Prove that if a square matrix commutes with every matrix of the same size, then it is a scalar multiple of the identity matrix.

Step-by-step solution

1. 1

   Pick the right test matrices. Rather than an arbitrary $B$, use the matrix units $E_{ij}$ — the matrix with a $1$ in row $i$, column $j$, and $0$ everywhere else. Since $A$ commutes with everything, it commutes with each $E_{ij}$.

2. 2

   Compute both products. Writing $A = (a_{kl})$:

   $AE_{ij} \text{ has the } i\text{-th column of } A \text{ placed in column } j$
   $E_{ij}A \text{ has the } j\text{-th row of } A \text{ placed in row } i$

3. 3

   Force the off-diagonal entries to vanish. Setting $AE_{ij} = E_{ij}A$ and comparing entries gives $a_{ki} = 0$ for every $k \ne i$. Since $i$ and $j$ were arbitrary, every off-diagonal entry of $A$ is zero, so $A$ is diagonal.

4. 4

   Force the diagonal entries to agree. Comparing the remaining entries gives $a_{ii} = a_{jj}$ for all $i, j$. So all diagonal entries share a common value $\lambda$.

5. 5

   Conclude. A diagonal matrix whose diagonal entries are all equal to $\lambda$ is exactly $\lambda I_n$. (The converse is immediate: $\lambda I$ commutes with everything.) This says the centre of the matrix ring $M_n(F)$ is precisely the scalar matrices.