14/08/08 00:42:19.02 5QKqY+aN.net
>>834
行列式ではなく固有値
トレース（対角成分の和）＝固有値の和
でマルコフ行列だから固有値１を持つのはすぐわかるから
.90+.85=1+λが成り立つ
これを解いてλ=.75
マルコフ行列（列を足して１になる非負行列）なら固有値1を持つよ
証明は解りやすいのは例えばこれ
A Markov matrix A always has an eigenvalue 1. All other eigenvalues are in absolute
value smaller or equal to 1.
Proof. For the transpose matrix AT , the sum of the row vectors is equal to 1. The matrix
AT therefore has the eigenvector [111......1]T.
Because A and AT have the same determinant also A . In and AT . In have the same
determinant so that the eigenvalues of A and AT are the same. With AT having an eigenvalue
1 also A has an eigenvalue 1.
Assume now that v is an eigenvector with an eigenvalue || > 1. Then Anv = ||nv has
exponentially growing length for n ! 1. This implies that there is for large n one coefficient
[An]ij which is larger than 1. But An is a stochastic matrix (see homework) and has all entries
1. The assumption of an eigenvalue larger than 1 can not be valid.