## Center of Matrix Algebra / Matrix Ring is Scalar Matrices

We will prove that the center $Z(M_n(A))=Z(A)I_n$, where $A$ is an $R$-algebra (or ring with unity).

One direction is pretty clear. Let $X\in Z(A)I_n, Y\in M_n(A)$. Then $X=zI_n$ for some $z\in Z(A)$. $XY=zI_n Y=zY$, $YX=YzI_n=zY$, so $X$ is in the center $Z(M_n(A))$.

The other direction will require the use of $E_{ij}$ matrices, which is a n by n matrix with (i,j) entry 1, and rest zero.

Let $X=(x_{ij})\in Z(M_n(A))$. We can write $X=\sum x_{ij}E_{ij}$. Our key step is compute $E_{pq}X=\sum x_{qj}E_{pj}=XE_{pq}=\sum x_{ip}E_{iq}$. Thus we may conclude that

$(E_{pq}X)_{ij}=\begin{cases}x_{qj}&\text{if}\ i=p\\ 0&\text{otherwise} \end{cases}$

$(XE_{pq})_{ij}=\begin{cases}x_{ip}&\text{if}\ j=q\\ 0&\text{otherwise} \end{cases}$

Plugging in some convenient values like $i=p, j=q$, we can conclude that $x_{qq}=x_{pp}$ for all $p,q$, i.e. all diagonal entries are equal.

Plug in $i=p, j\neq q$ gives us $x_{qj}=0$ for all $j\neq q$.

Thus $X$ is a scalar matrix, i.e. $X=\alpha I_n$ for some $\alpha\in A$.

Observing that $(\beta I_n)X=\beta\alpha I_n=X(\beta I_n)=\alpha\beta I_n$, we conclude that $\beta\alpha=\alpha\beta$ for all $\beta$ in $A$. So $\alpha$ has to be in the center $Z(A)$.