Recall that if *T* ∈ *L*(*x*) and, for some *m* ∈ ℕ, *T*^{m} is a Riesz operator (see, e.g., [15, p. 943] for a definition), then *T* is a Riesz operator too (see, e.g., [12, 3.2.24, p. 147]). In particular, if *T*^{m} is compact, then all non-zero spectral values λ(*T*) ∈ sp (*T*) are eigenvalues of finite (algebraic) multiplicity and have no limit point except possibly zero. Also, in this case the eigenvalue sequences of *T* and *T*^{m} can be arranged in such a way that {λ_{n}(*T*)^{m}} = {λ_{n}(*T*^{m})} (see [12, 3.2.24, p. 147]). Recall that in this case we denote by sp (*T*) (resp., by sp (*T*^{m}) the sequence {λ_{n}(*T*)^{m}} (resp., {λ_{n}(*T*^{m})}).

We are going to present a short proof of the theorem of B. Mityagin from [4, 5]. To clarify our idea of the proof, let us consider firstly the simplest case where *d* = 2.

#### Theorem 5.1

*Let X be a Banach space and T* ∈ *L*(*x*). *Suppose that some power of T is nuclear*. *The spectrum of T is central*-*symmetric if and only if there is an integer K* ≥ 0 *such that for every l* > *K the value* trace *T*^{l} is well defined and trace *T*^{2l+1} = 0 *for all l* > *K*.

#### Proof

Suppose that *T* ∈ *L*(*x*) and there is an *M* ∈ ℕ so that *T*^{M} ∈ *N*(*x*). Fix an odd *N*_{0}, *N*_{0} > *M*, with the property that *T*^{N0} ∈ *N*_{2/3}(*x*) (it is possible since a product of three nuclear operators is 2/3-nuclear) and trace *T*^{N0+2k} = 0 for all *k* = 0, 1, 2, …. By Corollary 3.6, the spectra of all *T*^{N0+2k} are central-symmetric (since, e.g., trace *T*^{N0} = trace (*T*^{N0})^{3} = trace (*T*^{N0})^{5} = … = 0 by assumption). Assume that the spectrum of *T* is not central-symmetric. Then there exists an eigenvalue λ_{0} ∈ sp (*T*) such that –λ_{0} ∉ sp (*T*).

Now,
$\begin{array}{}{\lambda}_{0}^{{N}_{0}}\end{array}$ ∈ sp (*T*^{N0}), so
$\begin{array}{}-{\lambda}_{0}^{{N}_{0}}\end{array}$ ∈ sp (*T*^{N0}). Hence, there exist *μ*_{N0} ∈ sp (*T*) and *θ*_{N0} so that |*θ*_{N0}| = 1,
$\begin{array}{}{\mu}_{{N}_{0}}^{{N}_{0}}=-{\lambda}_{0}^{{N}_{0}}\end{array}$ and *μ*_{N0} = *θ*_{N0} λ_{0}, *θ*_{N0} ≠ –1. Analogously,
$\begin{array}{}{\lambda}_{0}^{{N}_{0}+2}\in \mathrm{sp}\phantom{\rule{thinmathspace}{0ex}}({T}^{{N}_{0}+2}),\text{so}-{\lambda}_{0}^{{N}_{0}+2}\in \mathrm{sp}\phantom{\rule{thinmathspace}{0ex}}({T}^{{N}_{0}+2}).\end{array}$ Hence, there exist *μ*_{N0+2} ∈ sp (*T*) and *θ*_{N0+2} so that |*θ*_{N0+2}| = 1,
$\begin{array}{}{\mu}_{{N}_{0}+2}^{{N}_{0}+2}=-{\lambda}_{0}^{{N}_{0}+2}\end{array}$ and *μ*_{N0+2} = *θ*_{N0+2}λ_{0}, *θ*_{N0+2} ≠ –1 etc. By induction we get the sequences
$\begin{array}{}\{{\mu}_{{N}_{0}+2k}{\}}_{k=0}^{\mathrm{\infty}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\{{\theta}_{{N}_{0}+2k}{\}}_{k=0}^{\mathrm{\infty}}\end{array}$ with the properties that *μ*_{N0+2k} ∈ sp (*T*), |*θ*_{N0+2k}| = 1,
$\begin{array}{}{\mu}_{{N}_{0}+2k}^{{N}_{0}+2k}=-{\lambda}_{0}^{{N}_{0}+2k}\end{array}$ and *μ*_{N0+2k} = *θ*_{N0+2k} λ_{0}, *θ*_{N0+2k} ≠ –1. Since *μ*_{N0+2k} ∈ sp (*T*) and |*μ*_{N0+2k}| = |λ_{0}| > 0, the sequence {*μ*_{N0+2k} is finite as a set, i.e., we have that *μ*_{N0+2K} = *μ*_{N0+2K+2} = … for some *K* > 1. It follows that *θ*_{N0+2K} = *θ*_{N0+2K+2} = …. But
$\begin{array}{}{\theta}_{{N}_{0}+2k}^{{N}_{0}+2k}=-1\end{array}$ for all *k*. Thus
$\begin{array}{}{\theta}_{{N}_{0}+2K}^{l}=-1\end{array}$ for every odd *l* ≥ *N*_{0} + 2*K*. Therefore, *θ*_{N0+2K} = –1. A contradiction. □

Now we are going to consider a general case of a notion of ℤ_{d}-symmetry of a spectra, introduced and investigated by B. Mityagin in [4, 5]. Let *T* be an operator in *X*, all non-zero spectral values which are eigenvalues of finite multiplicity and have no limit point except possibly zero. Recall that we denote by sp (*T*) the corresponding unordered eigenvalue sequence for *T* (possibly, including zero). For a fixed *d* = 2, 3, … and for the operator *T*, the spectrum of *T* is called ℤ_{d}-symmetric, if λ ∈ sp (*T*) implies *t*λ ∈ sp (*T*) for every
$\begin{array}{}t\in \sqrt[d]{1}.\end{array}$

Let *r* ∈ (0, ∞], 𝔻 := {*z* ∈ ℂ: |*z*| < *r*}, *f* : 𝔻 → ℂ. and *d* ∈ ℕ ∖ {1}. We say that *f* is *d*-even if *f*(*tz*) = *f*(*z*) for every
$\begin{array}{}t\in \sqrt[d]{1}.\end{array}$

#### Lemma 5.2

Let *Φ*(*X*) *be a linear subspace of* *X*^{*}⊗̂*X of spectral type l*_{1}, *i*.*e*., *for every v* ∈ *Φ*(*X*) *the series* ∑_{λ∈sp(v͠)} |λ| *is convergent*. *Let d* ∈ ℕ, *d* > 1. *If u* ∈ *Φ*(*X*), *then the Fredholm determinant* det (1 – *zu*) *is d*-*even if and only if the eigenvalue sequence of **u͠* is ℤ_{d}-*symmetric if and only if* trace *u*^{kd+r} = 0 *for all k* = 0, 1, 2, … *and r* = 1, 2, …, *d* – 1.

#### Proof

If the function det (1 – *zu*) is *d*-even, then the eigenvalue sequence of *u͠* is ℤ_{d}-symmetric, since this sequence coincides with the sequence of inverses of zeros of det (1 – *zu*) (according to their multiplicities).

If the eigenvalue sequence of *u͠* is ℤ_{d}-symmetric, then trace *u* = ∑_{λ∈sp (u͠)} λ = 0 (since *Φ*(*X*) is of spectral type *l*_{1} and
$\begin{array}{}\sum _{t\in \sqrt[d]{1}}t=0).\end{array}$ Also, by the same reason trace *u*^{kd+r} = 0 for all *k* = 0, 1, 2, … and *r* = 1, 2, …, *d* – 1, since the spectrum of *u͠*^{l} is absolutely summable for every *l* ≥ 2 and we may assume that {λ_{m}(*u͠*^{l})} = {λ_{m}(*u͠*)^{l}} (for every fixed *l*) [12, 3.2.24, p. 147].

Now, let trace *u*^{kd+r} = 0 for all *k* = 0, 1, 2, … and *r* = 1, 2, …, *d* – 1. By (1), det (1 – *zu*) =
$\begin{array}{}\mathrm{exp}(-\sum _{m=1}^{\mathrm{\infty}}\frac{1}{md}\phantom{\rule{thinmathspace}{0ex}}{z}^{md}\mathrm{trace}\phantom{\rule{thinmathspace}{0ex}}{u}^{md})\end{array}$ in a neighborhood *U* of zero. Therefore, this function is *d*-even in the neighborhood *U*. By the uniqueness theorem det (1 – *zu*) is *d*-even in ℂ. □

#### Corollary 5.3

*For any Banach space X and for every u* ∈ *X*^{*}⊗̂_{2/3}*X the conclusion of Lemma 5.2 holds*.

#### Corollary 5.4

*Let Y be a subspace of a quotient of an L*_{p}-*space*, 1 ≤ *p* ≤ ∞. *For any u* ∈ *Y*^{*}⊗̂_{s}Y, *where* 1/*s* = 1 + |1/2 – 1/*p*|, *the conclusion of Lemma 5.2 holds*.

Now we are ready to present a short proof of the theorem of B. Mityagin [4, 5]. Note that the theorem in [4, 5] is formulated and proved for compact operators, but the proof from [4, 5] can be easily adapted for the general case of linear operators.

#### Theorem 5.5

*Let X be a Banach space and T* ∈ *L*(*X*). *Suppose that some power of T is nuclear*. *The spectrum of T is* ℤ_{d}-*symmetric if and only if there is an integer K* ≥ 0 *such that for every l* > *Kd the value* trace *T*^{l} is well defined and trace *T*^{kd+r} = 0 *for all k* = *K*, *K* + 1, *K* + 2, … *and r* = 1, 2, …, *d* – 1.

#### Proof

Fix *N*_{0} ∈ ℕ such that *T*^{N0} is 2/3-nuclear (it is possible by a composition theorem from [7, Chap, II, Theor. 3, p. 10]). Note that, by A. Grothendieck, the trace of *T*^{l} is well defined for all *l* ≥ *N*_{0}.

Suppose that the spectrum of *T* is ℤ_{d}-symmetric. Take an integer *l* : = *kd* + *r* ≥ *N*_{0} with 0 < *r* < *d*. Since the spectrum of *T*^{l} is absolutely summable, trace *T*^{l} = ∑_{λ∈sp (Tl)} λ and we may assume that {λ_{m}(*T*^{l}) = {λ_{m}(*T*)^{l}}, we get that trace *T*^{kd+r} = 0.

In proving the converse, we may (and do) assume that *Kd* > *N*_{0}. Consider an infinite increasing sequence {*p*_{m}} of prime numbers with *p*_{1} > (*K* + 1)*d*. Assuming that trace *T*^{kd+r} = 0 if *k* = *K*, *K* + 1, *K* + 2, … and *r* = 1, 2, …, *d* – 1, for a fixed *p*_{m} we get from Lemma 5.2 (more precisely, from Corollary 5.3) that the function det (1 – *zT*^{pm}) is *d*-even. Suppose that the spectrum of *T* is not ℤ_{d}-symmetric. Then there exist an eigenvalue λ_{0} ∈ sp (*T*) and a root
$\begin{array}{}\theta \in \sqrt[d]{1}\end{array}$ so that *θ*λ_{0} ∉ sp (*T*). On the other hand, again by Lemma 5.2, the spectrum of *T*^{pm} is ℤ_{d}-symmetric. Since
$\begin{array}{}{\lambda}_{0}^{{p}_{m}}\in \mathrm{sp}\phantom{\rule{thinmathspace}{0ex}}({T}^{{p}_{m}}),\end{array}$ there exists *μ*_{m} ∈ sp (*T*) such that
$\begin{array}{}{\mu}_{m}^{{p}_{m}}={\theta}^{{p}_{m}}{\lambda}_{0}^{{p}_{m}}\in \mathrm{sp}\phantom{\rule{thinmathspace}{0ex}}({T}^{{p}_{m}});\end{array}$ hence, *μ*_{m} = *θ*_{m} λ_{0} for some *θ*_{m} with |*θ*_{m}| = 1. But |λ_{0}| > 0. Therefore, the set {*μ*_{m}} is finite and it follows that there is an integer *M* > 1 such that *θ*_{M} = *θ*_{M+1} = *θ*_{M+2} = …. Hence,
$\begin{array}{}{\theta}^{{p}_{m}}={\theta}_{M}^{{p}_{m}}\end{array}$ for all *m* ≥ *M*. Thus, *θ*_{M} = *θ*. A contradiction. □

Let us give some examples in which we can apply Theorem 5.5, but the main result of [4, 5] does not work.

#### Example 5.6

*Let Π*_{p} be the ideal of absolutely p-*summing operators* (*p* ∈ [1, ∞); *see [10] for a definition and related facts)*. *Then for some n one has*
$\begin{array}{}{\mathit{\Pi}}_{p}^{n}\subset N.\end{array}$ *In particular*,
$\begin{array}{}{\mathit{\Pi}}_{2}^{2}(C[0,1])\subset N(C[0,1]),\end{array}$ *but not every absolutely 2*-*summing operator in C*[0, 1] *is compact*. *Another interesting example*:
$\begin{array}{}{\mathit{\Pi}}_{3}^{3}\end{array}$ *is of spectral type l*_{1} [16]. *We do not know (maybe it is unknown to everybody)*, *whether the finite rank operators are dense in this ideal*. *However*, *Theorem 5.5* *may be applied*. *Moreover*, *it can be seen that*, *for example*, *the spectrum of an operator* *T from*
$\begin{array}{}{\mathit{\Pi}}_{3}^{3}\end{array}$ *is central*-*symmetric if and only if the spectral traces of the operators T*^{2k–1} *are zero for all k* > 0.

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