The Reversed Algebra of an Algebra

# The Reversed Algebra of an Algebra

Proposition 1: Let $X$ be an algebra with multiplication denoted by $\cdot$. Then $X$ with the same operations of addition and scalar multiplication, but with the new operation $* : X \times X \to X$ defined for all $x, y \in X$ by $x * y = y \cdot x$ is an algebra. |

**Proof:**We are given that $X$ with the operations of addition and scalar multiplication is a linear space.

**1. $x * (y * z) = (x * y) * z$:**Let $x, y, z \in X$. Then by the associativity of $\cdot$ we have that:

\begin{align} \quad x * (y * z) = x * (z \cdot y) = (z \cdot y) \cdot x = z \cdot (y \cdot x) = (y \cdot x) * z = (x * y) * z \end{align}

**2. $x * (y + z) = x * y + x * z$:**Let $x, y, z \in X$. Then by the distributivity of $\cdot$ over $+$ we have that:

\begin{align} \quad x * (y + z) = (y + z) \cdot x = y \cdot x + z \cdot x = x * y + x * z \end{align}

**3. $(\alpha x) * y = \alpha (x * y) = x * (\alpha y)$:**Let $x, y \in X$ and let $\alpha \in \mathbf{F}$. Then:

\begin{align} \quad (\alpha x) * y = y \cdot (\alpha x) = \alpha (y \cdot x) = \alpha (x * y) \end{align}

- And also:

\begin{align} \quad \alpha (x * y) = \alpha (y \cdot x) = (\alpha y) \cdot x = x * (\alpha y) \end{align}

- So indeed $X$ with the operations of addition, scalar multiplication, and this new form of multiplication $*$ is an algebra. $\blacksquare$

We give the above algebra a special name.

Definition: Let $X$ be an algebra with addition, scalar multiplication, and vector multiplication $\cdot$. The Reversed Algebra of $X$ is the same space $X$ with addition, scalar multiplication, and the new vector multiplication $* : X \times X \to X$ defined for all $x, y \in X$ by $x * y = y \cdot x$. |