Invertible and Singular Elements in an Algebra

Definition: Let $\mathfrak{A}$ be an algebra. A point $e \in \mathfrak{A}$ is said to be a Unit Element or a (Multiplicative) Identity Element of $\mathfrak{A}$ if $e \neq 0$ and if for every $a \in \mathfrak{A}$ we have that $ea = a = ae$. An Algebra with Unit (or Unital Algebra) is an algebra $\mathfrak{A}$ that has a unit.

The following proposition tells us that if $\mathfrak{A}$ is an algebra with unit then the unit element is unique.

Proposition 1 (Uniqueness of Units): Let $\mathfrak{A}$ be an algebra. If $e, e' \in \mathfrak{A}$ are both units then $e = e'$.

• Proof: Suppose that $e, e' \in \mathfrak{A}$ 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 unique, it is sometimes conventional to use the symbol $1$ to denote the unit in an algebra with unit.

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

Observe that if the operation of multiplication on the algebra $\mathfrak{A}$ is commutative then the existence of a left multiplicative inverse implies the existence of a right multiplicative inverse. In general though, multiplication on $\mathfrak{A}$ is NOT assumed to be commutative.

Definition: Let $\mathfrak{A}$ be an algebra with unit. A point $a \in \mathfrak{A}$ is said to be Invertible in $\mathfrak{A}$ if $a$ has an inverse in $\mathfrak{A}$, and $\mathrm{Inv}(\mathfrak{A})$ is the set of all invertible elements in $\mathfrak{A}$. A point $a \in \mathfrak{A}$ is said to be Singular in $\mathfrak{A}$ if it has no inverse in $\mathfrak{A}$, and $\mathrm{Sing}(\mathfrak{A})$ is the set of all singular elements in $\mathfrak{A}$.

The following proposition tells us that if $a \in \mathfrak{A}$ is invertible then its inverse is unique.

Proposition 2 (Uniqueness of Inverses): Let $\mathfrak{A}$ be an algebra with unit and let $a \in \mathfrak{A}$. If $a$ is invertible and $x, y$ are both inverses of $a$ then $x = y$.

• Proof: Suppose that $x$ and $y$ are both inverses of $a$. Then in particular, $1 = ay$ and $xa = 1$. So:

(2)

\begin{align} \quad x = x1 = x(ay) = (xa)y = 1y = y \end{align}

• Therefore $x = y$. $\blacksquare$