Center of a matrix ring

**Greg98** · March 28th 2011, 09:20 AM

Hello,
the task is to prove that center of a matrix ring $R=Mat_2(\mathbb{R})$ is the scalar matrices $\alpha I, \alpha \in \mathbb{R}$ .

I found two more general proofs:
Center of an Algebra « Abstract Nonsense
The center of a matrix ring over a commutative ring is precisely the scalar matrices « Project Crazy Project

However I didn't understand those totally, so I coudn't reduce them to this simpler case. But I think the main point is:
Let's assume that $M \in Z(Mat_2(\mathbb{R}))$ and $E=\left(\begin{array}{cc}1&0\\0&0\end{array}\right )$ .
Matrix multiplication $ME$ is commutative ( $ME=EM$ ) iff $M$ is in form $\left(\begin{array}{cc}a_1&0\\0&a_2\end{array}\rig ht)$ .

I don't know how to continue etc., so any help is appreciated. Thanks!

**zoek** · March 28th 2011, 11:00 AM

[QUOTE=Greg98;634437]Hello,
the task is to prove that center of a matrix ring $R=Mat_2(\mathbb{R})$ is the scalar matrices $\alpha I, \alpha \in \mathbb{R}$ .

If $M=\left(\begin{array}{cc}m_{1}&m_{2}\\m_{3}&m_{4}\ end{array}\right) \in Z(Mat_2(\mathbb{R}))$ then $M \cdot E = E\cdot M, \forall E \in Mat_2(\mathbb{R})$ .

If you take $E=\left(\begin{array}{cc}1&0\\0&0\end{array}\right )$ then you have: $M \cdot E = \left(\begin{array}{cc}m_{1}&0\\m_{3}&0\end{array} \right)$ and $E \cdot M= \left(\begin{array}{cc}m_{1}&m_{2}\\0&0\end{array} \right)$ .

So, $m_{2} = m_{3} = 0$ .

If $E=\left(\begin{array}{cc}0&1\\0&0\end{array}\right )$ then you have $m_{1} = m_{4} =m$ $\Rightarrow$ $M=\left(\begin{array}{cc}m&0\\0&m\end{array}\right ) = m \cdot I.$

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