Quasi-Invertible and Quasi-Singular Elements in an Algebra

# Quasi-Invertible and Quasi-Singular Elements in an Algebra

Definition: Let $X$ be an algebra and let $x, y \in X$. The Quasi-Product of $x$ and $y$ is defined as $x \circ y = x + y - xy$. |

Proposition 1: Let $X$ be an algebra and let $x, y, z \in X$. Then:a) The quasi-product is associative, that is, $x \circ (y \circ z) = (x \circ y) \circ z$, and so $X$ with the quasi-product is a semigroup.b) The zero vector is a multiplicative identity for the quasi-product, that is, $x \circ 0 = 0 \circ x = x$. |

**Proof of a)**Let $x, y, z \in X$. Then:

\begin{align} \quad \quad x \circ (y \circ z) = x \circ (y + z - yz) = x + (y + z - yz) - x(y + z - yz) = x + y + z -xy - xz - yz + xyz \end{align}

- Meanwhile:

\begin{align} \quad \quad (x \circ y) \circ z = (x + y - xy) \circ z = (x + y - xy) + z - (x + y - xy)z = x + y + z -xy - xz -yz + xyz \end{align}

- We see that $x \circ (y \circ z) = (x \circ y) \circ z$. Thus, the quasi-product is associative. $\blacksquare$

**Proof of b)**Let $x \in X$. Then:

\begin{align} \quad x \circ 0 = x + 0 - x0 = x \quad \mathrm{and} \quad 0 \circ x = 0 + x - 0x = x \end{align}

- Thus $x \circ 0 = 0 \circ x = x$ for all $x \in X$. So $0$ is a multiplicative identity for the quasi-product. $\blacksquare$

Definition: Let $X$ be an algebra and let $x \in X$.1) A point $y \in X$ is said to be a Left Quasi-Inverse of $x$ if $y \circ x = 0$ and $x$ is said to be Left Quasi-Invertible.2) A point $y \in X$ is said to be a Right Quasi-Inverse of $x$ if $x \circ y = 0$ and $x$ is said to be Right Quasi-Invertible. |

Definition: Let $X$ be an algebra and let $x \in X$. A point $y \in X$ is said to be the Quasi-Inverse of $x$ if it is both a left quasi-inverse of $x$ and a right quasi-inverse of $X$, and $x$ is said to be Quasi-Invertible. The quasi-inverse of $x$ is denoted by $x^0$. The set of all quasi-invertible elements of $X$ is denoted by $\mathrm{q-Inv}(X)$. A point $x \in X$ is said to be Quasi-Singular if it is not quasi-invertible, and the set of all quasi-singular elements of $X$ is denoted by $\mathrm{q-Sing}(X)$. |

In the definition above we have stated that if $x$ is quasi-invertible then there is a unique quasi-inverse $x^0$ of $x$. We will now formally prove this.

Proposition 2: Let $X$ be an algebra. If $x$ is quasi-invertible and $y, y' \in X$ are both quasi-inverses of $x$ then $y = y'$. In otherwords, if $x$ is quasi-invertible then its quasi-inverse is unique. |

**Proof:**Let $x \in X$ and suppose that $y$ and $y'$ are both quasi-inverses of $x$. Then $x \circ y = y \circ x = 0$ and $x \circ y' = y' \circ x = 0$. We by proposition 1 that:

\begin{align} \quad y = y \circ 0 = y \circ (x \circ y') = (y \circ x) \circ y' = 0 \circ y' = y' \quad \blacksquare \end{align}

Proposition 3: Let $X$ be an algebra. If $x, y \in \mathrm{q-Inv}(X)$ then $x \circ y \in \mathrm{q-Inv}(X)$. |

**Proof:**Let $x, y \in \mathrm{q-Inv}(X)$. Then observe that:

\begin{align} \quad (x \circ y) \circ (y^0 \circ x^0) = x \circ ((y \circ y^0) \circ x^0) = x \circ (0 \circ x^0) = x \circ x^0 = 0 \end{align}

- And also:

\begin{align} \quad (y^0 \circ x^0) \circ (x \circ y) = (y^0 \circ (x^0 \circ x)) \circ y = (y^0 \circ 0) \circ y = y^0 \circ y = 0 \end{align}

- Therefore $x \circ y$ is quasi-invertible and moreover, $(x \circ y)^0 = y^0 \circ x^0$. $\blacksquare$

Corollary 4: Let $X$ be an algebra. Then the set of quasi-invertible elements of $X$, $\mathrm{q-Inv}(X)$ with the quasi-product is a group. |

**Proof:**This follows immediately by proposition 1 and proposition 3. $\blacksquare$