n{\mathrm{\parallel }}_1\le \mathrm{\parallel }T{\mathrm{\parallel }}_1\mathrm{\parallel }T{\mathrm{\parallel }}^{n-1}\mathrm{.}$ In particular, $T^n$ is compact and the trace decomposition [A9] applies to $T^n$.

Fix $n\ge 1$ and $\mu \mathrm{\ne }0$ in $\sigma (T^n)$. By [A5], $\mu ={\lambda }^n$ for some $\lambda \in \sigma (T)$. The roots $\alpha$ of $x^n-\mu$ are all distinct: if ${\alpha }^n=\mu \mathrm{\ne }0$, then the derivative $n{\alpha }^{n-1}$ is nonzero. By [A7], $x^n-\mu ={\prod }_{{\alpha }^n=\mu }(x-\alpha ),(T^n-\mu I{)}^k={\prod }_{{\alpha }^n=\mu }(T-\alpha I{)}^k\mathrm{.}$ The factors $(x-\alpha {)}^k$ are pairwise coprime. Put $V:=\ker {\prod }_{{\alpha }^n=\mu }(T-\alpha I{)}^k$. Since its defining polynomial commutes with $T$, $V$ is $T$-invariant. On $V$ that product annihilates $T{\mathrm{\mid }}_V$. Apply [A8] first to one factor and the product of the rest, then repeat on the remaining product-kernel. A vector in any one factor-kernel is already in $V$, so this gives $\ker (T^n-\mu I{)}^k={⨁}_{{\alpha }^n=\mu }\ker (T-\alpha I{)}^k\mathrm{.}$ Choose $k$ at least the finitely many stabilization exponents for $T^n$ at $\mu$ and for $T$ at the roots $\alpha$. By [A6], this proves $G_{\mu }(T^n)={⨁}_{{\alpha }^n=\mu }{G}_{\alpha }(T),m_{\mathrm{a}\mathrm{l}\mathrm{g}}(\mu ;T^n)={\sum }_{\begin{array}{c}{\alpha }^n=\mu \\ \alpha \in \sigma (T)\end{array}}{m}_{\mathrm{a}\mathrm{l}\mathrm{g}}(\alpha ;T)\mathrm{.}$ Roots outside $\sigma (T)$ have zero generalized eigenspace and are omitted. This gives the collision multiplicities for every nonzero eigenvalue of $T^n$.

Apply [A9] to $T^n$ and group by the finitely many roots of each $\mu$. The grouping is legitimate because ${\sum }_{j\ge 1}\mathrm{\mid }{\lambda }_j(T){\mathrm{\mid }}^n\le \mathrm{\parallel }T{\mathrm{\parallel }}^{n-1}{\sum }_{j\ge 1}\mathrm{\mid }{\lambda }_j(T)\mathrm{\mid }<\mathrm{\infty }\mathrm{.}$ Using the multiplicity identity established above gives, for every $n\ge 1$, $\mathrm{tr}(T^n)={\sum }_{\mu \in \sigma (T^n)\setminus \{0\}}{m}_{\mathrm{a}\mathrm{l}\mathrm{g}}(\mu ;T^n)\mu ={\sum }_{j\ge 1}{\lambda }_j(T{)}^n\mathrm{.}$ If the nonzero eigenvalue list is empty, both sides are zero by [A9].

On $\mathrm{\mid }z\mathrm{\mid }<r$, the Neumann expansion from [A4] and the ideal estimate [A2] give convergence in trace norm: $T(I+zT{)}^{-1}={\sum }_{k\ge 0}(-z{)}^k{T}^{k+1},{\sum }_{k\ge 0}\mathrm{\parallel }{z}^k{T}^{k+1}{\mathrm{\parallel }}_1\le \mathrm{\parallel }T{\mathrm{\parallel }}_1{\sum }_{k\ge 0}(\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }T\mathrm{\parallel }{)}^k<\mathrm{\infty }\mathrm{.}$ Trace linearity and its trace-norm bound [A11] therefore permit taking traces term by term. Dividing the logarithmic-derivative identity [A12] by the nonvanishing determinant established above, and using the trace-power identity already established, yields $\frac{D_T^{\mathrm{\prime }}(z)}{D_T(z)}={\sum }_{k\ge 0}(-1{)}^k{z}^k\mathrm{tr}(T^{k+1})={\sum }_{k\ge 0}(-1{)}^k{z}^k{\sum }_{j\ge 1}{\lambda }_j(T{)}^{k+1}\mathrm{.}$

For each finite $N$, the product rule [A13] gives on $\mathrm{\mid }z\mathrm{\mid }<r$ $\frac{P_N^{\mathrm{\prime }}(z)}{P_N(z)}={\sum }_{j=1}^N\frac{{\lambda }_j(T)}{1+z{\lambda }_j(T)}\mathrm{.}$ The denominators are bounded below by $1-r\mathrm{\parallel }T\mathrm{\parallel }>0$. On each closed disk $\mathrm{\mid }z\mathrm{\mid }\le \rho <r$, the right side converges uniformly as $N\to \mathrm{\infty }$, because its tail is bounded by $(1-\rho \mathrm{\parallel }T\mathrm{\parallel }{)}^{-1}{\sum }_{j>N}\mathrm{\mid }{\lambda }_j(T)\mathrm{\mid }$. By [A13], $P_N\to \mathrm{\Phi }$ and $P_N^{\mathrm{\prime }}\to {\mathrm{\Phi }}^{\mathrm{\prime }}$ locally uniformly. Since $\mathrm{\Phi }$ has no zeros on $\mathrm{\mid }z\mathrm{\mid }<r$, taking the limit gives $\frac{{\mathrm{\Phi }}^{\mathrm{\prime }}(z)}{\mathrm{\Phi }(z)}={\sum }_{j\ge 1}\frac{{\lambda }_j(T)}{1+z{\lambda }_j(T)}\mathrm{.}$ For $\mathrm{\mid }z\mathrm{\mid }\le \rho <r$, expand each denominator geometrically. The double series is absolutely convergent because ${\sum }_{j\ge 1}{\sum }_{k\ge 0}{\rho }^k\mathrm{\mid }{\lambda }_j(T){\mathrm{\mid }}^{k+1}\le \frac{S}{1-\rho \mathrm{\parallel }T\mathrm{\parallel }}<\mathrm{\infty }\mathrm{.}$ Thus it may be rearranged, and the trace-power identity gives $\frac{{\mathrm{\Phi }}^{\mathrm{\prime }}(z)}{\mathrm{\Phi }(z)}={\sum }_{k\ge 0}(-1{)}^k{z}^k{\sum }_{j\ge 1}{\lambda }_j(T{)}^{k+1}={\sum }_{k\ge 0}(-1{)}^k{z}^k\mathrm{tr}(T^{k+1})=\frac{D_T^{\mathrm{\prime }}(z)}{D_T(z)}\mathrm{.}$

The disk $U=\{z:\mathrm{\mid }z\mathrm{\mid }<r\}$ is a complex domain. Since $\mathrm{\Phi }$ is nonzero there, $Q:=D_T\mathrm{/}\mathrm{\Phi }$ is holomorphic on $U$. The quotient rule [A14] and the derivative equality above give $Q^{\mathrm{\prime }}=0$ on $U$. Hence [A14] makes $Q$ constant; as $D_T(0)=\mathrm{\Phi }(0)=1$, $Q\equiv 1$ on $U$. Thus $D_T=\mathrm{\Phi }$ on a nonempty open disk. Both functions are entire by [A12] and the locally uniform product construction. The identity theorem [A14] extends their equality to all of $\mathbb{C}$, and the product convergence is locally uniform by construction.

If there is exactly one nonzero eigenvalue in the list, the finite product is the single factor $1+z{\lambda }_1(T)$ and the product and trace-power calculations above still apply. In particular, on a one-dimensional $H$ with $T=tI$, the local determinant construction reduces to ${\det }_H(I+zT)=1+zt$, which is exactly the product when $t\mathrm{\ne }0$, and is the empty product when $t=0$. The zero operator and zero-dimensional space were handled above; finite lists stabilize in the product construction; no finite-dimensionality or nonzero-eigenvalue assumption is made for the general case. All disks used above have positive radius and every estimate is on a compact disk strictly inside the chosen radius; global equality includes every complex endpoint. AC is the exact assumption [A1], inherited by the spectral and trace suppliers, with no additional choice made. The conclusion is an equality, not an iff statement, so both iff directions are inapplicable.