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Where in the history of linear algebra did we pick up on referring to invertible matrices as 'non-singular'? In fact, since

  • the null space of an invertible matrix has a single vector
  • an invertible matrix has a single solution for every possible $b$ in $AX=b$

it's easy to imagine that that invertible matrices would be called 'singular'. What gives?

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If you take an $n\times n$ matrix "at random" (you have to make this very precise, but it can be done sensibly), then it will almost certainly be invertible. That is, the generic case is that of an invertible matrix, the special case is that of a matrix that is not invertible.

For example, a $1\times 1$ matrix (with real coefficients) is invertible if and only if it is not the $0$ matrix; for $2\times 2$ matrices, it is invertible if and only if the two rows do not lie in the same line through the origin; for $3\times 3$, if and only if the three rows do not lie in the same plane through the origin; etc.

So here, "singular" is not being taken in the sense of "single", but rather in the sense of "special", "not common". See the dictionary definition: it includes "odd", "exceptional", "unusual", "peculiar".

The noninvertible case is the "special", "uncommon" case for matrices. It is also "singular" in the sense of being the "troublesome" case (you probably know by now that when you are working with matrices, the invertible case is usually the easy one).

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    $\begingroup$ Interesting. I always assumed that a singular matrix was called such because it didn't have a partner! $\endgroup$ – JavaMan Jun 1 '11 at 20:26
  • $\begingroup$ @DJC: Whatever helps you remember works! $\endgroup$ – Arturo Magidin Jun 1 '11 at 20:27
  • $\begingroup$ Just to slightly nitpick, if a $3\times 3$ matrix is made of the same row three times, then the rows do not lie in a single plane, but rather on a single line (and infinitely many planes). You might want to choose better words to give out your exact meaning there. $\endgroup$ – Asaf Karagila Jun 1 '11 at 20:28
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    $\begingroup$ @Asaf Karagila: If the three points lie on a single line, then there is certanly a plane (in fact, infinitely many planes) that contain all three points, no? But I'll reword slightly. $\endgroup$ – Arturo Magidin Jun 1 '11 at 20:29
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    $\begingroup$ The term "singularity" for a place where a function blows up, often because you can't divide by $0$, is much older than "singular" for matrices. Maybe there is an analogical connection. $\endgroup$ – André Nicolas Jun 1 '11 at 21:24
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The singular matrices are the singular locus of the algebraic variety of matrices.

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  • $\begingroup$ I don't understand this answer. The variety of $n \times n$ matrices is isomorphic to $\mathbb{A}^{n^2}$, which has no singular points. $\endgroup$ – André 3000 Feb 11 '18 at 22:10
  • $\begingroup$ @Quasicoherent Yeah now I'm not sure this makes any sense. There is a stratification of $n\times n$ matrices by rank, and a stratification by singular loci. Probably there is a morphism (the singular locus of a stratum is of lower rank), but the full rank matrices have no singular locus... $\endgroup$ – ziggurism Feb 12 '18 at 19:35

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