I wrote an answer to this question based on determinants, but subsequently deleted it because the OP is interested in non-square matrices, which effectively blocks the use of determinants and thereby undermined the entire answer. However, it can be salvaged if there exists a function $\det$ defined on all real-valued matrices (not just the square ones) having the following properties.

  1. $\det$ is real-valued
  2. $\det$ has its usual value for square matrices
  3. $\det(AB)$ always equals $\det(A)\det(B)$ whenever the product $AB$ is defined.
  4. $\det(A) \neq 0$ iff $\det(A^\top) \neq 0$

Does such a function exist?

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    $\begingroup$ You could have a look at this paper: static.bsu.az/w24/pp.163-175.pdf $\endgroup$ – kjetil b halvorsen Jul 2 '14 at 11:15
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    $\begingroup$ I'm not sure whether there is a term for this, but you might consider the vector formed by all minors of maximal size. So for an $n\times m$ matrix, let $k=\min(n,m)$ then compute all determinants of $k\times k$ submatrices, perhaps with alternating sign. The result generalizes both the determinant and the cross product. It is however vector-valued, not real-valued, except for the square case. It also doesn't satisfy 3. either. But it is multilinear, so it might be useful for some applications of determinants. $\endgroup$ – MvG Jul 3 '14 at 7:24
  • $\begingroup$ @kjetilbhalvorsen Interesting article! $\endgroup$ – Billy Rubina Oct 4 '16 at 18:57

Such a function cannot exist. Let $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$. Then, since both $AB$ and $BA$ are square, if there existed a function $D$ with the properties 1-3 stated there would hold \begin{align} \begin{split} 1 &= \det \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \det(BA) = D(BA) = D(B)D(A) \\ &= D(A)D(B) = D(AB) = \det(AB) = \det \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0. \end{split} \end{align}

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    $\begingroup$ +1. It's worth pointing out that the components don't really matter here, $\mathrm{det}(AB)=0$ whenever $A$ has more rows than $B$. $\endgroup$ – Nikolaj-K Jul 2 '14 at 13:43
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    $\begingroup$ @NikolajK: Well, it is important for the answer that $\det(BA)\neq0$, so the entries do matter a bit. However the first example that came to my mind (honestly!) is taking $A$ to be the $n\times0$ matrix and $B$ the $0\times n$ matrix, for some $n>0$; then $AB$ is a $n\times n$ zero matrix so $\det(AB)=0$, while $BA$ is the $0\times0$ (identity) matrix, so $\det(BA)=1$. In this example there aren't even any entries of $A$ or $B$ to worry about. $\endgroup$ – Marc van Leeuwen Jul 3 '14 at 4:29
  • $\begingroup$ What about when B has more rows than A? $\endgroup$ – Aaron Franke Mar 23 '19 at 1:55
  • $\begingroup$ @Nikolaj-K What do you mean? I tried multiplying some matrices, and what you said seems to be true, but I'm not sure why. $\endgroup$ – Anurag Nov 29 '19 at 17:43
  • $\begingroup$ @AnuragB. In computing $ABv$, the vector$Bv$ has a smaller dimension than the final result, so the spanned spaces of $A$ and $B$ can't be in bijection. $\endgroup$ – Nikolaj-K Nov 30 '19 at 22:24

This extension of determinants has all 4 properties if A is a square matrix, and retains some attributes of determinants otherwise.


If you're willing to break the rules a little bit, this has a valid and useful geometric interpretation. If you have a space defined in a dimension higher than its own, this can still return the area it defines.

Since the square of the determinant of a matrix can be found with the above formula, and because this multiplication is defined for nonsquare matrices, we can extend determinants to nonsquare matrices. For example, take the 3 wide matrix A defined with column vectors, x y and z, where each have n components:


You can dot each of the vectors with each other by right multiplying A by its transpose:

$$A^{T}A=\begin{pmatrix}x\\y\\z\end{pmatrix}\begin{pmatrix}x&y&z\end{pmatrix}=\begin{pmatrix} x\cdot x & x\cdot y & x\cdot z\\ x\cdot y & y\cdot y & y\cdot z\\ x\cdot z & y\cdot z & z\cdot z \end{pmatrix}$$

Taking the determinant of this, you get the square of A's determinant: $$2 (x\cdot y) (x\cdot z) (y\cdot z)+(x\cdot x) (y\cdot y) (z\cdot z)-(x\cdot z)^2 (y\cdot y) - (x\cdot x )(y\cdot z)^2 - (x\cdot y)^2 (z\cdot z)$$

In this 3 vector example, the equation above returns the value of the volume defined by vectors x y and z.

You may take the positive square root of this to be the absolute value of the determinant. It's always positive because it doesn't make sense to define positive and negative areas for spaces defined in dimensions higher than the space itself. Depending on the perspective, a positive area can become a negative area if looked at from behind.

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The product of all nonzero singular values satisfies most of the properties you are looking for, with the exception of 3. It will compute the hypervolume of the space spanned by the matrix.

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  • $\begingroup$ Without condition 3, one would not be asking for anything very spectacular. Just define the function to have any constant value for all non-square matrices, and of course the usual value for square matrices, and it satisfies 1, 2, 4. $\endgroup$ – Marc van Leeuwen Jul 2 '14 at 18:44
  • $\begingroup$ Right; but that wouldn't have any conceptual relationship to a determinant anymore. Not that my partial answer is guaranteed to be of much use; im mostly just guessing as to what motivated this question in the first place. $\endgroup$ – Eelco Hoogendoorn Jul 2 '14 at 20:00
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    $\begingroup$ Also what you say in this answer is not correct, since the product of all nonzero singular values fails to have property$~$2. Being nonzero by definition, the product does not for square matrices detect the condition of being singular, which is by far the most basic application of determinants. And even if the matrix is non-singular, the product of the (nonzero) singular values only gives the absolute value of the determinant. $\endgroup$ – Marc van Leeuwen Jul 3 '14 at 4:22
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    $\begingroup$ You are correct; property 2 does not hold either. It is still the closest approximation of the concept of a determinant for rank-deficient matrices that I can think of, though. $\endgroup$ – Eelco Hoogendoorn Jul 3 '14 at 4:48
  • $\begingroup$ I think you should not restrict the product to the non-zero singular values, but instead multiply all singular values. (All values on the diagonal of the diagonal rectangular matrix in the singular value decomposition.) This will fix condition 2. $\endgroup$ – Lemming Mar 9 '18 at 8:51

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