To state our results we need to defined some graphs. The categorical product G × H of two graphs G and H is the graph with vertex set V (G) × V (H) in which two vertices (u_{1}, v_{1}) and (u_{2}, v_{2}) are adjacent if u_{1}u_{2} ∈ E(G) and v_{1}v_{2} ∈ E(H).
Let A ∈R^{m}^{×}^{n}, B ∈ R^{p}^{×}^{q}. The Kronecker product (or tensor product) of A and B is defined as the matrix
$$A\otimes B=\left[\begin{array}{ccc}{a}_{11}B& \cdots & {a}_{1n}B\\ \vdots & \ddots & \vdots \\ {a}_{m1}B& \cdots & {a}_{mn}B\end{array}\right]$$It is well known and easy to show that for graphs G_{1} and G_{2} for which loops are allowed, we have
$$A({G}_{1}\times {G}_{2})=A({G}_{1})\otimes A({G}_{2})$$The following is useful for determining the spectrum of a product of graphs from those of its factors.
Proposition 2.1.([9])
Let A ∈M^{m} and B ∈M^{n}. Furthermore, let σ be eigenvalue of matrix A with corresponding eigenvector x and μ be eigenvalue of matrix B with corresponding eigenvector y. Then σμ is an eigenvalue of A ⊗ B with corresponding eigenvector x ⊗ y.
Let G^{r} be the graph obtained from G by adding loops on every vertex and G^{ur} be the graph obtained from G by removing all loops. For any simple graph G, if we take ${G}_{1}={K}_{m}^{r}$ and G_{2} = G, using the facts that A(G^{r}) = A(G) + I, A(G^{ur}) = A(G) − I,$A({K}_{m}^{r})=J$ and writing A for A(G), we have,
$$A({K}_{m}^{r}\times G)=J\otimes A$$and
$$A{({K}_{m}^{r}\times {G}^{r})}^{ur}=J\otimes (A+I)-I\otimes I$$We introduce two convenient notations. Let ${D}_{m}(G)={K}_{m}^{r}\times G$ and ${D}_{m}^{*}(G)={({K}_{m}^{r}\times {G}^{r})}^{ur}$. It is easy to verify that if G is a graph with n vertices then both D_{m}(G) and ${D}_{m}^{*}(G)$ are graphs with mn vertices. The graphs D_{2}(C_{5}) and ${D}_{2}^{*}({C}_{5})$ are shown in Figure 1.
Lemma 2.2
If the Seidel spectrum of G is {σ_{1}, σ_{2}, · · ·, σ_{n}}then the Seidel spectrum of D_{m}(G) is
$$Spe{c}_{s}({D}_{m}(G))=\{m{\sigma}_{i}+(m-1)\mid i=1,2,\cdots ,n\}\cup \{-{1}^{mn-n}\}$$ProofAs $A({K}_{m}^{r})=J={J}_{m}$ has spectrum {m, 0^{m}^{−1}}, we have by Proposition 2.1 that an eigenvalue σ_{i} of S(G) yields in $A({K}_{m}^{r})\times (S(G)+I)-I$ an eigenvalue m(σ_{i} + 1) − 1 = mσ_{i} + (m − 1) and mn − n eigenvalue 0(σ_{i} + 1) − 1 = −1. It is enough to show that $S({K}_{m}^{r}\times G)=A({K}_{m}^{r})\times (S(G)+I)-I$. Writing A for A(G) and S(M) = −2M − I + J for any matrix M, We have to show that S(J ⊗ A) = J ⊗ (S(A) + I) − I. Observe that first,
$$S(B\otimes A)=-2(B\otimes A)-I\otimes I+J\otimes J$$and so taking B = J and moving I × I to the other side, $S({K}_{m}^{r}\times G)+I$ is
$$S(J\otimes A)+I\otimes I=-2(J\otimes A)+J\otimes J=J\otimes (-2A+J)=J\otimes (S(A)+I).$$ Lemma 2.3
If the Seidel spectrum of G is {σ_{1}, σ_{2}, · · ·, σ_{n}} then the Seidel spectrum of${D}_{m}^{*}(G)$is
$$Spe{c}_{s}({D}_{m}^{*}(G))=\{m{\sigma}_{i}-(m-1)\mid i=1,2,\cdots ,n\}\cup \{{1}^{mn-m}\}$$ProofMuch like above,
$$\begin{array}{l}[S(J\otimes (A+I)-I\otimes I]+I=[-2(J\otimes (A+I)-I\otimes I+J\otimes J-I\otimes I]+I\otimes I\\ =J\otimes (-2(A+I))+J\otimes J\\ =J\otimes (-2(A+I)+J)=J\otimes (S(A)-I).\end{array}$$ Theorem 2.4
Let G be a graph with Seidel eigenvalues σ_{1}, σ_{2}, · · ·, σ_{n} with$\mid {\sigma}_{i}\mid \ge {\scriptstyle \frac{m-1}{m}}$, for all 1 ≤ i ≤ n then, D_{m}(G) and${D}_{m}^{*}(G)$are Seidel non co-spectral equienergetic graphs if and only if G have equal numbers of positive and negative Seidel eigenvalues.
ProofIt is not hard to see that the sums of the spectrums in these two lemma are the same. To say the same about the sums of the absolute values, (so that the Seidel energies are the same) we observe that we need that
$$\sum _{i}\mid m{\sigma}_{i}+(m-1)\mid =\sum _{i}\mid m{\sigma}_{i}-(m-1)\mid $$Assuming that $\mid {\sigma}_{i}\mid >\frac{m-1}{m}$, we get that |mσ_{i} + (m − 1) | − |mσ_{i} − (m − 1)| is 2(m − 1) if σ_{i} is positive and −2(m − 1) if σ_{i} is negative for each i = 1, 2, · · ·, n. It follows that with these restrictions on the eigenvalues that the Seidel energies of D_{m}(G) and ${D}_{m}^{*}(G)$ are the same if and only if G has the same number of positive and negative eigenvalues. This gives Theorem 2.4.
Observe that if G is a graph with n vertices then both ${D}_{m}({D}_{m}^{*}(G))$ and ${D}_{m}^{*}({D}_{m}(G))$ are graphs with m^{2}n vertices. We next prove following theorem for Seidel equienergetic graphs.
Theorem 2.5
Let G be a graph with Seidel eigenvalues σ_{1}, σ_{2}, · · ·, σ_{n} with$\mid {\sigma}_{i}\mid \ge {\left(\frac{m-1}{m}\right)}^{2}$, for all 1 ≤ i ≤ n then, ${D}_{m}^{*}({D}_{m}(G))$and${D}_{m}({D}_{m}^{*}(G))$are Seidel non co-spectral equienergetic graphs if and only if G has equal numbers of positive and negative Seidel eigenvalues.
ProofBy Lemma 2.2, D_{m}(G) has spectrum {mσ_{i} + (m − 1) |i = 1, 2, · · ·, n}∪{−1^{mn}^{−}^{n}}and by Lemma 2.3 spectrum of ${D}_{m}^{*}({D}_{m}(G))$ is {m^{2}σ_{i} + (m − 1)^{2} |i = 1, 2, · · ·, n}∪{(1−2m)^{mn}^{−}^{n}}∪{1^{m2}^{n}^{−}^{mn}} and again by Lemma 2.2 and Lemma 2.3, spectrum of ${D}_{m}({D}_{m}^{*}(G))$ is {m^{2}σ_{i} −(m − 1)^{2} |i = 1, 2, · · ·, n} ∪ {(2m−1)^{mn}^{−}^{n}}∪ {−1^{m2}^{n}^{−}^{mn}}. To prove ${D}_{m}^{*}({D}_{m}(G))$ and ${D}_{m}({D}_{m}^{*}(G))$ are equienergetic, it is enough to prove
$$\sum _{i}\mid {m}^{2}{\sigma}_{i}+{(m-1)}^{2}\mid =\sum _{i}\mid {m}^{2}{\sigma}_{i}-{(m-1)}^{2}\mid .$$If $\mid {\sigma}_{i}\mid >\frac{{(m-1)}^{2}}{{m}^{2}}$, we have |m^{2}σ_{i} + (m + 1)^{2}| − |m^{2}σ_{i} − (m − 1)^{2}| is 2(m − 1)^{2} if σ_{i} is positive and −2(m − 1)^{2} if σ_{i} is negative. It follows that ${D}_{m}^{*}({D}_{m}(G))$ and ${D}_{m}({D}_{m}^{*}(G))$ are Seidel equienergetic if and only if G has equal numbers of positive and negative Seidel eigenvalues.