Abstract Nonsense

Crushing one theorem at a time

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


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December 16, 2010 - Posted by | Algebra, Halmos, Linear Algebra | , , ,

6 Comments »

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