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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 $\endgroup$ Commented 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
    Commented Jul 3, 2014 at 7:24
  • $\begingroup$ @kjetilbhalvorsen Interesting article! $\endgroup$
    – Red Banana
    Commented Oct 4, 2016 at 18:57

2 Answers 2

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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
    Commented 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. $\endgroup$ Commented Jul 3, 2014 at 4:29
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    $\begingroup$ What about when B has more rows than A? $\endgroup$ Commented 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. $\endgroup$ Commented 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
    Commented Nov 30, 2019 at 22:24
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This extension of determinants has all 4 properties if A is a square matrix (except it loses the sign), and retains some attributes of determinants otherwise.

If A has more rows than columns, then $$|A|^2=|A^{T}A|$$ If A has more columns than rows, then $$|A|^2=|AA^{T}|$$

This has a valid and useful geometric interpretation. If you have a transformation $A$, this can still return how the transformation will scale an input transformation, limited to the smaller of the dimensions of the input and output space.

You may take this to be the absolute value of the determinant. It's always positive because, when looking at a space embedded in a higher dimensional space, a positive area can become a negative area when looked at from behind.

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    $\begingroup$ Why $A^TA$ and not $AA^T$? $\endgroup$
    – MJD
    Commented Jul 2, 2023 at 16:45
  • $\begingroup$ @MJD Thanks. Reading back through this answer was a little embarrassing. Though clearly some people have gotten use from it. Thanks for drawing my attention to this so that I could update and improve my answer. $\endgroup$ Commented Jul 4, 2023 at 21:58
  • $\begingroup$ Thanks! I had another question but your update answers it. $\endgroup$
    – MJD
    Commented Jul 5, 2023 at 0:15

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