In this section, we give S-type Z-eigenvalue inclusion sets of the tensor \(\mathcal{A}\) by dividing N into disjoint subsets S and S̄, where S̄ is the complement of S in N. Furthermore, we establish comparisons among different Z-eigenvalue inclusion sets.
In what follows, we introduce a lemma for a general tensor.
Lemma 1
Theorem 3.1 of [1]
Let
\(\mathcal{A}\)
be a tensor with order
m
and dimension
\(n\geq2\). Then all
Z-eigenvalues of
\(\mathcal{A}\)
are located in the union of the following sets:
$$ \sigma(\mathcal{A})\subseteq\mathcal{K}(\mathcal {A})=\bigcup _{i\in N}\mathcal{K}_{i}(\mathcal{A}), $$
where
\(\mathcal{K}_{i}(\mathcal{A})=\{z\in\mathcal{C}:|z|\leq R_{i}(\mathcal{A})\}\)
and
\(R_{i}(\mathcal{A})=\sum_{i_{2},\ldots ,i_{m}\in N}|a_{{i}{i_{2}}\ldots{i_{m}}}|\).
By using the partition technique in [16], we present the following notations. Let \(\mathcal{A}\) be an mth order n-dimensional tensor and S be a nonempty proper subset of N. Set
$$\begin{gathered} \Delta^{N} :=\bigl\{ (i_{2},i_{3}, \ldots,i_{m}): \text{each }i_{j} \in N\text{ for }j = 2,\ldots,m\bigr\} , \\\Delta^{S} :=\bigl\{ (i_{2},i_{3}, \ldots,i_{m}): \text{each } i_{j} \in S\text{ for }j = 2,\ldots,m\bigr\} , \\\overline{\Delta^{S}} = \Delta^{N} \setminus \Delta^{S}. \end{gathered}$$
Then
$$ R_{i}(\mathcal{A})=\sum_{i_{2},\ldots,i_{m} \in N}|a_{{i}{i_{2}}\ldots {i_{m}}}|=R_{i}^{\Delta^{S}}( \mathcal{A})+ R_{i}^{\overline{\Delta ^{S}}}(\mathcal{A}),\quad \forall i \in S, $$
where
$$ R_{i}^{\Delta^{S}}(\mathcal{A}) = \sum _{ ( i_{2},\ldots,i_{m}) \in{\Delta^{S}} } |a_{{i}{i_{2}}\ldots{i_{m}}}|,\qquad R_{i}^{\overline{\Delta^{S}}}( \mathcal {A}) = \sum_{ (i_{2},\ldots,i_{m}) \in\overline{\Delta^{S}} }|a_{{i}{i_{2}}\ldots{i_{m}}}|. $$
Theorem 1
Let
\(\mathcal{A}\)
be a tensor with order
m
and dimension
\(n\geq2\)
and
S
be a nonempty proper subset of
N. Then all
Z-eigenvalues of
\(\mathcal{A}\)
are located in the union of the following sets:
$$ \sigma(\mathcal{A})\subseteq\mathcal{G}^{S}(\mathcal{A})= \biggl( \bigcup_{i\in S,j\in\bar{S}}\mathcal{G}^{S}_{i,j}( \mathcal{A}) \biggr) \cup \biggl( \bigcup _{i\in\bar{S},j\in S}\mathcal{G}^{\bar{S}}_{i,j}(\mathcal {A}) \biggr), $$
where
$$ \begin{gathered} \mathcal{G}_{i,j} ^{S}(\mathcal{A}) = \bigl\{ z\in \mathcal{C}:|z| \bigl(|z|- R_{j}^{\overline{\Delta^{S}}}(\mathcal{A}) \bigr) \leq R_{i} (\mathcal {A})R_{j}^{\Delta^{S}}(\mathcal{A}) \bigr\} , \\ \mathcal{G}_{i,j} ^{\bar{S}}(\mathcal{A}) = \bigl\{ z\in\mathcal {C}:|z| \bigl(|z|-R_{j}^{\overline{\Delta^{ \bar{S}}}}(\mathcal{A}) \bigr) \leq R_{i} (\mathcal{A})R_{j}^{\Delta^{\bar{S}}}(\mathcal{A}) \bigr\} . \end{gathered}$$
Proof
Let λ be a Z-eigenvalue of \(\mathcal{A}\) with corresponding eigenvector x, i.e.,
$$ \mathcal{A}x^{m-1}=\lambda x,\qquad x^{T}x=1. $$
(1)
Let \(|x_{t}|={\max\{|x_{i}|: i\in S\}}\), \(|x_{s}|= {\max\{ |x_{i}|: i\in\bar{S}\}}\). Then at least one of \(|x_{t}|\) and \(|x_{s}|\) is nonzero. We next divide the proof into three parts.
(i) If \(x_{t}x_{s}\neq0\) and \(|x_{s}|\geq|x_{t}|\), then \(|x_{s}|={\max\{|x_{i}|: i\in N\}} > 0\). From equality (1), we have
$$\lambda x_{s} = \sum_{ (i_{2},\ldots, i_{m}) \in\Delta^{S} } a_{s i_{2}\ldots i_{m}} x_{i_{2}}\cdots x_{i_{m}}+\sum _{(i_{2},\ldots, i_{m} ) \in\overline{\Delta^{S}} } a_{s i_{2}\ldots i_{m}} x_{i_{2}} \cdots x_{i_{m}}. $$
Noting that \(|x_{t}|^{m-1}\leq|x_{t}| \leq1\), \(|x_{s}|^{m-1}\leq|x_{s}| \leq 1\) and taking modulus in the above equation, one has
$$ \begin{aligned}[b]|\lambda||x_{s}| &\leq\sum _{(i_{2},\ldots, i_{m}) \in\Delta^{S} }|a_{s i_{2}\ldots i_{m}}||x_{i_{2}}| \cdots|x_{i_{m}}| + \sum_{(i_{2},\ldots, i_{m} ) \in\overline{\Delta^{S}} }|a_{s i_{2}\ldots i_{m}}||x_{i_{2}}| \cdots |x_{i_{m}}| \\ &\leq\sum_{(i_{2},\ldots, i_{m}) \in\Delta^{S} }|a_{s i_{2}\ldots i_{m}}||x_{t}|^{m-1} + \sum_{(i_{2},\ldots, i_{m} ) \in\overline{\Delta ^{S}} }|a_{s i_{2}\ldots i_{m}}||x_{s}|^{m-1} \\ &\leq R_{s}^{\Delta^{S}}(\mathcal{A})|x_{t}| + R_{s}^{\overline{\Delta ^{S}}}(\mathcal{A})|x_{s}|. \end{aligned}$$
(2)
Dividing both sides by \(|x_{s}|\) in (2), we get
$$ |\lambda| \leq R_{s}^{\Delta^{S}}(\mathcal{A}) \frac{|x_{t}|}{|x_{s}|} + R_{s}^{\overline{\Delta^{S}}}(\mathcal{A}). $$
(3)
On the other hand, by (1), we obtain
$$ |\lambda||x_{t}| \leq\sum_{i_{2},\ldots,i_{m}\in N}|a_{{t}{i_{2}}\ldots {i_{m}}}||x_{i_{2}}| \cdots|x_{i_{m}}|\leq\sum_{i_{2},\ldots,i_{m}\in N}|a_{{t}{i_{2}}\ldots{i_{m}}}||x_{s}|^{m-1}. $$
Dividing both sides by \(|x_{t}|\) in the above inequality and from \(|x_{s}|^{m-1}\leq|x_{s}|\), one has
$$ |\lambda|\leq\sum_{i_{2},\ldots,i_{m}\in N}|a_{{t}{i_{2}}\ldots{i_{m}}}| \frac {|x_{s}|^{m-1}}{|x_{t}|}\leq \sum_{i_{2},\ldots,i_{m}\in N}|a_{{t}{i_{2}}\ldots {i_{m}}}| \frac{|x_{s}|}{|x_{t}|}= R_{t}(\mathcal{A}) \frac{|x_{s}|}{|x_{t}|}. $$
(4)
Multiplying (3) by (4), we see
$$ |\lambda| \bigl(|\lambda|- R_{s}^{\overline{\Delta^{S}}}(\mathcal{A}) \bigr) \leq R_{t}(\mathcal{A})R_{s}^{\Delta^{S}}(\mathcal{A}), $$
thus, \(\lambda\in\mathcal{G}_{t,s} ^{S}(\mathcal{A})\subseteq \mathcal {G}^{S}(\mathcal{A})\).
(ii) If \(x_{t}x_{s}\neq0\) and \(|x_{t}|\geq|x_{s}|\), then \(|x_{t}|={\max\{|x_{i}|: i\in N \}}\). Similar to the proof of (i), we can get that
$$ |\lambda|- R_{t}^{\overline{\Delta^{\bar{S}}}}(\mathcal{A})\leq R_{t}^{\Delta^{\bar{S}}}( \mathcal{A}) \frac{|x_{s}|}{|x_{t}|} $$
and
$$ |\lambda|\leq R_{s}(\mathcal{A}) \frac{|x_{t}|}{|x_{s}|}, $$
which implies
$$ |\lambda| \bigl(|\lambda|- R_{t}^{\overline{\Delta^{\bar{S}}}}(\mathcal{A}) \bigr) \leq R_{s} (\mathcal{A}) R_{t}^{\Delta^{\bar{S}}}(\mathcal{A}), $$
that is, \(\lambda\in \mathcal{G}_{s,t} ^{\bar{S}}(\mathcal {A})\subseteq \mathcal{G}^{S}(\mathcal{A})\).
(iii) If \(x_{t}x_{s} = 0\), without loss of generality, let \(|x_{t}| = 0\) and \(|x_{s}|\neq0\). It follows from (3) that
$$ |\lambda|- R_{s}^{\overline{\Delta^{S}}}(\mathcal{A})\leq 0. $$
For any \(i\in S\), we have
$$ |\lambda| \bigl(|\lambda|- R_{s}^{\overline{\Delta^{S}}}(\mathcal{A}) \bigr) \leq0 \leq R_{i}(\mathcal{A})R_{s}^{\Delta^{S}}( \mathcal{A}), $$
that is, \(\lambda\in\mathcal{G}_{i,s} ^{S}(\mathcal{A})\subseteq \mathcal{G}^{S}(\mathcal{A})\).
The result follows from (i), (ii) and (iii). □
Corollary 1
Let
\(\mathcal{A}\)
be a tensor with order
m
and dimension
\(n\geq2\), and
S
be a nonempty proper subset of
N. Then
$$ \sigma(\mathcal{A})\subseteq\mathcal{G}^{S}(\mathcal{A})\subseteq \mathcal{K}(\mathcal{A}), $$
where
\(\mathcal{K}(\mathcal{A})\)
is a
Z-eigenvalue inclusion set in Lemma
1.
Proof
Let z be a point of \(\mathcal{K}(\mathcal{A})\). Two cases are discussed as follows:
(i) There exist \(t\in S \) and \(s \in\bar{S} \) such that \(z \in\mathcal {G}_{t,s} ^{S}(\mathcal{A})\), i.e.,
$$ |z| \bigl(|z|- R_{s}^{\overline{\Delta^{S}}}(\mathcal{A}) \bigr) \leq R_{t} (\mathcal {A})R_{s}^{\Delta^{S}}( \mathcal{A}). $$
(5)
If \(R_{t} (\mathcal{A})R_{s}^{\Delta^{S}}(\mathcal{A}) = 0 \), then \(z = 0\) or \(|z|- R_{s}^{\overline{\Delta^{S}}}(\mathcal{A}) \leq0\). Hence, \(z\in\mathcal{K}_{t} (\mathcal{A})\cup\mathcal{K}_{s} (\mathcal{A})\). Otherwise, it follows from (5) that
$$ \frac{|z|}{R_{t}(\mathcal{A})}\frac{|z|- R_{s}^{\overline{\Delta ^{S}}}(\mathcal{A})}{R_{s}^{\Delta^{S}}(\mathcal{A})}\leq1. $$
(6)
Furthermore,
$$ \frac{|z|}{R_{t}(\mathcal{A})}\leq1\quad \text{or} \quad\frac{|z|- R_{s}^{\overline{\Delta^{S}}}(\mathcal{A})}{R_{s}^{\Delta^{S}}(\mathcal {A})}\leq1, $$
that is, \(z\in\mathcal{K}_{t} (\mathcal{A})\) or \(z\in\mathcal{K}_{s} (\mathcal{A})\). This implies \(z\in\mathcal{K}_{t}(\mathcal{A})\cup \mathcal{K}_{s}(\mathcal{A})\subseteq\mathcal{K}(\mathcal{A})\).
(ii) There exist \(s \in\bar{S} \) and \(t\in S \) such that \(z \in\mathcal {G}_{s,t} ^{\bar{S}}(\mathcal{A})\), i.e.,
$$ |z| \bigl(|z|-R_{t}^{\overline{\Delta^{ \bar{S}}}}(\mathcal{A}) \bigr) \leq R_{s} (\mathcal{A})R_{t}^{\Delta^{\bar{S}}}(\mathcal{A}), $$
similar to (i), we obtain \(z\in\mathcal{K}_{s}(\mathcal {A})\cup\mathcal{K}_{t}(\mathcal{A})\subseteq\mathcal{K}(\mathcal {A})\). So, the result holds. □
Based on an exact characterization of (1), another S-type Z-eigenvalue localization set involved with a proper subset S of N is given below.
Theorem 2
Let
\(\mathcal{A}\)
be a tensor with order
m
and dimension
\(n\geq2\)
and
S
be a nonempty proper subset of
N. Then
$$ \sigma(\mathcal{A})\subseteq\Omega^{S}(\mathcal {A})= { \biggl( \bigcup_{i\in S,j\in\bar{S}} \bigl(\Omega _{i,j} ^{S}(\mathcal{A})\cup\Phi_{i,j}^{S}( \mathcal{A}) \bigr) \biggr) \cup \biggl( \bigcup _{i\in\bar{S},j\in S} \bigl(\Omega_{i,j} ^{\bar {S}}(\mathcal{A}) \cup\Phi_{i,j}^{\bar{S}}(\mathcal{A}) \bigr) \biggr) }, $$
where
$$ \begin{gathered} \Omega_{i,j} ^{S}(\mathcal{A}) = \bigl\{ z\in\mathcal{C}: \bigl(|z|- R_{i}^{\Delta ^{S}}(\mathcal{A}) \bigr) \bigl(|z|-R_{j}^{\overline{\Delta^{S}}}(\mathcal{A}) \bigr) \leq R_{i}^{\overline{\Delta^{S}}}(\mathcal{A})R_{j}^{\Delta^{S}}( \mathcal {A}) \bigr\} , \\ \Omega_{i,j} ^{\bar{S}}(\mathcal{A}) = \bigl\{ z\in\mathcal {C}: \bigl(|z|-R_{i}^{\Delta^{\bar{S}}}(\mathcal{A}) \bigr) \bigl(|z|-R_{j}^{\overline {\Delta^{ \bar{S}}}}( \mathcal{A}) \bigr) \leq R_{i}^{\overline{\Delta^{ \bar {S}}}}(\mathcal{A})R_{j}^{\Delta^{\bar{S}}}( \mathcal{A}) \bigr\} , \\ \Phi_{i,j} ^{S}(\mathcal{A}) = \bigl\{ z\in\mathcal{C}: |z| \leq R_{i}^{\Delta^{S}}(\mathcal{A}),|z|\leq R_{j}^{\overline{\Delta ^{S}}}( \mathcal{A}) \bigr\} , \\ \Phi_{i,j} ^{\bar{S}}(\mathcal{A}) = \bigl\{ z\in\mathcal{C}: |z| \leq R_{i}^{\Delta^{\bar{S}}}(\mathcal{A}), |z|\leq R_{j}^{\overline{\Delta^{ \bar{S}}}}( \mathcal{A}) \bigr\} . \end{gathered}$$
Proof
Let λ be a Z-eigenvalue of \(\mathcal{A}\) with corresponding eigenvector x. Let \(|x_{t}|=\max_{i\in S}|x_{i}|\) and \(|x_{s}|=\max_{i\in\bar{S}}|x_{i}|\). Similar to the proof of Theorem 1, we also divide the proof into three cases as follows.
(i) If \(x_{t}x_{s}\neq0\) and \(|x_{s}|\geq|x_{t}|\), then \(|x_{s}|={\max\{|x_{i}|: i\in N\}}\). By an exact characterization of (1), one has
$$\begin{aligned} |\lambda||x_{t}|&\leq\sum_{(i_{2},\ldots, i_{m}) \in\Delta^{S} }|a_{t i_{2}\ldots i_{m}}||x_{i_{2}}| \cdots|x_{i_{m}}|+\sum_{(i_{2},\ldots, i_{m} ) \in\overline{\Delta^{S}}}|a_{t i_{2}\ldots i_{m}}||x_{i_{2}}| \cdots|x_{i_{m}}| \\ &\leq R_{t}^{\Delta^{S}}(\mathcal{A}) |x_{t}|^{m-1}+ R_{t}^{\overline {\Delta^{S}}}(\mathcal{A})|x_{s}|^{m-1} \leq R_{t}^{\Delta^{S}}(\mathcal{A}) |x_{t}|+ R_{t}^{\overline{\Delta ^{S}}}(\mathcal{A})|x_{s}|, \end{aligned}$$
since \(|x_{t}|^{m-1}\leq|x_{t}| \leq1\), \(|x_{s}|^{m-1}\leq|x_{s}| \leq1\) hold. Furthermore,
$$ \bigl(|\lambda|-R_{t}^{\Delta^{S}}(\mathcal{A}) \bigr)|x_{t}| \leq R_{t}^{\overline {\Delta^{S}}}( \mathcal{A})|x_{s}|. $$
(7)
When \(|\lambda| > R_{s}^{\overline{\Delta^{S}}}(\mathcal{A}) \) or \(|\lambda|> R_{t}^{\Delta^{S}}(\mathcal{A})\) holds, multiplying (2) by (7), we see
$$ \bigl(|\lambda|-R_{t}^{\Delta^{S}}(\mathcal{A}) \bigr) \bigl(| \lambda|-R_{s}^{\overline {\Delta^{S}}}(\mathcal{A}) \bigr) \leq R_{t}^{\overline{\Delta^{S}}}(\mathcal {A})R_{s}^{\Delta^{S}}( \mathcal{A}). $$
This shows \(\lambda\in\Omega_{t,s} ^{S}(\mathcal{A})\subseteq\Omega ^{S}(\mathcal{A})\). Otherwise, when \(|\lambda|\leq R_{s}^{\overline{\Delta^{S}}}(\mathcal{A}) \) and \(|\lambda|\leq R_{t}^{\Delta^{S}}(\mathcal{A})\) hold, one has \(\lambda\in\Phi_{t,s} ^{S}(\mathcal{A})\subseteq \Omega^{S}(\mathcal{A})\).
(ii) If \(x_{t}x_{s}\neq0\) and \(|x_{t}|\geq|x_{s}|\), then \(|x_{t}|={\max\{|x_{i}|: i\in N \}}\). Similarly, by equality (1), we get
$$ \bigl(|\lambda|-R_{t}^{\overline{\Delta^{\bar{S}}}}(\mathcal{A}) \bigr)|x_{t}| \leq R_{t}^{\Delta^{\bar{S}}}( \mathcal{A})|x_{s}| $$
and
$$ \bigl(|\lambda|- R_{s}^{\Delta^{\bar{S}}}(\mathcal{A}) \bigr)|x_{s}| \leq R_{s}^{\overline{\Delta^{\bar{S}}}}( \mathcal{A})|x_{t}|. $$
When \(|\lambda|- R_{s}^{\Delta^{\bar{S}}}(\mathcal{A})> 0 \) or \(|\lambda |-R_{t}^{\overline{\Delta^{\bar{S}}}}(\mathcal{A})>0\) holds, we obtain
$$ \bigl(|\lambda|- R_{s}^{\Delta^{\bar{S}}}(\mathcal{A}) \bigr) \bigl(| \lambda|- R_{t}^{\overline{\Delta^{\bar{S}}}}(\mathcal{A}) \bigr) \leq R_{t}^{\Delta^{\bar {S}}}(\mathcal{A})R_{s}^{\overline{\Delta^{\bar{S}}}}( \mathcal{A}), $$
which implies \(\lambda\in\Omega_{s,t} ^{\bar{S}}(\mathcal{A})\subseteq \Omega^{S}(\mathcal{A})\). When \(|\lambda|- R_{s}^{\Delta^{\bar{S}}}(\mathcal{A}) \leq0 \) and \(|\lambda|-R_{t}^{\overline{\Delta^{\bar{S}}}}(\mathcal{A})\leq0\) hold, one has \(\lambda\in\Phi_{s,t} ^{\bar{S}}(\mathcal{A})\subseteq \Omega ^{S}(\mathcal{A})\).
(iii) If \(|x_{t}||x_{s}| = 0\), we could assume that \(|x_{s}| = 0\) and \(|x_{t}|\neq0\). It follows from (7) that
$$ |\lambda|-R_{t}^{\Delta^{S}}(\mathcal{A})\leq0. $$
For any \(j\in \bar{S}\), when \(|\lambda|-R_{j}^{\overline{\Delta ^{S}}}(\mathcal{A})> 0\) holds, we get
$$ \bigl(|\lambda|-R_{t}^{\Delta^{S}}(\mathcal{A}) \bigr) \bigl(| \lambda|- R_{j}^{\overline{\Delta^{S}}}(\mathcal{A}) \bigr)\leq R_{t}^{\overline{\Delta ^{S}}}(\mathcal{A})R_{j}^{\Delta^{S}}( \mathcal{A}), $$
that is, \(\lambda\in\Omega_{t,j} ^{S}(\mathcal{A})\subseteq\Omega ^{S}(\mathcal{A})\); otherwise, when \(|\lambda|-R_{j}^{\overline{\Delta ^{S}}}(\mathcal{A})\leq0\) holds, \(\lambda\in\Phi_{t,j} ^{S}(\mathcal {A})\subseteq \Omega^{S}(\mathcal{A})\). It follows from (i), (ii) and (iii) that the results hold. □
Corollary 2
Let
\(\mathcal{A}\)
be a tensor with order
m
and dimension
\(n\geq2\).
-
(I)
If there exists
\(S\subseteq N\)
such that
-
(i)
for all
\(i \in S\), \(j\in\bar{S}\), \(R_{j}^{\overline{\Delta ^{S}}}(\mathcal{A})\leq|z|\leq R_{i}(\mathcal{A})\)
and
\(R_{i}^{\overline {\Delta^{S}}}(\mathcal{A})R_{j}^{\Delta^{S}}(\mathcal{A})> 0\)
hold;
-
(ii)
for all
\(i \in\bar{S}\), \(j\in S\), \(R_{j}^{\overline {\Delta^{\bar{S}}}}(\mathcal{A}) \leq|z|\leq R_{i}(\mathcal{A})\)
and
\(R_{i}^{\overline{\Delta^{\bar{S}}}}(\mathcal{A}) R_{j}^{\Delta^{\bar {S}}}(\mathcal{A}) > 0\)
hold, then
$$ \mathcal{G}^{S}(\mathcal{A})\subseteq\Omega^{S}( \mathcal{A}). $$
-
(II)
If there exists
\(S\subseteq N\)
such that
-
(i)
for all
\(i \in S\), \(j\in\bar{S}\), \(|z|\leq\min\{ R_{i}^{\Delta^{S}}(\mathcal{A}), R_{j}^{\overline{\Delta^{S}}}(\mathcal {A})\}\)
holds; or
\(|z|\geq\max\{R_{i}(\mathcal{A}), R_{j}^{\overline {\Delta^{S}}}(\mathcal{A})\}\)
and
\(R_{i}^{\overline{\Delta^{S}}}(\mathcal{A})R_{j}^{\Delta^{S}}(\mathcal {A})> 0\)
are satisfied;
-
(ii)
for all
\(i \in\bar{S}\), \(j\in S\), \(|z|\leq\min\{ R_{i}^{\Delta^{\bar{S}}}(\mathcal{A}), R_{j}^{\overline{\Delta^{ \bar {S}}}}(\mathcal{A})\}\)
holds; or
\(|z|\geq\max\{ R_{i}(\mathcal{A}), R_{j}^{\overline{\Delta^{\bar{S}}}}(\mathcal{A})\}\)
and
\(R_{i}^{\overline {\Delta^{\bar{S}}}}(\mathcal{A})R_{j}^{\Delta^{\bar{S}}}(\mathcal{A}) > 0\)
are satisfied, then
$$ \Omega^{S}(\mathcal{A})\subseteq\mathcal{G}^{S}( \mathcal{A}). $$
Proof
(I) Let \(z\in\mathcal{G}^{S}(\mathcal{A})\), then \(z\in\mathcal{G}^{S}_{i,j}(\mathcal{A})\) or \(z\in\mathcal{G}^{\bar {S}}_{i,j}(\mathcal{A})\). We divide the proof into two parts.
(i) Suppose that \(z\in\mathcal{G}^{S}_{i,j}(\mathcal{A})\), then there exist \(t\in S \) and \(s \in\bar{S} \) such that \(z\in\mathcal {G}_{t,s} ^{S}(\mathcal{A})\).
If \(R_{t} (\mathcal{A}) = 0 \), then \(R_{t}^{\Delta^{S}}(\mathcal{A})= R_{t}^{\overline{\Delta^{S}}}(\mathcal{A})=0\), we have \(z = 0\) or \(|z|- R_{s}^{\overline{\Delta^{S}}}(\mathcal{A}) \leq0\). Hence, \(z\in\Omega _{t,s} ^{S}(\mathcal{A})\).
If \(R_{t} (\mathcal{A})R_{s}^{\Delta^{S}}(\mathcal{A}) >0 \), by (6), we have
$$ \frac{|z|}{R_{t}(\mathcal{A})}\leq1 \quad\text{or} \quad\frac{|z|- R_{s}^{\overline{\Delta^{S}}}(\mathcal{A})}{R_{s}^{\Delta^{S}}(\mathcal {A})}\leq1. $$
When \(\frac{|z|- R_{s}^{\overline{\Delta^{S}}}(\mathcal{A})}{R_{s}^{\Delta ^{S}}(\mathcal{A})}\geq0\) and \(\frac{|z|}{R_{t}(\mathcal{A})}\leq1 \), letting \(a=|z|\), \(b=R_{t}^{\Delta^{S}}(\mathcal{A})\), \(c=0\), \(d=R_{t}^{\overline{\Delta^{S}}}(\mathcal{A})>0\), from Lemma 5 in [16] and (6), we get
$$ \frac{|z|-R_{t}^{\Delta^{S}}(\mathcal {A})}{R_{t}^{\overline{\Delta^{S}}}(\mathcal{A})}\frac{|z|- R_{s}^{\overline{\Delta^{S}}}(\mathcal{A})}{R_{s}^{\Delta^{S}}(\mathcal {A})}\leq\frac{|z|}{R_{t}(\mathcal{A})}\frac{|z|- R_{s}^{\overline{\Delta ^{S}}}(\mathcal{A})}{R_{s}^{\Delta^{S}}(\mathcal{A})}\leq1. $$
Furthermore,
$$ \bigl(|z|-R_{t}^{\Delta^{S}}(\mathcal{A}) \bigr) \bigl(|z|-R_{s}^{\overline {\Delta^{S}}}(\mathcal{A}) \bigr) \leq R_{t}^{\overline{\Delta^{S}}}(\mathcal {A})R_{s}^{\Delta^{S}}( \mathcal{A}), $$
which implies \(z\in\Omega_{t,s} ^{S}(\mathcal{A})\). So,
$$ z\in\mathcal{G}^{S}_{t,s}(\mathcal{A}) \subseteq\Omega _{t,s} ^{S}(\mathcal{A}) \quad\mbox{and} \quad\mathcal{G}^{S}( \mathcal {A})\subseteq\Omega^{S}(\mathcal{A}). $$
(ii) Suppose that \(z\in\mathcal{G}^{\bar{S}}(\mathcal {A})\), then there exist \(s \in\bar{S} \) and \(t\in S \) such that \(z \in\mathcal {G}_{s,t} ^{\bar{S}}(\mathcal{A})\). Similar to the proof of (i), the conclusion holds.
(II) Let \(z\in\Omega^{S}(\mathcal{A})\), then \(z\in\bigcup_{i\in S,j\in\bar{S}}\Omega_{i,j} ^{S}(\mathcal {A})\cup\Phi_{i,j}^{S}(\mathcal{A})\) or \(z\in\bigcup _{i\in \bar{S},j\in S}\Omega_{i,j} ^{\bar{S}}(\mathcal{A})\cup\Phi _{i,j}^{\bar{S}}(\mathcal{A})\). We also divide the proof into two parts.
(i) Suppose that \(z\in\bigcup_{i\in S,j\in\bar {S}}\Omega_{i,j} ^{S}(\mathcal{A})\cup\Phi_{i,j}^{S}(\mathcal{A})\), then there exist \(t\in S \) and \(s \in\bar{S} \) such that \(z\in\Omega _{t,s} ^{S}(\mathcal{A})\) or \(z\in\Phi_{t,s} ^{S}(\mathcal{A})\).
If \(z\in\Phi_{t,s} ^{S}(\mathcal{A})\), that is, \(|z| \leq R_{t}^{\Delta ^{S}}(\mathcal{A})\) and \(|z|\leq R_{s}^{\overline{\Delta^{S}}}(\mathcal {A})\), then it is easy to get that \(\Omega^{S}(\mathcal{A})\subseteq \mathcal{G}^{S}(\mathcal{A})\).
If \(z\in\Omega_{t,s} ^{S}(\mathcal{A})\), that is,
$$ \bigl(|z|-R_{t}^{\Delta^{S}}(\mathcal{A}) \bigr) \bigl(|z|-R_{s}^{\overline{\Delta ^{S}}}(\mathcal{A}) \bigr) \leq R_{t}^{\overline{\Delta^{S}}}(\mathcal {A})R_{s}^{\Delta^{S}}( \mathcal{A}). $$
(8)
We assume \(R_{t}^{\overline{\Delta^{S}}}(\mathcal{A})R_{s}^{\Delta ^{S}}(\mathcal{A})>0 \), it follows from (8) that
$$ \frac{|z|-R_{t}^{\Delta^{S}}(\mathcal{A})}{R_{t}^{\overline{\Delta ^{S}}}(\mathcal{A})}\frac{|z|- R_{s}^{\overline{\Delta^{S}}}(\mathcal {A})}{R_{s}^{\Delta^{S}}(\mathcal{A})}\leq1. $$
(9)
When \(\frac{|z|- R_{s}^{\overline{\Delta^{S}}}(\mathcal{A})}{R_{s}^{\Delta ^{S}}(\mathcal{A})}\geq0\) and \(\frac{|z|}{R_{t}(\mathcal{A})}\geq1 \), letting \(a=|z|\), \(b=R_{t}^{\Delta^{S}}(\mathcal{A})\), \(c=0\), \(d=R_{t}^{\overline{\Delta^{S}}}(\mathcal{A})>0\), from Lemma 5 in [16] and (9), we obtain
$$ \frac{|z|}{R_{t}(\mathcal{A})}\frac{|z|- R_{s}^{\overline {\Delta^{S}}}(\mathcal{A})}{R_{s}^{\Delta^{S}}(\mathcal{A})}\leq\frac {|z|-R_{t}^{\Delta^{S}}(\mathcal{A})}{R_{t}^{\overline{\Delta ^{S}}}(\mathcal{A})}\frac{|z|- R_{s}^{\overline{\Delta^{S}}}(\mathcal {A})}{R_{s}^{\Delta^{S}}(\mathcal{A})} \leq1. $$
Moreover,
$$ |z| \bigl(|z|-R_{s}^{\overline{\Delta^{S}}}(\mathcal{A}) \bigr) \leq R_{t}(\mathcal{A})R_{s}^{\Delta^{S}}(\mathcal{A}), $$
which implies \(z\in\mathcal{G}_{t,s} ^{S}(\mathcal{A})\). Hence,
$$ z\in\Omega_{t,s} ^{S}(\mathcal{A}) \subseteq \mathcal {G}^{S}_{t,s}(\mathcal{A}) \quad \mbox{and} \quad\Omega^{S}( \mathcal{A})\subseteq \mathcal{G}^{S}(\mathcal{A}). $$
(ii) Suppose that \(z\in\bigcup_{i\in\bar {S},j\in S}{ (\Omega_{i,j} ^{\bar{S}}(\mathcal {A})\cup\Phi_{i,j}^{\bar{S}}(\mathcal{A}) )}\). Similar to the proof of (i), we arrive at the result. □
Owing to the uncertainty of S, we cannot compare \(\mathcal {G}^{S}(\mathcal{A})\) with \(\Omega^{S}(\mathcal{A})\) theoretically without the conditions of Corollary 2. Example 1 shows that they are different, since \(\mathcal{G}^{S}_{i,j}\) (\(\mathcal{A})(\mathcal{G}^{\bar {S}}_{i,j}(\mathcal{A})\)) and \(\Omega_{i,j} ^{S}\) (\(\mathcal{A})(\Omega _{i,j} ^{\bar{S}}(\mathcal{A}) \)) do not include each other.
Example 1
Let \(\mathcal{A}=(a_{ijk})\in\mathcal{R}^{[3,3]}\) be a tensor with elements defined as follows:
$$ a_{ijk}=\left \{ \textstyle\begin{array}{ll} a_{111}=1;\qquad a_{121}=-1; \qquad a_{122}=1;\qquad a_{133}=-1;\\ a_{211}=-1; \qquad a_{213}=1;\qquad a_{221}=2;\qquad a_{233}=-1;\\ a_{311}=3; \qquad a_{322}=-1;\qquad a_{332}=-1;\qquad a_{333}=1;\\ a_{ijk}=0, \quad \text{otherwise}. \end{array}\displaystyle \right . $$
According to Lemma 1, we have
$$ \mathcal{K}(\mathcal{A})=\bigcup_{i\in N}\mathcal {K}_{i}(\mathcal{A})=\bigl\{ \lambda\in C: |\lambda|\leq6\bigr\} . $$
Let \(S=\{{1}\}\). Obviously, \(\bar{S}=\{{2,3}\}\). From Theorem 1, one has
$$ \sigma(\mathcal{A})\subseteq\mathcal{G}^{S}(\mathcal {A})= \biggl\{ \lambda\in C: |\lambda|\leq\frac{3+\sqrt{57}}{2} \biggr\} , $$
where
$$\begin{gathered}\mathcal{G}^{S}_{1,2}(\mathcal{A})=\bigl\{ \lambda\in C: |\lambda| \leq 2+2\sqrt{2}\bigr\} , \qquad\mathcal{G}^{S}_{1,3}(\mathcal{A})= \biggl\{ \lambda\in C: |\lambda|\leq\frac{3+\sqrt{57}}{2}\biggr\} , \\ \mathcal{G}^{\bar{S}}_{2,1}(\mathcal{A})=\bigl\{ \lambda\in C: |\lambda |\leq1+\sqrt{11}\bigr\} , \qquad \mathcal{G}^{\bar{S}}_{3,1}( \mathcal{A})=\bigl\{ \lambda\in C: |\lambda|\leq1+\sqrt{13}\bigr\} .\end{gathered} $$
And it follows from Theorem 2 that
$$ \sigma(\mathcal{A})\subseteq\Omega^{S}(\mathcal{A})=\bigl\{ \lambda\in C: | \lambda|\leq 2+\sqrt{10}\bigr\} , $$
where
$$\begin{gathered}\Omega^{S}_{1,2}(\mathcal{A})=\biggl\{ \lambda\in C: |\lambda| \leq\frac {5+\sqrt{21}}{2}\biggr\} , \qquad \Omega^{S}_{1,3}( \mathcal{A})=\bigl\{ \lambda\in C: |\lambda|\leq2+\sqrt{10}\bigr\} , \\ \Omega^{\bar{S}}_{2,1}(\mathcal{A})=\biggl\{ \lambda\in C: |\lambda|\leq \frac{3+\sqrt{33}}{2}\biggr\} , \qquad \Omega^{\bar{S}}_{3,1}( \mathcal{A})=\bigl\{ \lambda\in C: |\lambda|\leq5\bigr\} .\end{gathered} $$