# Thread: Center of a matrix ring

1. ## Center of a matrix ring

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!

2. [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 $m_{1} = m_{4} =m$ $\Rightarrow$ $M=\left(\begin{array}{cc}m&0\\0&m\end{array}\right ) = m \cdot I.$