The Reversed Algebra of an Algebra

Definition: Let $\mathfrak{A}$ be an algebra over $\mathbf{F}$ with addition, scalar multiplication, and vector multiplication $\cdot$. The Reversed Algebra of $\mathfrak{A}$ over $\mathbf{F}$ is defined to be the set $\mathfrak{A}$ with the same operations of addition and scalar multiplication, with the vector multiplication $*$ defined by $a * b = b \cdot a$ for all $a, b \in \mathfrak{A}$.

Proposition 1: Let $\mathfrak{A}$ be an algebra over $\mathbf{F}$ with multiplication denoted by $\cdot$. Then the reversed algebra of $\mathfrak{A}$ is an algebra over $\mathbf{F}$.

• Proof: We are given that $\mathfrak{A}$ with the operations of addition and scalar multiplication is a linear space. All that needs to be checked is that the new product $*$ on $\mathfrak{A}$ satisfies the required properties.

• 1. $a * (b * c) = (a * b) * c$: Let $a, b, c \in \mathfrak{A}$. Then by the associativity of $\cdot$ we have that:

(1)

\begin{align} \quad a * (b * c) = (b * c) \cdot a = (c \cdot b) \cdot a = c \cdot (b \cdot a) = (b \cdot a) * c = (a * b) * c \end{align}

• 2. $a * (b + c) = a * b + a * c$: Let $a, b, c \in \mathfrak{A}$. Then by the distributivity of $\cdot$ over $+$ we have that:

(2)

\begin{align} \quad a * (b + c) = (b + c) \cdot a = b \cdot a + c \cdot a = a * b + a * c \end{align}

• 3. $(\alpha a) * b = \alpha (a * b) = a * (\alpha b)$: Let $a, b \in \mathfrak{A}$ and let $\alpha \in \mathbf{F}$. Then:

(3)

\begin{align} \quad (\alpha a) * b = b \cdot (\alpha a) = \alpha (b \cdot a) = \alpha (a * b) \end{align}

• And also:

(4)

\begin{align} \quad \alpha (a * b) = \alpha (b \cdot a) = (\alpha b) \cdot a = a * (\alpha b) \end{align}

• So indeed $\mathfrak{A}$ with the operations of addition, scalar multiplication, and multiplication given by $*$ is an algebra. $\blacksquare$