For a ring $R$, I must show that if R has a left identity that is not a right identity, then there is more than one left identity.
Could I do this by contradiction, using 2x2 matrices and show that a given set has two left identities that are not right identities?
Let $u$ be a left identity that is not a right identity. It suffices to find an element $v \ne 0$ such that $v y = 0$ for all $y \in R$, for then $u + v$ is a left identity. Since $u$ is not a right identity, there exists an element $x \in R$ such that $v = x u -x \ne 0$. Check that $v y = 0$ for all $y \in R$. Thus $v$ has the desired properties.