0

I know that a matrix with a determinant of zero means the matrix is invertible. But I have difficulties with grasping the concept of "no determinant". Does it also mean that only non-square matrices that have no determinant?

Aituusi
  • 11
  • 3
    The determinant is defined for square matrices. – cmk Jul 09 '19 at 18:24
  • 5
    Welcome to Mathematics Stack Exchange. A determinant of zero means the matrix is not invertible – J. W. Tanner Jul 09 '19 at 18:24
  • 5
    Determinant zero means not invertible. It has nothing to with not having a determinant. You are right that determinant is not defined for square matrices. Also, it isn't necessarily defined for square matrices with entries in a noncommutative ring, – Cheerful Parsnip Jul 09 '19 at 18:25
  • 4
    @CheerfulParsnip "You are right that determinant is not defined for square matrices." I think there's a typo there ... – Noah Schweber Jul 09 '19 at 18:26
  • 1
    @NoahSchweber: indeed I meant it is not defined for non-square matrices. Thanks. – Cheerful Parsnip Jul 09 '19 at 18:32
  • A small amendment to @CheerfulParsnip's statement "Also, it isn't necessarily defined for square matrices with entries in a noncommutative ring" - there are indeed notions of determinant for matrices with entries in non-commutative rings, but they're quite messy and should be ignored for now until the classical case (matrices with real numbers as entries - to the OP, in a bit of jargon which one will need to understand to fully "get" the topic, these represent linear transformations of finite-dimensional $\mathbb{R}$-spaces) is well-understood. – Noah Schweber Jul 09 '19 at 18:51

1 Answers1

4

First of all, your first sentence is incorrect: having a determinant of zero means the matrix is not invertible, and indeed the two are equivalent.

As to the difference between having determinant equal to zero and having no determinant, at the simplest level the determinant is only defined for square matrices: every square matrix has a determinant (possibly zero, possibly nonzero), and no non-square matrix has a determinant.

Now one might ask, "Why do we only define the determinant for square matrices?" I think the simplest answer is to remember that matrices are representations of linear maps - an $n$-by-$m$ matrix representing a linear map from an $m$-dimensional space to an $n$-dimensional space - and when $m=n$ it makes sense to ask how this map "distorts volume." E.g. in the setting $m=n=2$, the matrix $2I$ (where $I$ is the 2-dimensional identity matrix) represents the transformation which doubles the length (while not changing the direction) of every vector. This changes a rectangle of area $A$ into a rectangle of area $4A$, and the determinant is the ratio ${4A\over A}=4$.

But when $m\not=n$ this doesn't really make sense: if $m<n$ then every linear map transforms the whole domain into a set with zero $n$-dimensional volume (think about how a plane in $3$-space has no volume), and if $m>n$ then the $n$-dimensional volume of the image of an $m$-dimensional rectangle need not be determined by the $m$-dimensional volume of that rectangle (think about a projection map from $3$-space to $2$-space, and the non-difference between what happens to a "tall" prism versus a "short" prism). So if $m\not=n$, either the determinant is a silly notion (if $m<n$) or isn't even guaranteed to be well-defined (if $m>n$; indeed, if $m>n$ then this will only be well-defined in the case when our map sends everything to the zero vector!).

(See also this old answer of mine. Additionally, down the road you may be interested in further discussion of what the determinant "actually is," possibly in much more general contexts; there are a few discussions of this here and at mathoverflow.)

Note that here I'm omitting any discussion of what the entries of our matrix are. While important (to put it mildly!), I think bringing in that sort of generality at this point will only confuse matters. But it is important to note that $(1)$ we can consider matrices with entries not just from $\mathbb{R}$ but from an arbitrary "number system," like the complex numbers (specifically, I'm referring to fields), and $(2)$ we can go to even further levels of generality but then things break down (e.g. we can, and often do, consider matrices with entries from an arbitrary ring, but then the "determinant" idea gets much more complicated).

Noah Schweber
  • 245,398