# Abstract Nonsense

## Center of an Algebra

Point of post: In this post we define the center of an associative unital algebra and classify the center of $\text{Mat}_n\left(F\right)$.

Motivation

Often questions come up such as “do the linear transformations $T$ and $T'$ commute?” or “classify all linear transformations which commute with $T$“. An obvious question might be, for which transformations is the answer to this latter question “All of the endomorphsim algebra.” Said differently, which transformations have the property that they commute with every transformation on that space? This set, the set of all such ‘universal commuters’, is called the center of an algebra. It turns out that for the endomorphism algebra, the center is quite ‘small’.

Center of an Algebra

Let $\mathscr{A}$ be an associative unital algebra. We define the center of $\mathscr{A}$, denoted $\mathcal{Z}\left(\mathscr{A}\right)$, to be the set of all elements of $\mathscr{A}$ which commute with all the other elements of $\mathscr{A}$. More formally

$\mathcal{Z}\left(\mathscr{A}\right)=\left\{x\in\mathscr{A}:xy=yx, \text{ for every }y\in\mathscr{A}\right\}$

We prove one small theorem about the center of an algebra, namely that it is a subalgebra. Indeed:

Theorem: Let $\mathscr{A}$ be an associative unital $F$-algebra with identity $\mathbf{1}$. Then the center $\mathcal{Z}\left(\mathscr{A}\right)$ is an associative unital subalgebra of $\mathscr{A}$.

Proof: It’s clear that $\mathbf{0},\mathbf{1}\in\mathcal{Z}\left(\mathscr{A}\right)$ since for every $x\in\mathscr{A}$

$x(\alpha\mathbf{1})=\alpha(x\mathbf{1})=\alpha x=(\alpha\mathbf{1})x$

and

$x\bold{0}=\bold{0}=\bold{0}x$

Note next then that if $x,y\in\mathcal{Z}\left(\mathscr{A}\right)$, $\alpha,\beta\in F$, and $z\in\mathscr{A}$, then

$\left(\alpha x+\beta y\right)z=(\alpha x)z+(\beta y)z=z(\alpha x)+z(\beta y)=z(\alpha x+\beta y)$

and

$(xy)z=x(yz)=x(zy)=(xz)y=(zx)y=z(xy($

so that $\alpha x+\beta y,xy\in\mathcal{Z}\left(\mathscr{A}\right)$. The conclusion follows. $\blacksquare$

Center of $\text{Mat}_n\left(F\right)$

In this section we wish to describe, in full, $\mathcal{Z}\left(\text{Mat}_n\left(F\right)\right)$ for every $n\in\mathbb{N}$. Namely, we prove that:

Theorem: Let $F$ be a field and $n\in\mathbb{N}$. Then,

$\displaystyle \mathcal{Z}\left(\text{Mat}_n\left(F\right)\right)=\left\{\lambda I_n:\lambda\in F\right\}$

Proof: Let $E_{i_0,i_0}$ be the matrix whose general term is $\delta_{(i_0,i_0),(i,j)}$. In other words, $E_{i_0,i_0}$ has zeros for every entry except the $i_0,i_0$ entry where it’s one. Note then that if $M\in\mathcal{Z}\left(\text{Mat}_n\left(F\right)\right)$ then

$ME_{i_0,i_0}=E_{i_0,i_0}M$

Note though that the left hand side of the above equations is a matrix which is zero for all entries except it’s $i_0^{\text{th}}$ column and the right hand size has all zero entries except its $i_0^{\text{th}}$ row. Thus, for the two to be equal we realize that the only non-zero entry along the $i_0^{\text{th}}$ row and $i_0^{\text{th}}$ column occurs at the diagonal position $i_0,i_0$. Doing this for every $i_0\in[n]$ we may conclude that

$\displaystyle M=\begin{pmatrix}\alpha_1 & 0 & \cdots & 0 & 0\\ 0 & \alpha_2 & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \cdots & \alpha_{n-1} & 0\\ 0 & 0 & \cdots & 0 & \alpha_n\end{pmatrix}$

for some $\alpha_1,\cdots,\alpha_n\in F$ (note that this is a diagonal matrix, for those [almost surely everyone reading this] familiar with this notation).

But using the notation $E_{i_0,j_0}$ to represent the matrix whose general entry is $\delta_{(i_0,j_0),(i,j)}$ we see by assumption that

$M(E_{i_0,j_0}+E_{j_0,i_0})=(E_{i_0,j_0}+E_{j_0,i_0})M$

But, notice that the left hand side of the above is just $M$ except the $i_0^{\text{th}}$ and $j_0^{\text{th}}$ columns are interchanged and the right hand side is $M$ with the $i_0^{\text{th}}$ and $j_0^{\text{th}}$ rows interchanged. Doing this for every such combinations of $(i,j)$ gives that $\alpha_1=\cdots=\alpha_n=\lambda$ for some $\lambda\in F$. Thus, $M=\lambda I_n$ as desired.

Thus

$\mathcal{Z}\left(\text{Mat}_n\left(F\right)\right)\subseteq\left\{\lambda I_n:\lambda \in F\right\}$

and since the reverse inclusion is evidently true (in fact, it’s true via the theorem in the previous section) the conclusion follows. $\blacksquare$

From this, and a previous theorem we get the following corollary:

Corollary: Let $\mathscr{V}$ be an $n$-dimensional $F$ space, then

$\mathcal{Z}\left(\text{End}\left(\mathscr{V}\right)\right)=\left\{\lambda\mathbf{1}:\lambda\in F\right\}$

References:

1. Golan, Jonathan S. The Linear Algebra a Beginning Graduate Student Ought to Know. Dordrecht: Springer, 2007. Print.

2. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print

December 16, 2010 -

## 6 Comments »

1. […] The answer is no. Recall from a previous theorem that the center of is the set of all multiplies of . In particular, we see that if were to […]

Pingback by Halmos Sections 37 and 38: Matrices and Matrices of Linear Transformations(Pt. II) « Abstract Nonsense | December 19, 2010 | Reply

2. […] as was the case for the center of an algebra it is often fruitful to consider the center of a group. In essence, the center of a group is merely […]

Pingback by Review of Group Theory: The Center of a Group « Abstract Nonsense | January 5, 2011 | Reply

3. […] (for those who need a reminder conjugacy classes and centers of algebras are defined here and here respectively) […]

Pingback by Representation Theory: Class Functions « Abstract Nonsense | February 24, 2011 | Reply

4. […] Recall that is real the center of the group algebra . But, recall that the center of an algebra is a subalgebra. In particular since is a subalgebra of we have […]

Pingback by Representation Theory: A ‘Lemma’ (pt. II) « Abstract Nonsense | March 10, 2011 | Reply

5. […] to see that the matrix determines the corresponding automorphism up to scaling, so that (where we have proven before this last equality, where the stands for the center of a ring) and thus we have that where the […]

Pingback by Automorphisms of k(t)/k « Abstract Nonsense | February 28, 2012 | Reply