Why do we call this transformation non-singular?

Question

In linear algebra books, the authors call the linear transformation $T$ with the property

$$T(\alpha)=0\implies \alpha=0$$

non-singular.

What's the motivation behind the term "non-singular"?

Answer 1

Suppose $Tv=0$ implies $v=0$. It follows immediately that $T$ is injective since if $Tv_{1}=Tv_{2}$ then $Tv_{1}-Tv_{2}=T(v_{1}-v_{2})=0$ and hence $v_{1}-v_{2}=0$, or equivalently $v_{1}=v_{2}$. Therefore, we can unambiguously define $T^{-1}$ on the range of $T$.

The author might be calling $T$ nonsingular because it "comes with" a well-defined inverse (at least on the range of $T$).

Edit As QiaochuYuan points out, this could be considered bad terminology. Usually nonsingular is reserved for (in addition to $T$ being injective) when the range of $T$ is the whole codomain.