高代:同时可以对角化,另有证法吗?
已知,且可相似对角化,证明:可同时相似对角化。 证明: 因为可以相似对角化,不妨设,
其中是的不同特征值, 对应重数为.
对式子,左乘与右乘可得,
于是可以分块对角矩阵D可交换,因此可设, 其中是阶方阵.
且由于可对角化,从而也可对角化,
因此存在可逆分块矩阵,使得为对角阵.
于是可得为对角阵,为对角阵,
English Edition: Let and be matrix. If and can be diagonalized respectively by similarity, then and can be diagonalized by similarity together.
Proof: Since can be diagonalized by similarity, there exsits invertible matrix such that , where are these differents eigenvalues of . Their multiplicity is respectively. To equation ,left multiplicate and right multiplicate , then . Hence can be exchanged with block diagonal matrix D. So we let , where is a matrix. Now by the diagonalized property of , we get that are diagonalized. Then, there exsits block invertible matrix such that is a diagonal matrix. Hence and are both diagonal matrixs.
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