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There is a question in my textbook as follows:

Prove that for a square matrix $A$ to be commutable with any other square matrix $B$ (meaning that $AB=BA$), it is necessary and sufficient for $A$ to be a scalar matrix(i.e., to be of the form $cI$, where $c$ is a scalar and $I$ is the identity matrix).

I have proved the sufficiency part as below:

Given $A=cI$

Then $AB=cIB=cB$ and $BA=cBI=cB$.

Hence, it is proved that $AB=BA$.

But I cannot prove the necessity part, i.e., given $AB=BA$, I cannot prove that $A$ is of the form $cI$.

Please anyone help me solve it. Thanks in advance.

user587389
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Hint to get you started:

You know that $A$ commutes with all matrices $B$. It thus also commutes with elementary matrices (matrices that are $1$ on one position and $0$ elsewhere). What can you conclude from this?

J. De Ro
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  • Please explain a bit more. I can't understand. I have viewed all the other answers, but still it is not clear to me. – user587389 Nov 08 '19 at 12:19
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    Ok. Take your matrix $A$. Consider the matrix $E_{11}$ that is $1$ on position $(1,1)$ and $0$ elsewhere. What do you get when you calculate $AE_{11}$ and $E_{11}A$? – J. De Ro Nov 08 '19 at 12:23
  • Thanks a lot. Now it's clear to me. – user587389 Nov 08 '19 at 12:55
  • Yes, the key to finish is now to use that $AE_{ij} = E_{ij}A, i \neq j$, forcing all diagonal elements to be equal. Glad I could help. – J. De Ro Nov 08 '19 at 12:57
  • In terms of abstract algebra, this means that the center of the real matrices are the diagonal matrices with constant diagonal. – J. De Ro Nov 08 '19 at 13:01