In this section, we only consider *A* ∈*Ω*_{n}{0, 1} which corresponds to the adjacency matrix of a strict digraph. Recall that for an *n* × *n* matrix *A* = (*a*_{ij}), the *main diagonal* of *A* is the list of entries *a*_{11}, *a*_{22}, …, *a*_{nn}, and the *secondary diagonal* of *A* is the list of entries *a*_{1n}, *a*_{2, n − 1}, …, *a*_{n1}. Let *C*_{1} be the square matrix whose entries above the main diagonal are all 1’s, while other entries are all 0’s. The size of *C*_{1} will be clear from the context.

First we determine the possible values of *r*(*A*) ± *r*(*A*) in the general case.

#### Theorem 3.1

*Let n* ≥ 2 *be a positive integer*. *Then there exists A* ∈*Ω*_{n}{0, 1} *with r*(*A*) − *r*(*A*) = *k if and only if*

*k* = 0, ±2 *when n* = 2;

−*n* ≤ *k* ≤ *n when n* ≥ 3.

#### Proof

The case *n* = 2 is trivial.

The necessity is clear. Now we prove the sufficiency.

Let
$\begin{array}{}{B}_{1}=\left[\begin{array}{cc}{C}_{1}& {J}_{p,n-p}\\ {O}_{n-p,p}& {O}_{n-p}\end{array}\right]\in {\mathit{\Omega}}_{n}\{0,1\},\end{array}$ where 1 ≤ *p* ≤ *n* − 1. It is easy to verify that *r*(*B*_{1}) = *p*, *r*(*B*_{1}) = *n* − 1. Thus for every integer *k* with −*n*+2 ≤ *k* ≤ *n* − 2, there exists *A* ∈ *Ω*_{n}{0, 1} such that *r*(*A*) − *r*(*A*) = *k*.

Let
$\begin{array}{}{B}_{2}=\left[\begin{array}{c}{C}_{2}\\ {O}_{n-1,n}\end{array}\right]\in {\mathit{\Omega}}_{n}\{0,1\},\end{array}$ where *C*_{2} = [0, 1, …, 1, 0]. Then *r*(*B*_{2}) = 1, *r*(*B*_{2}) = *n*. Note that *r*(*O*_{n}) = 0, *r*(*O*_{n}) = *n*. Thus for *k* = ±(*n* − 1), ± *n*, there exists *A* ∈ *Ω*_{n}{0, 1} such that *r*(*A*) − *r*(*A*) = *k*. This proves the sufficiency. □

#### Theorem 3.2

*Let n* ≥ 2 *be a positive integer*. *Then there exists A* ∈ *Ω*_{n}{0, 1} *with r*(*A*)+*r*(*A*) = *k if and only if*

*k* = 2 *when n* = 2;

*n* ≤ *k* ≤ 2*n when n* ≥ 3.

#### Proof

Trivial.

*r*(*A*)+*r*(*A*) ≤ 2*n* is clear. Since *A*+*A* = *J*_{n} − *I*_{n}, *n* = *r*(*A*+*A*) ≤ *r*(*A*)+*r*(*A*). This proves the necessity.

For the sufficiency, first note that the matrix *B*_{1} ∈ *Ω*_{n}{0, 1} in the proof of Theorem 3.1, (ii) implies that for every integer *k* with *n* ≤ *k* ≤ 2*n* − 2, there exists *A* ∈ *Ω*_{n}{0, 1} such that *r*(*A*)+*r*(*A*) = *k*.

Let *B*_{3} = *C*_{1} − *E*_{1n} ∈ *Ω*_{n}{0, 1}. Then *r*(*B*_{3}) = *n* − 1, *r*(*B*_{3}) = *n*.

Let *B*_{4} = *B*_{3}+*E*_{n1} ∈ *Ω*_{n}{0, 1}. Then *r*(*B*_{4}) = *r*(*B*_{4}) = *n*.

Thus for *k* = 2*n* − 1, 2*n*, there exists *A* ∈ *Ω*_{n}{0, 1} such that *r*(*A*)+*r*(*A*) = *k*. This proves the sufficiency. □

Next we consider the case when *A* ∈ *Ω*_{n}{0, 1} is symmetric.

#### Lemma 3.3

*Let*
$\begin{array}{}G=\left[\begin{array}{cc}{G}_{1}& {J}_{p,n-p}\\ {J}_{n-p,p}& {J}_{n-p}-{I}_{n-p}\end{array}\right]\end{array}$ ∈ *Ω*_{n}{0, 1} *with* 1 ≤ *p* ≤ *n* − 1, *where G*_{1} *is the p* × *p matrix whose entries on the main diagonal*, *on the secondary diagonal and above the secondary diagonal are all 0’s*, *while other entries are all 1’s*. *Then r*(*G*) = *n if p is odd*, *and r*(*G*) = *n* − 1 *if p is even*.

#### Proof

First consider the case when *p* is odd. If *p* = *n* − 1, then *n* = *p*+1 is even and it is clear that det *G* ≠ 0. If *p* < *n* − 1, for *i* = *p*+1, *p*+2, …, *n* − 1, subtract the last row of *G* from the *i*-th row, and then add the *i*-th column to the last column. Using Laplace expansion formula, we deduce that det *G* ≠ 0.

When *p* is even, the
$\begin{array}{}\frac{p}{2}\end{array}$-th row (column) of *G* is identical to the (
$\begin{array}{}\frac{p}{2}\end{array}$+1)-th row (column). Denote by *G*_{2} the submatrix of *G* obtained by deleting the
$\begin{array}{}\frac{p}{2}\end{array}$-th row and the
$\begin{array}{}\frac{p}{2}\end{array}$-th column. Note that *G*_{2} =
$\begin{array}{}\left[\begin{array}{cc}{G}_{3}& {J}_{p-1,n-p}\\ {J}_{n-p,p-1}& {J}_{n-p}-{I}_{n-p}\end{array}\right]\in {\mathit{\Omega}}_{n-1}\{0,1\},\end{array}$ where *G*_{3} is the (*p* − 1) ×(*p* − 1) matrix which has the same form as *G*_{1}. Since *p* − 1 is odd, *r*(*G*) = *r*(*G*_{2}) = *n* − 1 by what we have just proved. □

#### Lemma 3.4

*Let H* ∈ *Ω*_{n}{0, 1} *be the matrix whose entries on the main diagonal*, *on the secondary diagonal and above the secondary diagonal are all 0’s*, *while other entries are all 1’s*. *Then*

$\begin{array}{}r(H+{E}_{1,\frac{n-1}{2}}+{E}_{\frac{n-1}{2},1})=n\end{array}$ *for odd n* ≥ 5,
$\begin{array}{}r(H+{E}_{1,\frac{n-1}{2}}+{E}_{\frac{n-1}{2},1}-{E}_{\frac{n+3}{2},n}-{E}_{n,\frac{n+3}{2}})=n\end{array}$ *for odd n* ≥ 7;
$\begin{array}{}r(\overline{H}-{E}_{1,\frac{n-1}{2}}-{E}_{\frac{n-1}{2},1}+{E}_{\frac{n+3}{2},n}+{E}_{n,\frac{n+3}{2}})=n\end{array}$

$\begin{array}{}r(H+{E}_{1,\frac{n}{2}}+{E}_{\frac{n}{2},1})=n\end{array}$ *for even n* ≥ 4,
$\begin{array}{}r(H+{E}_{1,\frac{n}{2}}+{E}_{\frac{n}{2},1}+{E}_{\frac{n}{2}-1,\frac{n}{2}+1}+{E}_{\frac{n}{2}+1,\frac{n}{2}-1})=n-1\end{array}$ *for even n* ≥ 6.

#### Proof

When *n* ≥ 5 is odd,
$\begin{array}{}\frac{n-1}{2}>1.\text{\hspace{0.17em}Then\hspace{0.17em}}{E}_{1,\frac{n-1}{2}}\ne {E}_{\frac{n-1}{2},1}.\end{array}$ Using Laplace expansion formula, it is easy to verify that
$\begin{array}{}det(H+{E}_{1,\frac{n-1}{2}}+{E}_{\frac{n-1}{2},1})\ne 0\end{array}$ for odd *n* ≥ 5 and
$\begin{array}{}det(H+{E}_{1,\frac{n-1}{2}}+{E}_{\frac{n-1}{2},1}-{E}_{\frac{n+3}{2},n}-{E}_{n,\frac{n+3}{2}})\ne 0\end{array}$ for odd *n* ≥ 7.

When *n* ≥ 7 is odd, subtract the second row of
$\begin{array}{}\overline{H}-{E}_{1,\frac{n-1}{2}}-{E}_{\frac{n-1}{2},1}+{E}_{\frac{n+3}{2},n}+{E}_{n,\frac{n+3}{2}}\end{array}$ from the first row, and then subtract the second column from the first column. Using Laplace expansion formula, we deduce that
$\begin{array}{}det(\overline{H}-{E}_{1,\frac{n-1}{2}}-{E}_{\frac{n-1}{2},1}+{E}_{\frac{n+3}{2},n}+{E}_{n,\frac{n+3}{2}})\ne 0.\end{array}$

When *n* ≥ 4 is even,
$\begin{array}{}\frac{n}{2}>1.\text{\hspace{0.17em}Then\hspace{0.17em}}{E}_{1,\frac{n}{2}}\ne {E}_{\frac{n}{2},1}.\end{array}$ Using Laplace expansion formula, it is easy to verify that
$\begin{array}{}det(H+{E}_{1,\frac{n}{2}}+{E}_{\frac{n}{2},1})\ne 0\end{array}$ for even *n* ≥ 4.

When *n* ≥ 6 is even,
$\begin{array}{}\frac{n}{2}\end{array}$ −1 > 1. Note that the (
$\begin{array}{}\frac{n}{2}\end{array}$)-th row (column) of
$\begin{array}{}H+{E}_{1,\frac{n}{2}}+{E}_{\frac{n}{2},1}+{E}_{\frac{n}{2}-1,\frac{n}{2}+1}+{E}_{\frac{n}{2}+1,\frac{n}{2}-1}\end{array}$ is the sum of the first row (column) and the (
$\begin{array}{}\frac{n}{2}\end{array}$ − 1)-th row (column). Then
$\begin{array}{}r(H+{E}_{1,\frac{n}{2}}+{E}_{\frac{n}{2},1}+{E}_{\frac{n}{2}-1,\frac{n}{2}+1}+{E}_{\frac{n}{2}+1,\frac{n}{2}-1})\end{array}$ ≤ *n* − 1. Using Laplace expansion formula and the fact that *J*_{3} − *I*_{3} is nonsingular, we can deduce that the submatrix of
$\begin{array}{}H+{E}_{1,\frac{n}{2}}+{E}_{\frac{n}{2},1}+{E}_{\frac{n}{2}-1,\frac{n}{2}+1}+{E}_{\frac{n}{2}+1,\frac{n}{2}-1}\end{array}$ obtained by deleting the (
$\begin{array}{}\frac{n}{2}\end{array}$)-th row and the ($\begin{array}{}\frac{n}{2}\end{array}$+2)-th column is nonsingular. Thus
$\begin{array}{}r(H+{E}_{1,\frac{n}{2}}+{E}_{\frac{n}{2},1}+{E}_{\frac{n}{2}-1,\frac{n}{2}+1}+{E}_{\frac{n}{2}+1,\frac{n}{2}-1})=n-1.\end{array}$

□

Finally we determine the possible values of *r*(*A*) ± *r*(*A*) in the symmetric case.

#### Theorem 3.5

*Let n* ≥ 2 *be a positive integer*. *Then there exists symmetric A* ∈ *Ω*_{n}{0, 1} *with r*(*A*) − *r*(*A*) = *k if and only if*

*k* = ±2 *when n* = 2;

*k* = 0, ±3 *when n* = 3;

−*n* ≤ *k* ≤ *n with k* ≠ ±(*n* − 1) *when n* ≥ 4.

#### Proof

(i) and (ii) are easy to verify.

(iii) First we prove the necessity. It is clear that −*n* ≤ *r*(*A*) − *r*(*A*) ≤ *n*. If *r*(*A*) − *r*(*A*) = −(*n* − 1), then either *r*(*A*) = 0, *r*(*A*) = *n* − 1, or *r*(*A*) = 1, *r*(*A*) = *n*. For the former case, *A* = *O*_{n} and thus *A* = *J*_{n} − *I*_{n} is nonsingular, a contradiction. For the later case, note that any nonzero symmetric *A* ∈ *Ω*_{n}{0, 1} always has a 2 × 2 submatrix
$\begin{array}{}\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right].\end{array}$ Then *r*(*A*) ≥ 2, a contradiction. Thus *r*(*A*) − *r*(*A*) ≠ −(*n* − 1). Likewise, *r*(*A*) − *r*(*A*) ≠ *n* − 1.

Next we prove the sufficiency. We will use the symmetric matrices *G*_{1} ∈ *Ω*_{p}{0, 1} and *G*, *H* ∈ *Ω*_{n}{0, 1} in Lemmas 3.3 and 3.4.

Note that *r*(*G*) = *r*(*G*_{1}) = *p* − 1 if *p* is odd, and *r*(*G*) = *r*(*G*_{1}) = *p* if *p* ≥ 2 is even. Then by Lemma 3.3, *r*(*G*) − *r*(*G*) = *n*+1 − *p* for odd *p*, and *r*(*G*) − *r*(*G*) = *n* − 1 − *p* for even *p* ≥ 2. When *n* ≥ 5 is odd, for odd *p* with 1 ≤ *p* ≤ *n* − 2, *r*(*G*) − *r*(*G*) can be 3, 5, 7, …, *n* − 2, *n*; for even *p* with 2 ≤ *p* ≤ *n* − 1, *r*(*G*) − *r*(*G*) can be 0, 2, 4, …, *n* − 5, *n* − 3. Thus *k* can be 0, ±2, ± 3, …, ± (*n* − 2), ± *n* for odd *n* ≥ 5. When *n* ≥ 4 is even, for odd *p* with 1 ≤ *p* ≤ *n* − 1, *r*(*G*) − *r*(*G*) can be 2, 4, 6, …, *n* − 2, *n*; for even *p* with 2 ≤ *p* ≤ *n* − 2, *r*(*G*) − *r*(*G*) can be 1, 3, 5, …, *n* − 5, *n* − 3. Thus *k* can be ±1, ±2, …, ±(*n* − 2), ± *n* for even *n* ≥ 4.

By Lemma 3.4 (i), for odd *n* ≥ 5,
$\begin{array}{}r(H+{E}_{1,\frac{n-1}{2}}+{E}_{\frac{n-1}{2},1})=n.\text{\hspace{0.17em}Since\hspace{0.17em}}r(\overline{H+{E}_{1,\frac{n-1}{2}}+{E}_{\frac{n-1}{2},1}})=r(\overline{H}-{E}_{1,\frac{n-1}{2}}-{E}_{\frac{n-1}{2},1})\end{array}$ = *n* − 1 when *n* is odd,
$\begin{array}{}r(H+{E}_{1,\frac{n-1}{2}}+{E}_{\frac{n-1}{2},1})-r(\overline{H}-{E}_{1,\frac{n-1}{2}}-{E}_{\frac{n-1}{2},1})=1.\end{array}$ Thus *k* can be ±1 for odd *n* ≥ 5. By Lemma 3.4 (ii), for even *n* ≥ 4,
$\begin{array}{}r(H+{E}_{1,\frac{n}{2}}+{E}_{\frac{n}{2},1})=n.\text{\hspace{0.17em}Since\hspace{0.17em}}r(\overline{H+{E}_{1,\frac{n}{2}}+{E}_{\frac{n}{2},1}})=r(\overline{H}-{E}_{1,\frac{n}{2}}-{E}_{\frac{n}{2},1})=n\end{array}$ when *n* is even, *k* can be 0 for even *n* ≥ 4.

Thus for *k* = 0, ± 1, ± 2, …, ±(*n* − 2), ± *n* with *n* ≥ 4, there exists symmetric *A* ∈ *Ω*_{n}{0, 1} such that *r*(*A*) − *r*(*A*) = *k*. This completes the proof. □

#### Theorem 3.6

*Let n* ≥ 2 *be a positive integer*. *Then there exists symmetric A* ∈ *Ω*_{n}{0, 1} *with r*(*A*)+*r*(*A*) = *k if and only if*

*k* = ±2 *when n* = 2;

*k* = 3, 4 *when n* = 3;

*k* = 4, 5, 6, 8 *when n* = 4;

*n* ≤ *k* ≤ 2*n when n* ≥ 5.

#### Proof

(i) and (ii) are easy to verify.

(iii) Denote by *f*(*A*) the number of 1’s in *A*. Then for symmetric *A* ∈ *Ω*_{4}{0, 1}, *f*(*A*) and *f*(*A*) are even with *f*(*A*)+*f*(*A*) = 12. Since *A* and *A* are mutually complementary, we may suppose *f*(*A*) ≤ 6.

If *f*(*A*) = 0, then *A* = *O*_{4} and thus *r*(*A*)+*r*(*A*) = 0 + 4 = 4.

If *f*(*A*) = 2, under permutation similarity, it suffices to consider the case *A* = *E*_{12}+*E*_{21}. A direct computation shows that *r*(*A*)+*r*(*A*) = 2 + 3 = 5.

If *f*(*A*) = 4, under permutation similarity, it suffices to consider the cases *A* = *E*_{12}+*E*_{21}+*E*_{13}+*E*_{31} and *A* = *E*_{12}+*E*_{21}+*E*_{34}+*E*_{43}. A direct computation shows that *r*(*A*)+*r*(*A*) = 2+4 = 6 in the first case and *r*(*A*)+*r*(*A*) = 4 + 2 = 6 in the second case.

If *f*(*A*) = 6, under permutation similarity, it suffices to consider the cases *A* = *E*_{12}+*E*_{21}+*E*_{13}+*E*_{31}+*E*_{23}+*E*_{32}, *A* = *E*_{12}+*E*_{21}+*E*_{13}+*E*_{31}+*E*_{14}+*E*_{41} and *A* = *E*_{12}+*E*_{21}+*E*_{13}+*E*_{31}+*E*_{24}+*E*_{42}. A direct computation shows that *r*(*A*)+*r*(*A*) = 3+2 = 5 in the first case, *r*(*A*)+*r*(*A*) = 2+3 = 5 in the second case and *r*(*A*)+*r*(*A*) = 4 + 4 = 8 in the third case.

Therefore, *r*(*A*)+*r*(*A*) can only be 4, 5, 6, 8 for symmetric *A* ∈ *Ω*_{4}{0, 1}.

(iv) The necessity has been proved in Theorem 3.2 (ii). Now we prove the sufficiency.

By Lemma 3.3 and the proof of Theorem 3.5 (iii), *k* can be *n* − 1+*p* for 1 ≤ *p* ≤ *n* − 1.

By Lemma 3.4 (i), for odd *n* ≥ 5,
$\begin{array}{}r(H+{E}_{1,\frac{n-1}{2}}+{E}_{\frac{n-1}{2},1})+r(\overline{H}-{E}_{1,\frac{n-1}{2}}-{E}_{\frac{n-1}{2},1})=n+(n-1)=2n-1.\end{array}$ By Lemma 3.4 (ii), for even *n* ≥ 6,
$\begin{array}{}r(H+{E}_{1,\frac{n}{2}}+{E}_{\frac{n}{2},1}+{E}_{\frac{n}{2}-1,\frac{n}{2}+1}+{E}_{\frac{n}{2}+1,\frac{n}{2}-1})+r(\overline{H}-{E}_{1,\frac{n}{2}}-{E}_{\frac{n}{2},1}-{E}_{\frac{n}{2}-1,\frac{n}{2}+1}-{E}_{\frac{n}{2}+1,\frac{n}{2}-1})\end{array}$ = (*n* − 1)+*n* = 2*n* − 1. Thus *k* can be 2*n* − 1 when *n* ≥ 5.

By Lemma 3.4 (i), for odd *n* ≥ 7,
$\begin{array}{}r(H+{E}_{1,\frac{n-1}{2}}+{E}_{\frac{n-1}{2},1}-{E}_{\frac{n+3}{2},n}-{E}_{n,\frac{n+3}{2}})+r(\overline{H}-{E}_{1,\frac{n-1}{2}}-{E}_{\frac{n-1}{2},1}+{E}_{\frac{n+3}{2},n}+{E}_{n,\frac{n+3}{2}})\end{array}$ = *n*+*n* = 2*n*. By Lemma 3.4 (ii), for even *n* ≥ 4,
$\begin{array}{}r(H+{E}_{1,\frac{n}{2}}+{E}_{\frac{n}{2},1})+r(\overline{H}-{E}_{1,\frac{n}{2}}-{E}_{\frac{n}{2},1})=n+n=2n.\end{array}$ Note that
$$\begin{array}{}r\left(\left[\begin{array}{ccccc}0& 0& 1& 1& 0\\ 0& 0& 0& 1& 1\\ 1& 0& 0& 0& 1\\ 1& 1& 0& 0& 0\\ 0& 1& 1& 0& 0\end{array}\right]\right)+r\left(\left[\begin{array}{ccccc}0& 1& 0& 0& 1\\ 1& 0& 1& 0& 0\\ 0& 1& 0& 1& 0\\ 0& 0& 1& 0& 1\\ 1& 0& 0& 1& 0\end{array}\right]\right)=5+5=10.\end{array}$$

Thus *k* can be 2*n* when *n* ≥ 4.

Then for symmetric *A* ∈ *Ω*_{n}{0, 1} with *n* ≥ 5, *r*(*A*)+*r*(*A*) can be *n*, *n*+1, …, 2*n*. □

## Comments (0)

General note:By using the comment function on degruyter.com you agree to our Privacy Statement. A respectful treatment of one another is important to us. Therefore we would like to draw your attention to our House Rules.