Originally Posted by

**ThePerfectHacker** If you pick a particular basis for $\displaystyle V$ then $\displaystyle A$ can be represented by a matrix $\displaystyle [A]$, and $\displaystyle B$ can be represented by a matrix $\displaystyle [b]$. We are told that $\displaystyle AB = BA$, so $\displaystyle [AB] = [BA] \implies [A][b] = [b][A]$. Therefore, $\displaystyle A: V\to V$ is such a linear trasformation so that its corresponding matrix commutes with all other matrices. In order to complete your problem we need to show that $\displaystyle [A]$ is a scalar multiple of the identity matrix. Let $\displaystyle [A] = (a_{ij})$. Define $\displaystyle E_{ij}$ the matrix so that the $\displaystyle ij$ entry is $\displaystyle 1$ and everything else is $\displaystyle 0$. Notice that $\displaystyle I + E_{ij}$ is always invertible. Thus, if $\displaystyle [A]$ commutes with every matrix it means $\displaystyle (I + E_{ij})[A] = [A](I + E_{ij})$. If $\displaystyle i\not = j$ the $\displaystyle ij$-entry in that matrix equation tells us that $\displaystyle a_{ij} + a_{jj} = a_{ij} + a_{ii} \implies a_{ii}= a_{jj}$ i.e. the matrix $\displaystyle [A]$ has all its diagnol entries equal. Now consider the matrix equation $\displaystyle (I + E_{ij})[A] = [A](I + E_{ij})$ the $\displaystyle ii$-entry tells us $\displaystyle a_{ii} + a_{ji} = a_{ii} \implies a_{ji} = 0$. Thus, $\displaystyle [A]$ is a scalar of a diagnol matrix.