Perron–Frobenius theorem:
证明方阵，如果其行列的元素为正值，则存在最大的特征值，并且特征向量的每个元素是正的。这个原理应用在统计推断，经济，人口统计学，搜索引擎的基础。
PF原理：说明了最大特征值的作用和实对称矩阵的作用。
如果临界矩阵A是个实对称矩阵，则具有下列性质：
1.There is a positive real number r, called the Perron root or the Perron–Frobenius eigenvalue, such that r is an eigenvalue of A and any other eigenvalue λ (possibly,complex) is strictly smaller than r in absolute value, |λ| < r. Thus, the spectral radiusρ(A) is equal to r.
2.The Perron–Frobenius eigenvalue is simple: r is a simple root of the characteristic polynomial of A. Consequently, eigenspace associated to r is one-dimensional. (Same is true for the left eigenspace i.e. eigenspace for AT.)
3.There exists an eigenvector v = (v1,…,vn) of A with eigenvalue r such that all components ofv are positive: A v = r v, vi > 0 for 1 ≤ i ≤ n. (Respectively exists positive left eigenvector w : wT A = r wT, wi > 0.)转置就可以了。
4.There are no other positive (moreover non-negative) eigenvectors except v (respectively the left eigenvectors except w). I.e. all other eigenvectors must have at least one negative or complex component.
5.
, where the left and right eigenvectors for A are normalized so thatwTv = 1. Moreover the matrix v wT is the spectral projection corresponding to r, such projection is called the Perron projection.
, where the left and right eigenvectors for A are normalized so thatwTv = 1. Moreover the matrix v wT is the spectral projection corresponding to r, such projection is called the Perron projection.6.Collatz–Wielandt formula: for all non-negative non-zero vectors x let f(x) be the minimum value of [Ax]i / xi taken over all those i such that xi ≠ 0. Then f is a real valued function whose maximum is the Perron–Frobenius eigenvalue.
7.Min-max Collatz–Wielandt formula is similar to the one above: for all strictly positive vectors x (pay attention that one above required only non-negativity here) let g(x) be the maximum value of [Ax]i / xi taken over i. Then g is a real valued function whose minimum is the Perron–Frobenius eigenvalue.
8.The Perron–Frobenius eigenvalue satisfies the inequalities：
因为在科研中，比如计算pagerank,希望向量的每个值都是正的，在这里我着重介绍定理1,4的证明：
定理四启发式证明：由于不同特征值对应的特征向量是正交的，所以如果有一个特征值对应的特征向量的每个元素大于零，则其余的特征向量的分量要不然是负数要不然就是复数。
数学证明：设 (λ, y)为A的一个特征值和特征向量，(r, x)为A的另一个特征值和特征向量。则r xt y = (xt A) y= xt (A y)=λ xt y。所以r=λ。
定理一证明：矩阵除以矩阵的普半径，最大特征值在半径为1的单位园上。假设有不止一个特征值在这个单位圆上，最后得出矛盾来证明。
