Basic Theorems Regarding Quasi-Invertible Elements in an Algebra

# Basic Theorems Regarding Quasi-Invertible Elements in an Algebra

Proposition 1: Let $X$ be an algebra with unit and let $x, y \in X$. Then $x$ is quasi-invertible with quasi-inverse $y$ if and only if $1 - x$ is invertible with inverse $1 - y$. |

*Note that in the above proposition uses both the terms "quasi-inverse" AND "inverse".*

**Proof:**$\Rightarrow$ Suppose that $x$ is quasi-invertible with quasi-inverse $y$. Then $x \circ y = y \circ x = 0$. We see that:

\begin{align} \quad \quad (1 - x)(1 - y) = 1 - y - x + xy = 1 - (x + y - xy) = 1 - (x \circ y) = 1 - 0 = 1 \end{align}

- And similarly we have that:

\begin{align} \quad \quad (1 - y)(1 - x) = 1 - x - y + yx = 1 - (y + x - yx) = 1 - (y \circ x) = 1 - 0 = 1 \end{align}

- Therefore $1 - x$ is invertible and the inverse of $1 - x$ is $1 - y$.

- $\Leftarrow$ Suppose that $1 - x$ is invertible with inverse $1 - y$. Then $(1 - x)(1 - y) = 1$ and $(1 - y)(1 - x) = 1$, or equivalently:

\begin{align} \quad 1 - x - y + xy = 1\quad \Leftrightarrow \quad x \circ y = 0 \end{align}

- And:

\begin{align} \quad 1 -y - x + yx = 1 \quad \Leftrightarrow \quad y \circ x = 0 \end{align}

- So $x$ is quasi-invertible with quasi-inverse $y$. $\blacksquare$

Proposition 2: Let $X$ be an algebra and let $x, y \in X$. Then $x$ has quasi-inverse $y$ if and only if in the unitization $X + \mathbf{F}$, $(0, 1) - (x, 0)$ has inverse $(0, 1) - (y, 0)$. |

*Note that in proposition 2 we do NOT require $X$ to have a unit. This is one reason why the unitization $X + \mathbf{F}$ is significant because the unitization is always an algebra with unit $(0, 1)$.*

**Proof:**Let $f : X \to X + \mathbf{F}$ be defined for all $x \in X$ by $f(x) = (x, 0)$. Then $f$ is a monomorphism. Let $x, y \in X$.

- $\Rightarrow$ Suppose that $x$ has quasi-inverse $y$. Then $x \circ y = y \circ x = 0$. Now observe that:

\begin{align} \quad \quad (0, 0) \overset{(*)} = f(0) = f(x \circ y) = f(x + y - xy) = (x + y - yx, 0) = (x, 0) + (y, 0) - (x, 0)(y, 0) = (x, 0) \circ (y, 0) \end{align}

- And similarly:

\begin{align} \quad \quad (0, 0) \overset{(*)}= f(0) = f(y \circ x) = f(y + x - yx) = (y + x - yx, 0) = (y, 0) + (x, 0) - (y, 0)(x, 0) = (y, 0) \circ (x, 0) \end{align}

- (Where the equalities at $(*)$ come from the fact that $f$ is a monomorphism).

- The above two equations show that $(x, 0)$ is quasi-invertible with quasi-inverse $(y, 0)$. Since $X + \mathbf{F}$ is an algebra with unit, by proposition 1 we have (since $(0, 1)$ is the unit in $X + \mathbf{F}$) that $(0, 1) - (x, 0)$ is invertible with inverse $(0, 1) - (y, 0)$.

- $\Leftarrow$ Suppose that $(0, 1) - (x, 0)$ is invertible and has inverse $(0, 1) - (y, 0)$. Since $X + \mathbf{F}$ is an algebra with unit we have by proposition $1$ that $(x, 0)$ is quasi-invertible with quasi-inverse $(y, 0)$.

- Now observe that:

\begin{align} \quad \quad f(0) = (0, 0) \overset{(**)} = (x, 0) \circ (y, 0) = (x, 0) + (y, 0) - (x, 0)(y, 0) =(x + y, 0) - (xy, 0) = (x + y - xy, 0) = f(x + y - xy) \end{align}

- (Where the equality at $(**)$ comes from the fact that $(x, 0)$ is quasi-invertible with quasi-inverse $(y, 0)$).

- So $f(0) = f(x + y - xy)$. Since $f$ is injective (as it is a monomorphism), $0 = x + y - xy = x \circ y$. A similar argument shows that $0 = y \circ x$. So $x$ is quasi-invertible with quasi-inverse $y$. $\blacksquare$

Proposition 3: Let $X$ be an algebra and let $x, y \in X$. Then:a) $xy$ is left quasi-invertible if and only if $yx$ is left quasi-invertible.b) $xy$ is right quasi-invertible if and only if $yx$ is right quasi-invertible. |

**Proof of a)**Suppose that $xy$ is left quasi-invertible. Let $z$ be the left quasi-inverse of $xy$. Then $z \circ (xy) = 0$, that is:

\begin{align} \quad z + xy - zxy = 0 \end{align}

- Multiplying the above equation on the left by $y$ gives us:

\begin{align} \quad yz + yxy - yzxy = 0 \end{align}

- And multiplying the above equation on the right by $x$ gives us:

\begin{align} \quad yzx + yxyx - yzxyx &= 0 \\ \quad yzx -yzx(yx) + yxyx &= 0 \quad (*) \end{align}

- We claim that $(yzx - yx)$ is a left quasi-inverse of $yx$. To see this, observe that:

\begin{align} \quad (yzx - yx) \circ (yx) &= (yzx - yx) + (yx) - (yxz - yx)(yx) \\ &=yzx -yxz(yx) + yx(yx) \\ & \overset{(*)} = 0 \end{align}

- Thus $yx$ is left quasi-invertible and $(yzx - yx)$ is a left quasi-inverse of $yx$. The converse argument is essentially identical. $\blacksquare$

**Proof of b)**Suppose that $xy$ is right quasi-invertible with quasi-inverse $z$. Then $xy \cdot z = 0$. It can be shown (similarly to above) that then $yzx - yx$ is a right quasi-inverse of $yx$ and of course, the converse follows immediately. $\blacksquare$