*Suppose that A* ≥ 0 *is a* (*p, q*)-*th order n × n dimensional rectangular tensor. Then*
$${\lambda}_{0}\left(\mathcal{A}\right)\ge {a}_{i\dots i\text{\hspace{0.17em}}i\dots i}.$$

*Proof*. Let *e*_{i} be the *n* dimension real vector, whose i th entry is 1 and others 0. Hence, from Theorem 5 in [4],
$${\lambda}_{0}\left(\mathcal{A}\right)\ge \text{\hspace{0.17em}}\frac{{\left(\mathcal{A}{e}_{i}^{p}{e}_{i}^{q-1}\right)}_{i}}{{\left({e}_{i}\right)}_{i}^{M-1}}={a}_{i\dots ii\dots i}$$

for any *i*. Thus, we complete the proof.

□

*Proof of Theorem 1.3*. Let *λ*_{0}(*A*) be the largest singular value of *A*. Then there exist two nonnegative vectors *x* = {*x*_{1}, *x*_{2}, …, *x*_{n})^{T} and *y* = {*y*_{1}, *y*_{2}, …, *y*_{n})^{T} such that
$$\{\begin{array}{c}\mathcal{A}{x}^{p-1}{y}^{q}={\lambda}_{0}\left(\mathcal{A}\right){x}^{\left[M-1\right]}\\ \mathcal{A}{x}^{p}{y}^{q-1}={\lambda}_{0}\left(\mathcal{A}\right){y}^{\left[M-1\right]}\end{array}.$$(1)

Denote
$${x}_{t}=\mathrm{max}\{{x}_{i},\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\le i\le n\},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}_{s}=\mathrm{max}\{{y}_{i},\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\le i\le n\}.$$

and *ω*_{i} = max{*x*_{i}, y_{i}}. Let *g* be an index such that *ω*_{g} = max{*ω*_{i}, *i* ∈ *N*}, *N* = {1,..., *n*}. Obviously, *ω*_{g} ≠ 0. Let *h* be an index such that *ω*_{h}, = max{*ω*_{i}, *i* ∈ *N*, *i* ≠ *g*}.

Case I: we suppose *ω*_{g} = *x*_{t}, *ω*_{h} = *x*_{s}, then, the *t*-th equations in (1) imply
$${\lambda}_{0}\left(\mathcal{A}\right){x}_{t}^{}-{a}_{t\dots t\text{\hspace{0.17em}}t\dots t}{x}_{t}^{p-1}{y}_{t}^{q}={\displaystyle \sum _{t{i}_{2}\dots {i}_{p}{j}_{1}\dots {j}_{q}\ne t\dots tt\dots t}{a}_{t{i}_{2}\dots {i}_{p}{j}_{1}\dots {j}_{q}}{x}_{{i}_{2}\dots}{x}_{{i}_{p}}{y}_{{j}_{1}\dots}{y}_{{j}_{q}.}}$$

Then, we can get
$${\lambda}_{0}\left(\mathcal{A}\right)-{a}_{t\dots t\text{\hspace{0.17em}}t\dots t}{\left(\frac{{y}_{t}}{{x}_{t}}\right)}^{q}{x}_{t}^{M-1}\le {\displaystyle \sum _{\begin{array}{l}t{i}_{2}\dots {i}_{p}{j}_{1}\dots {j}_{q}\ne t\dots tt\dots t\\ t{i}_{2}\dots {i}_{p}{j}_{1}\dots {j}_{q}\ne t\text{\hspace{0.17em}}s\dots ss\dots s\end{array}}{a}_{t{i}_{2}\dots {i}_{p}{j}_{1}\dots {j}_{q}}{x}_{t}^{M-1}+{a}_{ts\dots ss\dots s}}{x}_{s}^{M-1}.$$

By using $\frac{{y}_{t}}{{x}_{t}}\le 1$ and Lemma 2.1, we have
$$\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{t\dots t\text{\hspace{0.17em}}t\dots t}\right){x}_{t}^{M-1}\le \left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{t\dots t\text{\hspace{0.17em}}t\dots t}{\left(\frac{{y}_{t}}{{x}_{t}}\right)}^{q}\right){x}_{t}^{M-1}\le {r}_{t}^{s}\left(\mathcal{A}\right){x}_{t}^{M-1}+{a}_{ts\dots ss\dots s}{x}_{s}^{M-1}.$$

Then, we get
$$\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{t\dots t\text{\hspace{0.17em}}t\dots t}-{r}_{t}^{s}\left(\mathcal{A}\right)\right){x}_{t}^{M-1}\le {a}_{ts\dots ss\dots s}{x}_{s}^{M-1}.$$(2)

Similarly, the *s*-th equations in (1) imply
$$\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{s\dots s\text{\hspace{0.17em}}s\dots s}\right){x}_{s}^{M-1}\le {r}_{s}\left(\mathcal{A}\right){x}_{t}^{M-1}.$$(3)

Multiplying inequalities (2) with (3), we have
$$\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{t\dots t\text{\hspace{0.17em}}t\dots t}-{r}_{t}^{s}\left(\mathcal{A}\right)\right)\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{s\dots s\text{\hspace{0.17em}}s\dots s}\right)\le {a}_{ts\dots ss\dots s}{r}_{s}\left(\mathcal{A}\right).$$

Then, we can get
$${\lambda}_{0}(\mathcal{A})\le \underset{i,j\in N,j\ne i}{max}\frac{1}{2}\left\{{a}_{i\dots i\phantom{\rule{thickmathspace}{0ex}}i\dots i}+\phantom{\rule{thickmathspace}{0ex}}{a}_{j\dots j\phantom{\rule{thickmathspace}{0ex}}j\dots j}+{r}_{i}^{j}(\mathcal{A})+{\mathrm{\Delta}}_{1}^{\frac{1}{2}}..(\mathcal{A})\right\}={\mathrm{\Theta}}_{1}(\mathcal{A}).$$

where
$${\Delta}_{1}\left(\mathcal{A}\right)={\left({a}_{i\dots i\text{\hspace{0.17em}}i\dots i}-{a}_{j\dots j\text{\hspace{0.17em}}j\dots j}+{r}_{i}^{j}\left(\mathcal{A}\right)\right)}^{2}+4{a}_{ij\dots jj\dots j}{r}_{j}\left(\mathcal{A}\right).$$

Case II: we suppose *ω*_{g} = *y*_{t}, *ω*_{h} = *y*_{s}, then, the *t*-th equations in (1) imply
$${\lambda}_{0}\left(\mathcal{A}\right){y}_{t}^{}-{a}_{t\dots t\text{\hspace{0.17em}}t\dots t}{x}_{t}^{p}{y}_{t}^{q-1}={\displaystyle \sum _{{i}_{1}\dots {i}_{p}t{j}_{2}\dots {j}_{q}\ne t\dots tt\dots t}{a}_{{i}_{1}\dots {i}_{p}{j}_{2}\dots {j}_{q}}{x}_{{i}_{1}\dots}{x}_{{i}_{p}}{y}_{{j}_{2}\dots}{y}_{{j}_{q}.}}$$

Similar to the proof of Case I, we get
$$\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{t\dots t\text{\hspace{0.17em}}t\dots t}{c}_{t}^{s}\left(\mathcal{A}\right)\right){y}_{t}^{M-1}\le {a}_{s\dots sts\dots s}{y}_{s}^{M-1}.$$(4)

$$\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{s\dots s\text{\hspace{0.17em}}s\dots s}\right){y}_{s}^{M-1}\le {c}_{s}\left(\mathcal{A}\right){y}_{t}^{M-1}.$$(5)

Multiplying inequalities (4) with (5), we have
$$\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{t\dots t\text{\hspace{0.17em}}t\dots t}-{c}_{t}^{s}\left(\mathcal{A}\right)\right)\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{s\dots s\text{\hspace{0.17em}}s\dots s}\right)\le {a}_{ts\dots s\text{\hspace{0.17em}}s\dots s}{c}_{s}\left(\mathcal{A}\right).$$

Then, we can get
$${\lambda}_{0}(\mathcal{A})\le \underset{i,j\in N,j\ne i}{max}\frac{1}{2}\left\{{a}_{i\dots i\phantom{\rule{thickmathspace}{0ex}}i\dots i}+\phantom{\rule{thickmathspace}{0ex}}{a}_{j\dots j\phantom{\rule{thickmathspace}{0ex}}j\dots j}+{c}_{i}^{j}(\mathcal{A})+{\mathrm{\Delta}}_{2}^{\frac{1}{2}}..(\mathcal{A})\right\}={\mathrm{\Theta}}_{2}(\mathcal{A}),$$

where
$${\Delta}_{2}\left(\mathcal{A}\right)={\left({a}_{i\dots i\text{\hspace{0.17em}}i\dots i}-{a}_{j\dots j\text{\hspace{0.17em}}j\dots j}-{c}_{j}^{i}\left(\mathcal{A}\right)\right)}^{2}+4{a}_{j\dots jij\dots j}{c}_{j}\left(\mathcal{A}\right).$$

Case III: we suppose *ω*_{g} = *x*_{t}, *ω*_{h} = *y*_{s}, then, the *t*-th equations in (1) imply
$$\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{t\dots t\text{\hspace{0.17em}}t\dots t}-{r}_{t}^{s}\left(\mathcal{A}\right)\right){x}_{t}^{M-1}\le {a}_{ts\dots s\text{\hspace{0.17em}}s\dots s}{y}_{s}^{M-1}.$$(6)

the *s*-th equations in (1) imply
$$\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{s\dots s\text{\hspace{0.17em}}s\dots s}\right){y}_{s}^{M-1}\le {c}_{s}\left(\mathcal{A}\right){x}_{t}^{M-1}.$$(7)

Multiplying inequalities (6) with (7), we have
$$\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{t\dots t\text{\hspace{0.17em}}t\dots t}-{r}_{t}^{s}\left(\mathcal{A}\right)\right)\left({\lambda}_{0}\left(\mathcal{A}\right)-{a}_{s\dots s\text{\hspace{0.17em}}s\dots s}\right)\le {a}_{ts\dots s\text{\hspace{0.17em}}s\dots s}{c}_{s}\left(\mathcal{A}\right).$$

then, we can get
$${\lambda}_{0}(\mathcal{A})\le \underset{i,j\in N,j\ne i}{max}\frac{1}{2}\left\{{a}_{i\dots i\phantom{\rule{thickmathspace}{0ex}}i\dots i}+\phantom{\rule{thickmathspace}{0ex}}{a}_{j\dots j\phantom{\rule{thickmathspace}{0ex}}j\dots j}+{r}_{i}^{j}(\mathcal{A})+{\mathrm{\Delta}}_{3}^{\frac{1}{2}}..(\mathcal{A})\right\}={\mathrm{\Theta}}_{3}(\mathcal{A}),$$

where
$${\Delta}_{3}\left(\mathcal{A}\right)={\left({a}_{i\dots i\text{\hspace{0.17em}}i\dots i}-{a}_{j\dots j\text{\hspace{0.17em}}j\dots j}+{r}_{i}^{j}\left(\mathcal{A}\right)\right)}^{2}+4{a}_{ij\dots jj\dots j}{c}_{j}\left(\mathcal{A}\right),$$

Case IV: we suppose *ω*_{g} = *y*_{t}, *ω*_{h} = *y*_{s}, similar to the proof of Case III, we can get
$${\lambda}_{0}\left(\mathcal{A}\right)\le \underset{i,j\in N,j\ne i}{\mathrm{max}}\frac{1}{2}\{{a}_{i\dots i\text{\hspace{0.17em}}i\dots i}+\text{\hspace{0.17em}}{a}_{j\dots j\text{\hspace{0.17em}}j\dots j}+{c}_{i}^{j}\left(\mathcal{A}\right)+{\Delta}_{4}^{\frac{1}{2}}\left(\mathcal{A}\right)\}={\Theta}_{4}\left(\mathcal{A}\right).$$

where
$${\Delta}_{4}\left(\mathcal{A}\right)={\left({a}_{i\dots i\text{\hspace{0.17em}}i\dots i}-{a}_{j\dots j\text{\hspace{0.17em}}j\dots j}+{c}_{i}^{j}\left(\mathcal{A}\right)\right)}^{2}+4{a}_{j\dots jij\dots j}{r}_{j}\left(\mathcal{A}\right).$$

Thus, we complete the proof.

Remark 2.2. *If s*_{i}(*A*) = max{*c*_{i}(*A*), *r*_{i}(*A*)} = *r*_{j}(*A*), then *c*_{j}(*A*) ≤ *r*_{i}(*A*), *from Theorem 3.5 in [13], we know that*
$${\Theta}_{3}\left(\mathcal{A}\right)\le {\Theta}_{1}\left(\mathcal{A}\right)\le {r}_{i}\left(\mathcal{A}\right).$$

*Similarly, if s*_{j} (*A*) = max{*c*_{i}, (*A*), *r*_{i}, (*A*)} = *c*_{i} (*A*), *then* *r*_{i} (*A*) ≤ *c*_{i} (*A*), *from Theorem 3.5 in [13], we know that*
$${\Theta}_{4}\left(\mathcal{A}\right)\le {\Theta}_{2}\left(\mathcal{A}\right)\le {c}_{i}\left(\mathcal{A}\right).$$

*that is to say, our new bound in Theorem 1.3 is always better than the bound in Theorem 4 in [1]*.

**Example 2.3**. *Consider the nonnegative (2, 2)-th order* 2 × 2 *dimensional rectangular tensor with entries defined as follows*:
$${a}_{1111}=\frac{1}{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{a}_{2222}=3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}{a}_{ijkl}=\frac{1}{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}elsewhere$$

*By Theorem 1.2, we have*
$${\lambda}_{0}\left(\mathcal{A}\right)\le \text{\hspace{0.17em}}\mathrm{5.3333.}$$

*By Theorem 1.3, we have*
$${\lambda}_{0}\left(\mathcal{A}\right)\le \text{\hspace{0.17em}}\mathrm{5.1667.}$$

*Hence, the bound in Theorem 1.3 is tight and sharper*.

## Comments (0)