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 .

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