n阶方阵与任意n阶乘法可换那么这个方阵是某个纯量矩阵怎么证明 证明：与任意n阶方阵都乘法可交换的方阵一定是数量矩阵。
\u5982\u4f55\u8bc1\u660e\uff1a\u4e0e\u4efb\u610f\u4e00\u4e2an\u9636\u65b9\u9635\u76f8\u4e58\u90fd\u53ef\u4ea4\u6362\u7684\u65b9\u9635\u5fc5\u4e3a\u6570\u91cf\u77e9\u9635\uff1f\u8bc1:\u8bbea=(aij)\u4e0e\u4efb\u610f\u7684n\u9636\u77e9\u9635\u53ef\u4ea4\u6362,\u5219a\u5fc5\u662fn\u9636\u65b9\u9635.
\u8bbeeij\u662f\u7b2ci\u884c\u7b2cj\u5217\u4f4d\u7f6e\u4e3a1,\u5176\u4f59\u90fd\u662f0\u7684n\u9636\u65b9\u9635.
\u5219eija=aeij
eija\u662f\u7b2ci\u884c\u4e3aaj1,aj2,...,ajn,\u5176\u4f59\u884c\u90fd\u662f0\u7684\u65b9\u9635
aeij\u662f\u7b2cj\u5217\u4e3aa1i,a2i,...,ani,\u5176\u4f59\u5217\u90fd\u662f0\u7684\u65b9\u9635
\u6240\u4ee5\u5f53i\u2260j\u65f6,aij=0.
\u6240\u4ee5a\u662f\u4e00\u4e2a\u5bf9\u89d2\u77e9\u9635.
\u8bbee(i,j)\u662f\u5bf9\u6362i,j\u4e24\u884c\u7684\u521d\u7b49\u77e9\u9635.
\u7531e(i,j)a=ae(i,j)\u53ef\u5f97
aii=ajj
\u6240\u4ee5a\u662f\u4e3b\u5bf9\u89d2\u7ebf\u5143\u7d20\u76f8\u540c\u7684\u5bf9\u89d2\u77e9\u9635,\u5373\u6570\u91cf\u77e9\u9635.
\u8bc1: \u8bbe A=(aij) \u4e0e\u4efb\u610f\u7684n\u9636\u77e9\u9635\u53ef\u4ea4\u6362, \u5219A\u5fc5\u662fn\u9636\u65b9\u9635.
\u8bbeEij\u662f\u7b2ci\u884c\u7b2cj\u5217\u4f4d\u7f6e\u4e3a1,\u5176\u4f59\u90fd\u662f0\u7684n\u9636\u65b9\u9635.
\u5219EijA = AEij
EijA \u662f \u7b2ci\u884c\u4e3a aj1,aj2,...,ajn, \u5176\u4f59\u884c\u90fd\u662f0\u7684\u65b9\u9635
AEij \u662f \u7b2cj\u5217\u4e3a a1i,a2i,...,ani, \u5176\u4f59\u5217\u90fd\u662f0\u7684\u65b9\u9635
\u6240\u4ee5\u5f53i\u2260j\u65f6, aij=0.
\u6240\u4ee5A\u662f\u4e00\u4e2a\u5bf9\u89d2\u77e9\u9635.
\u8bbeE(i,j)\u662f\u5bf9\u6362i,j\u4e24\u884c\u7684\u521d\u7b49\u77e9\u9635.
\u7531E(i,j)A=AE(i,j)\u53ef\u5f97
aii=ajj
\u6240\u4ee5A\u662f\u4e3b\u5bf9\u89d2\u7ebf\u5143\u7d20\u76f8\u540c\u7684\u5bf9\u89d2\u77e9\u9635, \u5373\u6570\u91cf\u77e9\u9635.
设Eij是第i行第j列位置为1,其余都是0的n阶方阵.
则EijA = AEij
EijA 是 第i行为 aj1,aj2,...,ajn, 其余行都是0的方阵
AEij 是 第j列为 a1i,a2i,...,ani, 其余列都是0的方阵
所以当i≠j时, aij=0.
所以A是一个对角矩阵.
设E(i,j)是对换i,j两行的初等矩阵.
由E(i,j)A=AE(i,j)可得
aii=ajj
所以A是主对角线元素相同的对角矩阵, 即数量矩阵.
答：真的不会。
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