Algebras - xy is Quasi-Invertible IFF yx is Quasi-Invertible

# Algebras - xy is Quasi-Invertible IFF yx is Quasi-Invertible

Proposition 3: Let $\mathfrak{A}$ be an algebra and let $x, y \in \mathfrak{A}$. 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. $\blacksquare$