# The Unapologetic Mathematician

## The Center of an Algebra

Sorry I forgot to get this posted this morning.

Given an algebra $A$, it’s interesting to consider the “center” $Z_A$ of $A$. This is the collection of algebra elements that commute with all the others. That is,

$\displaystyle Z_A=\{a\in A\vert\forall b\in A, ab=ba\}$

It’s straightforward to see that sums, scalar multiples, and products of central elements — elements of $Z_A$ — are themselves central. That is, $Z_A$ is an algebra, and it’s a commutative one to boot. This gives us a construction that starts with an associative algebra and ends with a commutative algebra, and yet it turns out that it is not a functor! I don’t really want to get into that right now, though, but I wanted to mention it in passing, since it’s one of the few examples of a natural algebraic construction that isn’t functorial.

What I do want to get into right now, is calculating the center of the matrix algebra $\mathrm{Mat}_d(\mathbb{C})$. The answer is reminiscent of Schur’s lemma:

$Z_{\mathrm{Mat}_d(\mathbb{C})}=\{c I_d\vert c\in\mathbb{C}\}$

Suppose that $C$ is a central $d\times d$ matrix. Then in particular it commutes with the matrix $E_{i,i}$, which has a $1$ at the $i$th place along the diagonal and $0$s everywhere else. That is, $CE_{i,i}=E_{i,i}C$. But $CE_{i,i}$ zeroes out everything except the $i$th column of $C$, while $E_{i,i}C$ zeroes out everything except the $i$th row. For these two be equal, the $i$th column must be all zeroes except for the one spot along the diagonal, and similarly for the $i$th row. And so $C$ must be diagonal.

For $i\neq j$, $C$ must also commute with $E_{i,j}+E_{j,i}$ — the matrix with ones in the $j$th column of the $i$th row and the $i$th column of the $j$th row. That is, $C(E_{i,j}+E_{j,i})=(E_{i,j}+E_{j,i})C$. Multiplying on the right by $E_{i,j}+E_{j,i}$ swaps the $i$th and $j$th columns of $C$, while multiplying on the left swaps the $i$th and $j$th rows. Thus we can tell that not only is $C$ diagonal, but all the diagonal entries must be the same. And so $C=c I_d$ for some complex $c$.