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

 Definition: Let $X$ be an algebra. A point $e \in X$ is said to be a Unit Element or a (Multiplicative) Identity Element of $X$ if $e \neq 0$ and if for every $x \in X$ we have that $ex = x = xe$, and $X$ is said to be an Algebra with Unit.
 Proposition 1: Let $X$ be an algebra. If $e$ and $e'$ are both units then $e = e'$.
• Proof: Suppose that $e, e' \in X$ are both units. Then:
(1)
\begin{align} \quad e = ee' = e' \end{align}
• Where the first equality comes from the fact that $e'$ is a unit, and the second equality comes from the fact that $e$ is a unit. $\blacksquare$

Since units in an algebra are unit, it is conventional to use the symbol $1$ to denote the unit in an algebra with unit.

 Definition: Let $X$ be an algebra with unit $1$ and let $x \in X$. A point $a \in X$ is said to be a Left (Multiplicative) Inverse of $x$ if $ax = 1$. A point $b \in X$ is said to be a Right (Multiplicative) Inverse of $x$ if $xb = 1$. A point $c \in X$ is said to be a (Multiplicative) Inverse of $x$ if it is both a left and right inverse of $x$.

Observe that if the operation of multiplication on the algebra $X$ is commutative then the existence of a left multiplicative inverse implies the existence of a right multiplicative inverse. In general, multiplication on $X$ is not assumed to be commutative.

 Definition: Let $X$ be an algebra with unit. A point $x \in X$ is said to be Invertible if $x$ has an inverse in $X$, and $\mathrm{Inv}(X)$ is the set of all invertible elements in $X$. A point $x \in X$ is said to be Singular if it has no inverse in $X$, and $\mathrm{Sing}(X)$ is the set of all singular elements in $X$.
 Proposition 2: Let $X$ be an algebra with unit and let $x \in X$. If $x$ is invertible and $a, b$ are both inverses of $x$ then $a = b$.

Whenever $x$ is invertible (i.e., a left and right inverse to $x$ exists) then that inverse element is unique. In general, if $x$ is not invertible then $x$ may have many left inverses or many right inverses.

• Proof: Suppose that $a$ and $b$ are both inverses of $x$. Then:
(2)
\begin{align} \quad a = a1 = a(xb) = (ax)b = 1b = b \end{align}
• Therefore $a = b$. $\blacksquare$
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