Why do we call this transformation non-singular?

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"?

• This is bad terminology. The right terminology for this condition is injective or one-to-one. "Nonsingular" is typically used to mean bijective or invertible (en.wikipedia.org/wiki/Invertible_matrix). – Qiaochu Yuan Jul 3 '16 at 18:15
• @QiaochuYuan Kunze and Hoffman's book use both terms. This gives me an impression that there are other types of function such that these terms are not equivalent. – user42912 Jul 3 '16 at 18:18
• @user42912 They are equivalent on finite-dimensional spaces, but not for infinite-dimensional spaces. – Aweygan Jul 3 '16 at 18:26
• Historically, the term "singular" was used to describe a square matrix rather than a linear transformation. A singular matrix means a square matrix with zero determinant. As matrices usually have nonzero determinant and such matrices have behave more nicely, matrices with zero determinants were considered exceptional, hence the term "singular". – user1551 Jul 3 '16 at 19:57

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.