矩阵论与数值分析
1 from which we obtain an eigenvector 1 1 of 1 1. 0 0 Similarly, for 2 1, we have an eigenvector 2 1 . 2 for 3 3, we have an eigenvector 3 3 . 1
1
1 Let P 1
0 1
2 3 , then P 1 AP diag{1, 1,3}.
百度搜索“77cn”或“免费范文网”即可找到本站免费阅读全部范文。收藏本站方便下次阅读,免费范文网,提供经典小说教育文库第1章 矩阵运算与矩阵分解_3(18)在线全文阅读。
相关推荐: