作者herstein (翔爸)
看板Math
标题Re: [线代] 试证 此矩阵为hermitian
时间Mon May 2 05:24:53 2011
※ 引述《pennyleo (落日黄花)》之铭言：
: 考虑一有限维度n维矩阵
: 若此矩阵存在n个互为正交的eigenvector
: 且这些eigenvector对应到的eigenvale皆为实数
: 试证明
: 此矩阵为hermitian矩阵
: 谁能帮忙解答
: 谢谢
Assume that {v_i} is a set of orthonormal eigenvectors
of A, where A:V->V is the given linear operator.
Since V has dimension n, {v_i} is orthonormal, it forms
a basis for V. For every vector v in V, v can be expressed
in terms of linear combination of {v_i}:
v=Σ <v,v_i>v_i.
Then Av=Σ<v,v_i>Av_i==Σλ_i<v,v_i>v_i, where λ_i
is the corresponding eigenvalue of A w.r.t. v_i.
Hence
_____
<Av,w>=Σλ_i<v,v_i><w,i_i>
= <v,Aw>
Hence A is Hermitian.
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