perron-frobenius theory Flashcards
if A and B are mxn real matrices:
A>B if all aij>bij and A>=B if all aij>=bij
nonnegative:
A>=0 (so all elements more than or equal to 0)
positive:
A>0 (all elements more than 0)
vector of absolute values of another:
|v|=(|vi|)
matrix of absolute values of another:
|A|=(|aij|)
fun facts:
|Ax|<=|A||x|
A>0 and x>=0, x!=0 => Ax>0
A>=0 => A^k>=0 for all k>=1
A>0 => A^k>0 for all k>=1
A>0 and x>=y with x!=y => Ax>Ay
perron’s theorem:
if A is square and A>0, then p(A)>0 (the spectral radius), p(A) is an eigenvalue of A, there exists an eigenvector with x>0 and Ax=p(A)x, p(A) has algebraic multiplicity 1, p(A) is the eigenvalue of maximum modulus
powers of positive matrices:
if A>0, x is any positive eigenvector corresponding to p(A), and y is any positive eigenvector of A^T corresponding to p(A)=p(A^T) then lim(k->∞)(A/p(A))^k = (xy^T)/(y^(T)x) > 0
reducible matrix:
a square matrix where there exists a permutation matrix such that P^(T)AP=[ X Y
0 Z], where X and Z are both square, a matrix is irreducible otherwise
graphs and matrices!!:
a digraph of an nxn matrix is found by making n points P1,…,Pn, then drawing a directed line between Pi and Pj if aij!=0
strongly connected:
a graph is strongly connected if for any two points, there is a finite sequence of directed links between them
graphs connections and reducability (not a word lmao):
A>=0 is irreducible iff the digraph is strongly connected
perron-frobenius theorem:
if A>=0 is irreducible then p(A)>0, p(A) is an eigenvalue of A, there’s an eigenvector x>0 such that Ax=p(A)x, p(A) has algebraic multiplicity 1
perron root:
λmax(A)=p(A)
perron vector:
the unique vector p such that Ap=p(A)p, p>0, ||p||1=1