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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"?

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    $\begingroup$ 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). $\endgroup$ – Qiaochu Yuan Jul 3 '16 at 18:15
  • $\begingroup$ @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. $\endgroup$ – user42912 Jul 3 '16 at 18:18
  • $\begingroup$ @user42912 They are equivalent on finite-dimensional spaces, but not for infinite-dimensional spaces. $\endgroup$ – Aweygan Jul 3 '16 at 18:26
  • $\begingroup$ 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". $\endgroup$ – user1551 Jul 3 '16 at 19:57
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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.

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  • $\begingroup$ Why do we call a function with a well-defined inverse by "nonsingular"? thanks $\endgroup$ – user42912 Jul 3 '16 at 18:20
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    $\begingroup$ That's more etymology than anything. I would guess because "nonsingular" makes it sound like the operator "comes with" another operator, namely its inverse (i.e., singular = all alone, no other friendly operators to hang out with). $\endgroup$ – parsiad Jul 3 '16 at 18:21

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