Algebras over F

# Algebras over F

 Definition: An (Associative) Algebra over $\mathbb{F}$ is a linear space $\mathfrak{A}$ over the field $\mathbf{F}$ along with a binary function $\cdot \mathfrak{A} \times \mathfrak{A} \to \mathfrak{A}$, sometimes called (vector) multiplication or product, that satisfies the following three properties: 1) $a \cdot (b \cdot c) = (a \cdot b) \cdot c$ for all $a, b, c \in \mathfrak{A}$ (Associativity of multiplication). 2) $a \cdot (b + c) = a \cdot b + a \cdot c$ for all $a, b, c \in \mathfrak{A}$ (Distributivity of multiplication over addition). 3) $(\alpha a) \cdot b = \alpha (a \cdot b) = a \cdot (\alpha b)$ for all $a, b \in \mathfrak{A}$ and all $\alpha \in \mathbf{F}$. If $\mathbb{F} = \mathbb{R}$ then the above structure is called a Real Algebra, and if $\mathbf{F} = \mathbb{C}$ then the above structure is called a Complex Algebra.

Here, $\mathbf{F}$ denotes the field of real numbers $\mathbb{R}$ or the field of complex numbers $\mathbb{C}$ depending on the context.

The notation $\alpha x$ is used denote scalar multiplication in $X$, while $x \cdot y$ is used to denote vector multiplication in $X$. When no ambiguity arises, we will omit writing the "$\cdot$" and simply write "$xy$".

 Definition: Let $\mathfrak{A}$ be an algebra over $\mathbf{F}$. A Subalgebra of $\mathfrak{A}$ is a linear subspace $\mathfrak{B}$ of $\mathfrak{A}$ with the additional property that for all $b_1, b_2 \in \mathfrak{B}$ we have that $b_1 \cdot b_2 \in \mathfrak{B}$.

Equivalently, if $X$ is an algebra then a subset $Y$ of $X$ is a subalgebra of $X$ if it is closed under both addition, scalar multiplication, and multiplication.

## Example 1

The set of real numbers $\mathbb{R}$ is an algebra over $\mathbb{R}$ with the operations of standard number addition, scalar multiplication, and multiplication of real numbers (which is really the same as scalar multiplication in this case).

Similarly, the set of complex numbers $\mathbb{C}$ is an algebra over $\mathbb{C}$.

## Example 2

Let $X$ be a nonempty set and let $\mathfrak{A}$ be an algebra over $\mathbf{F}$. Consider the set of all functions from $X$ to $\mathfrak{A}$. We define the operations of pointwise function addition, pointwise scalar multiplication, and pointwise multiplication respectively for all $f, g : X \to \mathfrak{A}$ and for all $\alpha \in \mathbb{F}$,

(1)
\begin{align} \quad [f + g](x) &:= f(x) + g(x), \quad x \in X\\ \quad [\alpha f](x) &:= \alpha f(x), \quad x \in X \\ \quad [f \cdot g](x) &:= f(x) \cdot g(x), \quad x \in X \end{align}

(Note that $f(x), g(x) \in \mathfrak{A}$ and so $f(x) \cdot g(x)$ is a well-defined element of $\mathfrak{A}$.)

It is easy to check that this space is an algebra over $\mathbf{F}$.

## Example 3

Let $X$ be a nonempty set. The set of all real-valued (or complex-valued functions) on $X$ with the operations of pointwise function addition, pointwise scalar multiplication, and pointwise function multiplication as defined in example 2, is an algebra. (This is simply a special case of the class of algebras in example 2 with $\mathfrak{A} = \mathbb{F}$.

## Example 4

Let $X$ be a linear space. Let $\mathcal L(X, X)$ be the set of all linear operators from $X$ to $X$. We already know that $\mathcal L(X, X)$ is a linear space. We define multiplication on $\mathcal L(X, X)$ as composition, i.e., for all $S, T \in \mathcal L(X, X)$:

(2)
\begin{align} \quad (S \circ T)(x) = S(T(x)) \end{align}

Then $\mathcal L(X, X)$ with composition as the multiplication becomes an algebra