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Guo et al. Intell Robot 2023;3(4):596-613 I http://dx.doi.org/10.20517/ir.2023.32 Page 7 of 18

 . . . 
 1 2 3 
 
−1 1 0 . . . 0 
 
−1 = −1 0 1 . . . 0  ∈ R × (12)
 
 . . . . . . . . . . 
 . . . . . 
 
 −1 0 0 . . . 1 
 
Í
Thus, 1 = and = − 1, where = {2, 3, . . . , }. The state vector of the system is transformed
=1
into the error vector by linear transformations. System (4) is equivalent to the system as follows:

−1
⊗ + −1 ⊗ + ⊗ = 0 (13)
¥
¤
Lemma 4. For Laplacian matrix = [ ] × related to the strongly connected graph , there exists positive
vector = [ 1 , 2 , . . . , ] that is the eigenvector corresponding to the eigenvalue 0 of the matrix Λ , where
Λ = −1 > 0 and Í = 1.
=1
Proof: Define = [ ] × = [− ] × and = max ∈ | / |. Since Laplacian matrix is related to the
strongly connected graph, matrix Λ + is semi-positive definite. Based on the Gershgorin circle theorem,
the eigenvalues of the matrix Λ and Λ + satisfy conditions as follows:

(Λ ) ∈ { ∈ C k + | ≤ }
(14)
(Λ + ) ∈ { ∈ C k | ≤ }

where ∈ . Since the eigenvalues of the matrix will not be altered after transposition, for ∀ ∈

( Λ + ) = max {| 1 |, | 2 |, . . . , | |} = (15)

Since Λ + ≥ 0, there exists a non-negative vector corresponding to the right eigenvector of the

eigenvalues . Thus, one has that

Λ = ( Λ + − )

= ( Λ + ) − (16)

= − = 0
then
(17)
Λ = ( Λ ) = 0

Since = − , there exists a non-negative vector satisfies the condition as follows:

Λ = Λ(− ) = 0 (18)