令J是0對(duì)應(yīng)的n階Jordan塊,即
J=
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
0 0 0 0 0
那么題目里的矩陣就是
A=(x-1)I+(I-J)^{-1}
用二項(xiàng)式定理可得
A^n = sum_{k=0..n} C_n^k (x-1)^{n-k}(I-J)^{-k}
用Taylor公式展開(I-J)^{-k},并注意J^n=0得
(I-J)^{-k} = sum_{s=0..n} C_{s+k}^s J^s
交換一下求和次序得到
A^n = sum_{k=0..n} sum_{s=0..n} C_n^k C_{s+k}^s (x-1)^{n-k}J^s
= sum_{s=0..n} [sum_{k=0..n} C_n^k C_{s+k}^s (x-1)^{n-k}] J^s
J^s項(xiàng)的系數(shù)就是A^n的第s條對(duì)角線的元素大小
大致是這個(gè)方法,如果計(jì)算有問題的話你自己修正