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We know that if $A$ is $2\times 2$ square matrix\begin{bmatrix}a&b\\c&d\end{bmatrix}, such that $A$ is commutative over multiplication with any $2\times 2$ matrix, then $A$ is a scalar matrix. To prove that I tried to rely on the matrix multiplication and then I got $4$ equations to solve. I calculated the dot product of $A$ and $B$:\begin{bmatrix}e&f\\g&h\end{bmatrix} two times in both directions. So $$ce + dg = ga + hc \ \ \text and \ \ af + bh = eb + fd.$$ And for the diagonal elements we get $$gb = cf = ea.$$ So the result of the multiplication give us a matrix such that its diagonal elements are the same. That's all I could get.

I just needed to prove that the diagonal elements of $A$ are scalars and the rest are zeros, but I faced a dead end, because I got stuck in a circle.

Can you help?!

Rick
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Hoda Bibo
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  • Rather than having the second matrix be general with arbitrary coefficients, try using very specific second matrices. For example, if the second matrix was $\begin{bmatrix}1&1\0&0\end{bmatrix}$ and commutes with your matrix, what does that imply about the elements in your matrix? You are overloading yourself with too many variables at once. – JMoravitz May 26 '20 at 13:04
  • Right!! Thank you I get it :) – Hoda Bibo May 26 '20 at 13:14

2 Answers2

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The fact that $A$ is supposed to commute with all matrices means we don't have to be general about it. We can pick a few nice matrices and just work with those.

Consider multiplying $A$ with the matrix $$ E = \begin{bmatrix}1&0\\0&0\end{bmatrix} $$ We have $$ AE = \begin{bmatrix}a&0\\c&0\end{bmatrix}, \quad EA = \begin{bmatrix}a&b\\0&0\end{bmatrix} $$ so for $A$ to commute with $E$, we must have that $b = 0$ and $c = 0$, so $A$ is diagonal.

Now that we know that $A$ is diagonal, can you think of a matrix that in a similar way demonstrates that we must have $a = d$?

Arthur
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If a matrix $A $ is commute with every matrix then in particular it commute with \begin{bmatrix}0&1\\1&0\end{bmatrix} and :\begin{bmatrix}1&0\\0&0\end{bmatrix}
With first matrix you find condition on a,b,c,d as $a=d$ and $b=c$

And with 2nd matrix you find the condition $b= 0$ with these matrix $A $ becomes scalar .

ਮੈਥ
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