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We now express the L polynomials of signed cycles and

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Proof.
In general, we can give the following upper bound for the largest Laplacian eigenvalue of a signed graph. Other similar bounds can be found in [16].
Lemma 2.8.
          Let  Γ=(G,σ)Γ=(G,σ)be a signed graph with  Δ1Δ1and  Δ2Δ2being the first and second largest vertex degrees in G, and let  μ(Γ)μ(Γ)be its Laplacian spectral radius. Then  μ(Γ)≤Δ1+Δ2μ(Γ)≤Δ1+Δ2, with equality if and only if  Γ=K1,nΓ=K1,nor  Γ=(Kn,−)Γ=(Kn,−).Proof.
          For any given matrix A  , let A A be the absolute value matrix whose entries are obtained from A by replacing each entry with the corresponding absolute value. Recall that 4EGI-1 the largest eigenvalue of a square matrix A   is less than or equal to the largest eigenvalue of A A (due to the Perron–Frobenius theorem). Hence, we have scrotum μ(Γ)=μ(BB?)≤μ( BB? )=μ(G,−)μ(Γ)=μ(BB?)≤μ( BB? )=μ(G,−), namely the largest L-eigenvalue of a signed graph is bounded by the largest L  -eigenvalue of the corresponding all-negative edges signed graph. On the other hand, L(G,−)L(G,−) is the signless Laplacian of the underlying simple graph G  , for which the inequality μ(G,−)≤Δ1+Δ2μ(G,−)≤Δ1+Δ2 holds (e.g., Theorem 4.2 in [9]). The equality holds if and only if G is either the n  -star K1,nK1,n or the complete graph KnKn. This completes the proof.  □

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