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 $\endgroup$ Jul 2, 2014 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, 2014 at 7:24
  • $\begingroup$ @kjetilbhalvorsen Interesting article! $\endgroup$
    – Red Banana
    Oct 4, 2016 at 18:57

2 Answers 2


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, 2014 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. $\endgroup$ Jul 3, 2014 at 4:29
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    $\begingroup$ What about when B has more rows than A? $\endgroup$ Mar 23, 2019 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. $\endgroup$ Nov 29, 2019 at 17:43
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    $\begingroup$ @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, 2019 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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