s***e
发帖数: 911 | 1 【 以下文字转载自 Whisper 讨论区 】
【 原文由 space 所发表 】
For any vector b perpendicular to a, we have:
=a_ib_i=0(repeated index means summation)
Lets calculate the i-th component of (S.b):
(S.b)_i=2*a_i*-(Delta_{ik}*b_k)*a^2(repeated index k means summation)
So, (S.b)_1=a^2*b_1,(S.b)_2=a^2*b_2,(S.b)_3=a^2*b_3
Then we know b must be an eigenvecotr of S also, with the eigenvalue -a^2. |
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