On the Properties of Orthogonal Projection Method for Reaching Consensus

Abstract

The article is devoted to an asymptotic behavior of a multi-agent system with information links. We proved that the orthogonal projection method proposed for the regularization of the consensus protocol is characterized by a pseudoinverse matrix for the introduced auxiliary matrix for an arbitrary communication digraph of a multi-agent system. We cosidered the eigenprojection of the Laplacian matrix corresponding to the communication digraph, in which the influences on the fixed agent change proportionally. We obtained a number of results that are of independent importance and can be used in models of multi-agent systems with different protocols.

APPENDIX

Proof of Proposition 2. (1) The matrix U has the following form:

$$U = \left( {\begin{array}{*{20}{c}} 1&{{{l}_{{12}}}}& \cdots &{{{l}_{{1k'}}}}&{{{{\mathbf{0}}}_{{1,n - k}}}} \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 1&{{{l}_{{k2}}}}& \cdots &{{{l}_{{kk'}}}}&{{{{\mathbf{0}}}_{{1,n - k}}}} \\ {{{{\mathbf{1}}}_{{n - k,1}}}}&*&*&*&{{{L}_{R}}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&{{{U}_{1}}}&{}&{} \\ 1& \ddots &{}&{} \\ \vdots &{}&{{{U}_{{v}}}}&{} \\ {{{{\mathbf{1}}}_{{n - k,1}}}}&*&*&{{{L}_{R}}} \end{array}} \right),$$

where k' = k – ${v}$ + 1, k = $\sum\nolimits_{i = 1}^{v} {{{m}_{i}}}$ (${v}$ is a number of basic bicomponents), and the matrix L_R is square and nonsingular. Assume that the mentioned minor was obtained by deleting rows among which there is at least one row with a number from the set {k + 1, …, n}. Then the submatrix U $\left( \begin{gathered} {{i}_{1}}...{{i}_{r}} \\ 1...r \\ \end{gathered} \right)$ will have a block-triangular form, and its right lower square block L_R'  contains a zero row. Therefore det L_R' = 0 and det U $\left( \begin{gathered} {{i}_{1}}...{{i}_{r}} \\ 1...r \\ \end{gathered} \right)$ = 0.

(2) Let the set {i₂, …, i_r} contain all vertices {  j₁, …, ${{j}_{{{{m}_{s}}}}}$} of one sth basic bicomponent. Then the submatrix with rows {j₁, …, ${{j}_{{{{m}_{s}}}}}$} contains m_s – 1 nonzero columns. The minor detU $\left( \begin{gathered} {{i}_{2}}...{{i}_{r}} \\ 2...r \\ \end{gathered} \right)$ consists of terms, each of which is the product of r – 1 submatrix elements taken from different rows and columns. Therefore, each term contains a zero factor and the minor detU $\left( \begin{gathered} {{i}_{2}}...{{i}_{r}} \\ 2...r \\ \end{gathered} \right)$ is equal to zero.

(3) If the minor detU  $\left( \begin{gathered} {{i}_{2}}...{{i}_{r}} \\ 2...r \\ \end{gathered} \right)$ is different from zero, then according to item 2, the set {i₂, …, i_r} consists of m_s – 1 rows (s = 1, …, ${v}$) from each basic bicomponent and rows corresponding to vertices from $\bar {\mathcal{K}}$. Obviously, the minor is equal to the determinant of the block-diagonal matrix, i.e.,

$$\det U\left( \begin{gathered} {{i}_{2}}...{{i}_{r}} \\ 2...r \\ \end{gathered} \right) = \det \left( {\begin{array}{*{20}{c}} {U_{1}^{'}}&{}&{}&{} \\ {}& \ddots &{}&0 \\ {}&{}&{U_{{v}}^{'}}&{} \\ {}&*&{}&{{{L}_{R}}} \end{array}} \right),$$

where the matrix $U_{i}^{'}$ is obtained from U_i by deleting one i_kth row. According to the matrix tree theorem, the determinant of the matrix $U_{i}^{'}$ is equal to the minor of any element of the i_kth row of U_i and its absolute value is equal to the sum of the weights of all outgoing trees from the i_kth vertex of the ith basic bicomponent. It is true for any block $U_{t}^{'}$.

Thus, the absolute value of the non-zero minor detU $\left( \begin{gathered} {{i}_{2}}...{{i}_{r}} \\ 2...r \\ \end{gathered} \right)$ is equal to the product of det L_R and the weight of the set of all outgoing forests on $\mathcal{K}$ with roots {1, …, n}\{i₂, …, i_r}.

Proof of Proposition 3. (1) Without loss of generality, we assume that the vertices are numbered as j_p = p, p = 1, …, m_s. Let {i₁, …, i_r} be a subset {i₁, …, i_r'}, where r' = r – |$\bar {\mathcal{K}}$|, is a subset of the base vertex set. Consider the determinant det U$\left( \begin{gathered} 1...{{m}_{s}}...{{i}_{r}} \\ 1...r \\ \end{gathered} \right)$ and represent it in block form as

$$\det \left( {\begin{array}{*{20}{c}} 1&{{{l}_{{12}}}}& \cdots &{{{l}_{{1{{m}_{s}}}}}}&{{{0}_{{1,r' - {{m}_{s}}}}}}&0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 1&{{{l}_{{{{m}_{s}}2}}}}& \cdots &{{{l}_{{{{m}_{s}}{{m}_{s}}}}}}&{{{1}_{{1,r' - {{m}_{s}}}}}}&0 \\ {{{{\mathbf{1}}}_{{r' - p,1}}}}&{{{0}_{{r' - p,1}}}}& \cdots &{{{0}_{{r' - p,1}}}}&{{{Q}_{{r' - {{m}_{s}}}}}}&0 \\ *&*&*&*&*&{{{L}_{R}}} \end{array}} \right) = \det \left( {\begin{array}{*{20}{c}} {{{Q}_{{{{m}_{s}}}}}}&{\mathbf{0}}&{\mathbf{0}} \\ *&{{{Q}_{{r' - {{m}_{s}}}}}}&{\mathbf{0}} \\ *&*&{{{L}_{R}}} \end{array}} \right),$$

According to the matrix tree Theorem 2, the algebraic complement of the first element of any kth row of the block ${{Q}_{{{{m}_{s}}}}}$ is equal to the sum of the weights of the trees outgoing from the vertex k. Therefore, the determinant of the matrix ${{Q}_{{{{m}_{s}}}}}$ is equal to the sum of the weights of all outgoing trees of the basic bicomponent with the vertex set {1, …, m_s} and $\left| {\det U\left( \begin{gathered} 1...{{m}_{s}}...{{i}_{r}} \\ 1...r \\ \end{gathered} \right)} \right|$ the product of the sum of weights of all outgoing trees of the basic bicomponent with vertices {1, …, m_s} and |det${{Q}_{{r' - {{m}_{s}}}}}$|, which is equal to the sum of the weights of the maximum outgoing forests on $\mathcal{K}$\{1…m_s} with vertices: $\mathcal{K}$\{i₁, …, i_r'}, and det L_R. Thus, equality (8) is satisfied.

(2) We show that the signs of minors det U  $\left( \begin{gathered} {{j}_{p}}{{K}_{p}} \\ 1...r \\ \end{gathered} \right)$ ≠ 0 and det U  $\left( \begin{gathered} {{K}_{p}} \\ 2...r \\ \end{gathered} \right)$ ≠ 0, p = 1, …, m_s, is same as.

Note that det U$\left( \begin{gathered} {{i}_{1}}...{{i}_{r}} \\ 1...r \\ \end{gathered} \right)$ = $\sum\nolimits_{k = 1}^{{{m}_{s}}} ($–1)^1+k+ldet$U_{{k1}}^{s}$ξ = (–1)^lζξ, where l is a number of rows in U$\left( \begin{gathered} {{i}_{1}}...{{i}_{r}} \\ 1...r \\ \end{gathered} \right)$ up to sth basic bicomponent, U^s = (${{{\mathbf{1}}}_{{{{m}_{s}}}}}$U_s), ζ > 0 is a weight of the set of all outgoing trees in sth basic bicomponent, ξ is a product of determinants of other blocks. Since U$\left( \begin{gathered} {{j}_{p}}{{K}_{p}} \\ 1...r \\ \end{gathered} \right)$ differs from U$\left( \begin{gathered} {{i}_{1}}...{{i}_{r}} \\ 1...r \\ \end{gathered} \right)$ by a permutation of one row, their determinants can only differ in sign, and we have

$$\det U\left( \begin{gathered} {{j}_{p}}{{K}_{p}} \\ 1...r \\ \end{gathered} \right) = \det U\left( \begin{gathered} {{i}_{1}}...{{i}_{r}} \\ 1...r \\ \end{gathered} \right){{( - 1)}^{{l + p - 1}}} = {{( - 1)}^{{2l + p - 1}}}\zeta \xi ,$$

where p is the row number in the sth block.

So, the sign of det  U $\left( \begin{gathered} {{j}_{p}}{{K}_{p}} \\ 1...r \\ \end{gathered} \right)$ is equal to (–1)^{p – 1}ξ. It can be easily established that det  U  $\left( \begin{gathered} {{K}_{p}} \\ 2...r \\ \end{gathered} \right)$ has the sign (–1)^{p + 1}ξ.

Proof of Theorem 3. Due to (6), the sum of the elements of the first row U⁺ can be written as

$$\sum\limits_{{{i}_{1}} \in N}^{} {u_{{1{{i}_{1}}}}^{ + }} = \frac{{{{\Sigma }_{\mathcal{K}}} + {{\Sigma }_{{\bar {\mathcal{K}}}}}}}{D},$$

(A.1)

where ${{\Sigma }_{\mathcal{K}}}$ corresponds to elements from $\mathcal{K}$, ${{\Sigma }_{{\bar {\mathcal{K}}}}}$ to vertices from $\bar {\mathcal{K}}$ = N \$\mathcal{K}$ and

$$D = {{\sum\limits_{{{i}_{1}} < ... < {{i}_{r}}}^{} {\left( {\det U\left( \begin{gathered} {{i}_{1}}...{{i}_{r}} \\ 1...r \\ \end{gathered} \right)} \right)} }^{2}}.$$

According to item 1 of the Proposition 2, if the set {i₁, …, i_r} does not contain all vertices from $\bar {\mathcal{K}}$, then the corresponding term detU$\left( \begin{gathered} {{i}_{1}}...{{i}_{r}} \\ 1...r \\ \end{gathered} \right)$ in D representation is equal to zero. Also, by virtue of item 1 of the Proposition 2, we have

$${{\Sigma }_{{\bar {\mathcal{K}}}}} = \sum\limits_{{{i}_{1}} \in \bar {\mathcal{K}}}^{} {\sum\limits_{{{i}_{2}} < ... < {{i}_{r}}}^{} {\det U\left( \begin{gathered} {{i}_{2}}...{{i}_{r}} \\ 2...r \\ \end{gathered} \right)} \det U\left( \begin{gathered} {{i}_{1}}...{{i}_{r}} \\ 1...r \\ \end{gathered} \right)} = 0.$$

Note that ${{\Sigma }_{\mathcal{K}}}$ and D include the factor det L_R. The proof that $\frac{{{{\Sigma }_{\mathcal{K}}}}}{D}$ is equal to 1 is given in [11].

Proof of Proposition 4. Let us prove the assertion using the representation $u_{{1{{k}_{1}}}}^{ + }$. Indeed, all corresponding factors detU$\left( \begin{gathered} {{i}_{1}}...{{i}_{r}} \\ 1...r \\ \end{gathered} \right)$, detU$\left( \begin{gathered} {{j}_{1}}...{{j}_{r}} \\ 1...r \\ \end{gathered} \right)$ in the expression (6) for $u_{{1{{i}_{1}}}}^{ + }$ and $u_{{1{{j}_{1}}}}^{ + }$ differ only in sign, while detU$\left( \begin{gathered} {{i}_{2}}...{{i}_{r}} \\ 2...r \\ \end{gathered} \right)$ and detU$\left( \begin{gathered} {{j}_{2}}...{{j}_{r}} \\ 2...r \\ \end{gathered} \right)$ are weights of outgoing trees from i₁ and j₁.

Proof of Theorem 4. Let us prove the theorem constructively, i.e., we construct a diagonal matrix D. Without loss of generality, we assume that the Laplacian matrix L has a block-triangular form, and for L we also construct an auxiliary matrix B^L (A.2), which is obtained from L by replacing the first column in each block L_s corresponding to the basic bicomponent s per column of ones:

$${{B}^{L}} = \left( {\begin{array}{*{20}{c}} {B_{1}^{L}}&0& \cdots &0&0 \\ 0&{B_{2}^{L}}& \cdots &0&0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0&0& \cdots &{B_{{v}}^{L}}&0 \\ *&*& \cdots & \cdots &{{{L}_{R}}} \end{array}} \right),$$

(A.2)

the diagonal blocks $B_{s}^{L}$, s = 1, …, ${v}$, for the matrix B^L are defined as

$$B_{s}^{L} = \left( {\begin{array}{*{20}{c}} 1&{l_{{12}}^{s}}& \cdots &{l_{{1{{m}_{s}}}}^{s}} \\ 1&{l_{{22}}^{s}}& \cdots &{l_{{2{{m}_{s}}}}^{s}} \\ \vdots & \vdots & \ddots & \vdots \\ 1&{l_{{{{m}_{s}}2}}^{s}}& \cdots &{l_{{{{m}_{s}}{{m}_{s}}}}^{s}} \end{array}} \right).$$

(A.3)

Similarly to B^L and $B_{s}^{L}$, we define matrices B^M and $B_{s}^{M}$ for M.

Note that the ith row M is obtained by multiplying the analogous row L by τ_i. Let us define an eigenprojection ${{M}^{ \vdash }}$ according to item 1 of the Theorem 1:

$$m_{{ij}}^{ \vdash } = \frac{{\varepsilon ({{\mathcal{F}}^{{j \to i}}})}}{{\varepsilon (\mathcal{F})}}.$$

(A.4)

According to the Theorem 2, for each sth basic bicomponent, the sum of the weights of all outgoing trees (on the set of all vertices from the given basic bicomponent) is equal to det(B_s). Therefore ε($\mathcal{F}$) = det(B^M).

Let the vertex i be not reachable from  j  in any maximal outgoing forest. Since the graphs corresponding to the Laplacian matrices L and M = TL have the same structure, then $m_{{ij}}^{ \vdash }$ = $l_{{ij}}^{ \vdash }$ = 0.

We consider the case when i is reachable from j in at least one maximal out-forest and  j is a vertex from the sth basic bicomponent. Denote by {Γ_k(V, E_k)} the set of all spanning subgraphs, which is obtained from the set of all maximal outgoing spanning digraphs in which i is reachable from the root vertex j, with the addition of all missing arcs from basic bicomponents. Let ${{B}^{{{{L}^{k}}}}}$ and ${{B}^{{{{M}^{k}}}}}$ be the corresponding matrices of the resulting digraphs constructed by analogy with B^L and B^M. The diagonal blocks ${{B}^{{{{L}^{k}}}}}$ and ${{B}^{{{{M}^{k}}}}}$, which correspond to the basic bicomponents, match with the similar blocks of the matrices B^L and B^M, respectively. Obviously, the number ε(${{\mathcal{F}}^{{j \to i}}}$) for the digraph corresponding to the matrix M is equal to the sum of the algebraic complements of the elements (  j, i') of the ${{B}^{{{{M}^{k}}}}}$ matrices, i.e., ε(${{\mathcal{F}}^{{j \to i}}}$) = $\sum\nolimits_k^{} {B_{{ji'}}^{{{{M}^{k}}}}}$, where i' ∈ {1, …, n} is the column number ${{B}^{{{{M}^{k}}}}}$, which corresponds to the number of a column of ones in the submatrix corresponding to the basic bicomponent s.

Let j' ∈ {1, …, m_s} be the row number of the block with the number s that matches the row j.

Unlike i', the number j' points to the row of the block s. Then due to $B_{q}^{M}$ = $B_{q}^{{{{M}_{k}}}}$ for all q and k we get

$$\begin{gathered} m_{{ij}}^{ \vdash } = \frac{{\varepsilon ({{\mathcal{F}}^{{j \to i}}})}}{{\varepsilon (\mathcal{F})}} = \frac{{\sum\limits_k^{} {\det B_{{ji'}}^{{{{M}^{k}}}}} }}{{\det {{B}^{M}}}} = \frac{{\prod\limits_{q = 1,q \ne s}^{v} {\det B_{q}^{M}\left| {\det B_{s}^{M}\left( \begin{gathered} (1...{{m}_{s}}){{\backslash }}{\kern 1pt} j{{'}} \\ 2...{{m}_{s}} \\ \end{gathered} \right)} \right|\sum\limits_k^{} {\det M_{R}^{k}} } }}{{\det {{M}_{R}}\prod\limits_{q = 1}^{v} {\det B_{q}^{M}} }} \\ = \frac{{\left| {\det B_{s}^{M}\left( \begin{gathered} (1...{{m}_{s}}){{\backslash }}{\kern 1pt} j{{'}} \\ 2...{{m}_{s}} \\ \end{gathered} \right)} \right|\sum\limits_k^{} {\det M_{R}^{k}} }}{{\det {{M}_{R}}\det B_{s}^{M}}}. \\ \end{gathered} $$

Note that $L_{R}^{k}$ and $M_{R}^{k}$ are blocks corresponding to non-basic vertices in the matrices ${{B}^{{{{L}^{k}}}}}$ and ${{B}^{{{{M}^{k}}}}}$, respectively. Next, we represent the resulting expression through the matrix L:

$$m_{{ij}}^{ \vdash } = \frac{{\prod\limits_{i = 1,i \ne j'}^{{{m}_{s}}} {{{\tau }_{i}}\left| {\det L_{s}^{M}\left( \begin{gathered} (1...{{m}_{s}}){{\backslash }}{\kern 1pt} j{{'}} \\ 2...{{m}_{s}} \\ \end{gathered} \right)} \right|\sum\limits_k^{} {\det L_{R}^{k}} } }}{{\prod\limits_{i = 1}^{{{m}_{s}}} {{{\tau }_{i}}\sum\limits_{p = 1}^{{{m}_{s}}} {\frac{1}{{{{\tau }_{p}}}}\det B_{s}^{L}\left( \begin{gathered} (1...{{m}_{s}}){{\backslash }}p \\ 2...{{m}_{s}} \\ \end{gathered} \right)\det {{L}_{R}}} } }} = \frac{{\left| {\det L_{s}^{M}\left( \begin{gathered} (1...{{m}_{s}}){{\backslash }}{\kern 1pt} j{{'}} \\ 2...{{m}_{s}} \\ \end{gathered} \right)} \right|\sum\limits_k^{} {\det L_{R}^{k}} }}{{{{\tau }_{j}}\sum\limits_{p = 1}^{{{m}_{s}}} {\frac{1}{{{{\tau }_{p}}}}\det B_{s}^{L}\left( \begin{gathered} (1...{{m}_{s}}){{\backslash }}p \\ 2...{{m}_{s}} \\ \end{gathered} \right)\det {{L}_{R}}} }}.$$

Multiply the denominator and numerator by $\prod\nolimits_{i = 1}^{v} {\det } (B_{i}^{L})$. According to the matrix tree theorem, this number is nonzero and is equal to the weight of the set of all outgoing spanning forests in a digraph that consists of only basic bicomponents. Then:

$$m_{{ij}}^{ \vdash } = \frac{{\left| {\det L_{s}^{M}\left( \begin{gathered} (1...{{m}_{s}}){{\backslash }}{\kern 1pt} j{{'}} \\ 2...{{m}_{s}} \\ \end{gathered} \right)} \right|\sum\limits_k^{} {\det L_{R}^{k}\det B_{s}^{L}\prod\limits_{q \ne s}^{} {\det B_{q}^{L}} } }}{{{{\tau }_{i}}\sum\limits_{p = 1}^{{{m}_{s}}} {\frac{1}{{{{\tau }_{p}}}}\det B_{s}^{L}\left( \begin{gathered} (1...{{m}_{s}}){{\backslash }}{\kern 1pt} p \\ 2...{{m}_{s}} \\ \end{gathered} \right)\det {{L}_{R}}\prod\limits_{q = 1}^{v} {\det B_{q}^{L}} } }}.$$

Note that in the last fraction

$$l_{{ij}}^{ \vdash } = \frac{{\left| {\det L_{s}^{M}\left( \begin{gathered} (1...{{m}_{s}}){{\backslash }}{\kern 1pt} j{{'}} \\ 2...{{m}_{s}} \\ \end{gathered} \right)} \right|\sum\limits_k^{} {\det L_{R}^{k}\prod\limits_{q \ne s}^{} {\det B_{q}^{L}} } }}{{\det {{L}_{R}}\prod\limits_{q = 1}^{v} {\det B_{q}^{L}} }}.$$

Then

$$m_{{ij}}^{ \vdash } = l_{{ij}}^{ \vdash }\frac{{\det B_{s}^{L}}}{{{{\tau }_{j}}\det {{C}_{s}}}},$$

(A.5)

where the matrices C_s, s = 1, …, ${v}$, are defined as follows:

$${{C}_{s}} = \left( {\begin{array}{*{20}{c}} {\frac{1}{{\tau _{1}^{s}}}}&{l_{{12}}^{s}}& \cdots &{l_{{1{{m}_{s}}}}^{s}} \\ {\frac{1}{{\tau _{2}^{s}}}}&{l_{{22}}^{s}}& \cdots &{l_{{2{{m}_{s}}}}^{s}} \\ \vdots & \vdots & \ddots & \vdots \\ {\frac{1}{{\tau _{{{{m}_{s}}}}^{s}}}}&{l_{{{{m}_{s}}2}}^{s}}& \cdots &{l_{{{{m}_{s}}{{m}_{s}}}}^{s}} \end{array}} \right).$$

(A.6)

We construct two diagonal matrices F = diag(f₁, …, f_n) and H = diag(h₁, …, h_n) as follows: f_t = det$B_{s}^{L}$ and h_t = detC_s if the vertex t belongs to the sth basic bicomponent. For all other diagonal elements of the matrices F and H we set f_t = h_t = 1.

Then

$${{M}^{ \vdash }} = {{L}^{ \vdash }}{{T}^{{ - 1}}}F{{H}^{{ - 1}}} = {{L}^{ \vdash }}D.$$

(A.7)

Note that in (A.5) no requirements are imposed on the vertex i. In particular, it may belong to some basic bicomponent.