We will prove that the center , where is an -algebra (or ring with unity).
One direction is pretty clear. Let . Then for some . , , so is in the center .
The other direction will require the use of matrices, which is a n by n matrix with (i,j) entry 1, and rest zero.
Let . We can write . Our key step is compute . Thus we may conclude that
Plugging in some convenient values like , we can conclude that for all , i.e. all diagonal entries are equal.
Plug in gives us for all .
Thus is a scalar matrix, i.e. for some .
Observing that , we conclude that for all in . So has to be in the center .