If a $n \times n$ matrix $A$ satisfies $AB = BA$ for any $n \times n$ matrix $B$, then $A$ must be of the form $cI$ where $c$ is a scalar and $I$ is the identity matrix. I tried to use the definition of matrix multiplication, but I failed. I am wondering if I should use the inverse to solve the problem, but since now, I have no idea.
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4Please provide more details, for instance, what are your thoughts about the question, what you have tried, where you are stuck. – Sahiba Arora Jun 26 '17 at 15:10
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Thank you for your reply! I tried to use the definition of matrix multiplication, but I failed. I am wondering if I should use the inverse to solve the problem, but since now, I have no idea. – GraceZ Jun 26 '17 at 15:13
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1Possible duplicate of https://math.stackexchange.com/questions/1120202/matrices-that-commute-with-all-matrices – Crostul Jun 26 '17 at 15:13
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@GraceZ The matrices are not given to be invertible. So you can't use the inverse. – Sahiba Arora Jun 26 '17 at 15:15
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@Sahiba Arora You are right, thank you for your correction!!! – GraceZ Jun 26 '17 at 15:21
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@Crostul Opps! I did not find the question. Thank you very much!!! – GraceZ Jun 26 '17 at 15:24
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Are you looking for something relating to a form of Schur's Lemma? – snulty Jun 26 '17 at 15:40
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@snulty Thank you very much! But it seems that they do not have many relation. – GraceZ Jun 26 '17 at 15:51
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Take as $B$ the $n^2$ matrices $E_{ij}$, whose elements are $1$ in position $ij$ and $0$ elsewhere.
Francesco Polizzi
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Thank you very much!!! I know the solution now. I am sorry that my reputation is not enough, so I cannot change the upvote score. – GraceZ Jun 26 '17 at 16:24