Wednesday, August 7, 2019

Yuan method on adjacency matrix controllability



Yuan clearly used weighted matrix, and j->i as direction. So, column to row indicate direction? 
Wikipedia, “In directed graphs, the in-degree of a vertex can be computed by summing the entries of the corresponding column, and the out-degree can be computed by summing the entries of the corresponding row.” The question now is does i->j and j->i matters? It can be tested using a star shaped network with outward and inwarding arrows. A quick exam on these show both star networks should have the same number of minimal control nodes. Well, eigen values of a matrix and its transpose are the same, see  https://yutsumura.com/eigenvalues-of-a-matrix-and-its-transpose-are-the-same/.  
So, i->j and j->i does not matter! This is somewhat shocking to me. 


Eigenvalues of a Matrix and its Transpose are the Same



https://yutsumura.com/eigenvalues-of-a-matrix-and-its-transpose-are-the-same/

Recall that the eigenvalues of a matrix are roots of its characteristic polynomial.
Hence if the matrices A and AT have the same characteristic polynomial, then they have the same eigenvalues.
So we show that the characteristic polynomial pA(t)=det(AtI) of A is the same as the characteristic polynomial pAT(t)=det(ATtI) of the transpose AT.
We have
pAT(t)=det(ATtI)=det(ATtIT)since IT=I=det((AtI)T)=det(AtI)since det(BT)=det(B) for any square matrix B=pA(t).

Therefore we obtain pAT(t)=pA(t), and we conclude that the eigenvalues of A and AT are the same.

Remark: Algebraic Multiplicities of Eigenvalues

Remark that since the characteristic polynomials of A and the transpose AT are the same, it furthermore yields that the algebraic multiplicities of eigenvalues of A and AT are the same.