# Abstract Nonsense

## The Endomorphism Algebra

Point of post: In this post we discuss the equivalent of sections 34 and 35 if Halmos’s book.

Motivation

In our last post we discussed the concept of linear transformations from a $F$-space $\mathscr{V}$ to itself  and how we can form the vector space $\text{End}\left(\mathscr{V}\right)$ of all such linear transformations. It turns out though that this set of all linear transformation can be endowed with much more structure than that of a vector space. To motivate how special this property is consider the following question: given a vector space $\mathscr{V}$ what does $v_1\times v_2$ mean for vectors $v_1,v_2\in\mathscr{V}$? In general, it’s meaningless because a vector space need not have a “multiplication”. It turns out though that some vector spaces do have a canonical way to define multiplication in a way that “makes sense”, such vector spaces are invariably called associative algebras. In particular, it turns out that there is a canonical way to turn $\text{End}\left(\mathscr{V}\right)$ into an associative algebra. Namely, there is a natural way to define the product of two endomorphisms.

Algebras

Let $\mathscr{V}$ be a $F$-space. Then, $\mathscr{V}$ is called an associative algebra if there exists a map

$\mu:\mathscr{V}\times\mathscr{V}\to\mathscr{V}$

such that for all $x,y,z\in\mathscr{V}$

1.                                                                 $\mu\left(x,\mu(y,z)\right)=\mu\left(\mu(x,y),z\right)$

2.                                                                 $\mu\left(x,y+z\right)=\mu(x,y)+\mu(y,z)$

3.                                                                 $\mu\left(x+y,z\right)=\mu(x,z)+\mu(y,z)$

4.                                                                 $\alpha\mu(x,y)=\mu(\alpha x,y)=\mu(x\alpha y)$

for all $\alpha\in F$. If $\mathscr{V}$ also possess a non-zero element, denoted $\mathbf{1}$, such that

5.                                                                $\mu\left(x,\mathbf{1}\right)=\mu\left(\mathbf{1},x\right)=x$

for all $x\in\mathscr{V}$ then $\mathscr{V}$ is called an associative algebra with identity/unital algebra. In practice one doesn’t explicitly write out $\mu$ but denotes $\mu(x,y)$ by the concatenation $xy$.

We now proceed to give some examples:

Ex(1): Consider $C[\mathbb{R},\mathbb{R}]$ the space of all continuous functions  from $\mathbb{R}$ to itself. With the usual (pointwise) definitions of addition and multiplication this is an associative algebra (a Banach algebra if given the usual norm and the induced topology) with identity $1(x)$.

Ex(2): $\mathbb{R}^3$ with $\mu(x,y)=x\times y$ where $\times$ is the usual cross product is a non-associative algebra.

Ex(3): The set $\text{Mat}_n\left(F\right)$ of all $n\times n$ matrices over some field $F$ with the usual definitions of matrix multiplication and addition, with identity $I$.

Remark: Note that the last example gives an example of an associative unital algebra  that isn’t commutative in the sense that, in general, for two matrices it’s not true that $MN=NM$. Also, it shows in general that elements of an associative unital algebra need not have inverses, in the multiplicative sense.

While there are many theorems relating to general associative unital algebras we shall have need for only one, and it’s merely an observation. Namely, observe that for any $x\in\mathscr{V}$ we have that $x\bold{0}=x(\bold{0}+\bold{0})=x\bold{0}+x\bold{0}$ and so upon subtraction $x\bold{0}=\mathbf{0}$. Similarly, $\mathbf{0}x=\mathbf{0}$.

Polynomials

Given an associative algebra with identity $\mathscr{A}$ over $F$, there is a canonical way to define polynomials on $\mathscr{A}$.  Namely, we may start by inductively defining $v^n$ for $\mathscr{V}$ by the relation $v^0=\mathbf{1}$ and $v^{n+1}=vv^{n}$. Note that by associativity this definition is well-defined. From there we can extend the notion of a real valued polynomial $\displaystyle p(x)=\sum_{j=0}^{n}a_j x^j$ to a polynomial on $\mathscr{A}$ by

$\displaystyle p(v)=\sum_{j=0}^{n}a_jv^j=a_0\mathbf{1}+a_1v+\cdots+a_nv^n$

for $a_0,\cdots,a_n\in F$.

Endomorphism Algebra

The rest of this post will be devoted to showing how one can canonically impose a multiplication on $\text{End}\left(\mathscr{V}\right)$ which transforms it into an associative unital algebra. The surprising thing is the identity of this multiplication, namely, function composition. In essence we will show that we can consider

$\circ:\text{End}\left(\mathscr{V}\right)\times\text{End}\left(\mathscr{V}\right)\to\text{End}\left(\mathscr{V}\right):\left(T,T'\right)\mapsto T\circ T'$

as a multiplication map, with identity $\text{id}_\mathscr{V}\overset{\text{def.}}{=}\mathbf{1}$. But, this is just grunt work, so let’s get to it:

Theorem: $\text{End}\left(\mathscr{V}\right)$ with usual addition and function composition is an associative unital algebra with identity $\text{id}_\mathscr{V}\overset{\text{def.}}{=}\mathbf{1}$.

Proof: We have already established that $\text{End}\left(\mathscr{V}\right)$ is a $F$-space, and so it suffices to check axioms one through five as listed in the algebra section. We do this axiom by axiom.

1. This follows since function composition, in general, is associative

2. This follows since

\begin{aligned}\left(T_1\left(T_2+T_3\right)\right)(x)&= T_1\left(\left(T_2+T_3\right)(x)\right)\\ &=T_1\left(T_2(x)+T_3(x)\right)\\ &=T_1(T_2(x))+T_1(T_3(x))\end{aligned}

for all $x\in\mathscr{V}$, and thus $T_1\left(T_2+T_3\right)=T_1T_2+T_1T_3$.

3. This is done using the exact same method.

4. This follows since

\begin{aligned}\left(\alpha\left(T_1T_2\right)\right)(x) &=\alpha T_1(T_2(x))\\ &=\left(\alpha T_1\right)\left(T_2(x)\right)\\ &=T_1\left(\alpha T_2(x)\right)\\ &=T_1\left(\left(\alpha T_2\right)(x)\right)\end{aligned}

5. This follows directly from the definition of the identity function.

Thus, since all five axioms are satisfied the conclusion follows. $\blacksquare$

Note that while $\text{End}\left(\mathscr{V}\right)$ is an associative unital algebra, it is far, far from being as nice as a field. Explicitly the multiplication (for spaces of dimension greater than one) admits zero divisors as can be seen by considering that $T_{1,2}T_{1,2}=\bold{0}$. Moreover, the multiplication isn’t even commutative (for spaces of dimension greater than one) as can be seen by considering that $\left(T_{1,2}T_{1,1}\right)(x_1)=x_2$ but $\left(T_{1,1}T_{1,2}\right)(x_1)=\bold{0}$

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

3. Simmons, George Finlay. Introduction to Topology and Modern Analysis. Malabar, FL: Krieger Pub., 2003. Print.

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