Algebras over F

Algebras over F

Definition: An Algebra over $\mathbb{F}$ is a linear space $X$ over the field $\mathbf{F}$ along with a binary function $\cdot X \times X \to X$, sometimes called multiplication, that satisfies the following three properties:
1) $x \cdot (y \cdot z) = (x \cdot y) \cdot z$ for all $x, y, z \in X$ (Associativity of multiplication).
2) $x \cdot (y + z) = x \cdot y + x \cdot z$ for all $x, y, z \in X$ (Distributivity of multiplication over addition).
3) $(\alpha x) \cdot y = \alpha (x \cdot y) = x \cdot (\alpha y)$ for all $x, y \in X$ 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 set 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 "$\cdot$".

The above definition is sometimes referred to as an Associative Algebra over $\mathbb{F}$.

For example, the set of real numbers $\mathbb{R}$ and the set of complex numbers are both algebras with the operation $\cdot$ of standard number multiplication.

Definition: Let $X$ be a linear space over $\mathbb{F}$ and let $E \subseteq X$. Let $f, g : E \to X$ and $\alpha \in \mathbf{F}$.
1) The Sum (or Pointwise Sum) of the functions $f$ and $g$ is the function $f + g : E \to X$ defined for all $x \in E$ by $(f + g)(x) = f(x) + g(x)$.
2) The Scalar Multiple (or Pointwise Scalar Multiple of the function $f$ and the number $\alpha$ is the function $\alpha f : E \to X$ defined for all $x \in E$ by $(\alpha f)(x) = \alpha f(x)$.
Suppose that $X$ is also an algebra over $\mathbb{F}$.
3) The Product (or Pointwise Product) of the functions $f$ and $g$ is the function $f \cdot g :E \to X$ defined for all $x \in E$ by $(f \cdot g)(x) = f(x) \cdot g(x)$.

Example 1

If $E$ is a nonempty set then consider the collection of all functions $f : E \to X$ with the operations of pointwise addition, pointwise scalar multiplication with elements from $\mathbf{F}$, and pointwise multiplication defined above. It is easy to verify that this space with these first two operations is a linear space over $\mathbf{F}$, and with the inclusion of the third operation, this space because an algebra over $\mathbf{F}$.

Example 2

For another example, if $X$ is a linear space then recall that $\mathcal L (X, X)$ (or simply $\mathcal L(X)$ is the linear space of all linear operators from $X$ to $X$. We have already proven the more general result that $\mathcal L(X, Y)$ is a linear space whenever $X$ and $Y$ are linear spaces.

We will now turn $\mathcal L(X, X)$ into an algebra over $\mathbf{F}$ by introduction a multiplication operation which satisfies the three properties above.

Definition: Let $X$ be a linear space and let $S, T \in \mathcal L(X, X)$. The Composition of $S$ with $T$ is defined to be the function $S \circ T : X \to X$ defined for all $x \in X$ by $(S \circ T)(x) = S(T(x))$.

With the above definition, it is not hard to verify that $\mathcal L(X, X)$ with $\circ$ is an algebra over $\mathbf{F}$. To brutely verify this, let $S, T, U \in \mathcal L(X, X)$ and let $\alpha \in \mathbf{F}$. Then:

  • 1. $S \circ (T \circ U) = (S \circ T) \circ U$ since function compositions are always associative.
  • 2. Let $x \in X$. Then by the additivity of $S$ and the definition of function addition we have that:
(1)
\begin{align} \quad [S \circ (T + U)](x) = S([T + U](x)) = S(T(x) + U(x)) = S(T(x)) + S(U(x)) = (S \circ T)(x) + (S \circ U)(x) \end{align}
  • Since this holds true for all $x \in X$ we see that $S \circ (T + U) = S \circ T + S \circ U$.
  • 3. We have that for all $x \in X$:
(2)
\begin{align} \quad [(\alpha S) \circ (T)](x) = (\alpha S)(T(x)) = \alpha S(T(x)) = [\alpha (S \circ T)](x) \end{align}
  • And so $(\alpha S) \circ T = \alpha (S \circ T)$. Additionally:
(3)
\begin{align} \quad [\alpha (S \circ T)](x) = \alpha(S(T(x)) = S(\alpha T(x)) = [S \circ (\alpha T)](x) \end{align}
  • And so $\alpha (S \circ T) = S \circ (\alpha T)$.

So indeed, $\mathcal L(X, Y)$ with composition is an algebra over $\mathbf{F}$.

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