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I'm reading a book that claims that determinants and traces are only defined for square matrices, but doesn’t really explain why. From a calculation standpoint, this seems correct because I wouldn’t really know how to calculate the determinant of something like a column vector (with more than one row). However, the only justification I can think of is that using the meaning of a determinant, we’re trying to calculate the scaling factor of our unit cube, hypercube, whatever, while missing some “dimension” of it. I don’t really know if that makes sense, but it’s like if I were to ask you to find the volume of a box using only the dimensions of one of its faces. Is this a reasonable explanation, or is there a better reason why we can’t calculate the determinant of a non-square matrix ? And as for the trace, does that really just come down to the definition of the trace, or is there something more “meaningful” ?

componentvector
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    I like your analogy – J. W. Tanner May 13 '20 at 18:40
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    The problems we want to solve come first. The definitions come after and are written in such a way as to be useful for solving the problems we want to solve. In the case of determinants, one of the leading motivations was finding a way to determine whether or not a system of equations had a unique solution. This was in the 1600's by Leibniz before matrices had even been developed yet. – JMoravitz May 13 '20 at 18:43
  • Related: What is the origin of the determinant in linear algebra?. – JMoravitz May 13 '20 at 18:47
  • Thinking in terms of eigenvalues, the trace is the sum of the eigenvalues, and the determinant is the product of the eigenvalues. Are eigenvalues/eigenvectors defined for non-square matrices? – Josh B. May 13 '20 at 18:58
  • Note that one can, in fact, find the (lower-dimensional) volume of the parallelopiped spanned by the columns of an $m \times n$ matrix with $m \geq n$. In particular: if $A$ is such a matrix, then this volume is given by $V = \sqrt{\det(A^TA)}$. – Ben Grossmann May 13 '20 at 21:50
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    One possible line of justification is to take note of coordinate-free interpretations of determinant and trace that fail to generalize to "non-square" cases. For instance, the determinant can be seen as the unique $n$-linear alternating form on a dimension $n$ vector space, and the trace can be seen as the result of linearly extending the map $T:L(V) \to \mathbb F$ defined so that $T(vf(\cdot)) = f(v)$ for any $v \in V$ and $f:V \to \Bbb F$. Both of these interpretations require that the matrix in question be square. – Ben Grossmann May 13 '20 at 21:56
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    If you want to define determinants for non-square matrices, you have to make a big compromise: since $\det(AB)$ in general is different from $\det(BA)$ when $A$ and $B$ are non-square, if $\det(A)$ and $\det(B)$ are defined, one of the most important properties of determinant, namely, $\det(AB)=\det(A)\det(B)$, will no longer hold universally. – user1551 May 14 '20 at 12:21
  • @user1551 Alternatively you can drop the requirement that the determinant be a real number, make the result of $\det$ non-commutative, and you can gain back the property that $\det(AB)= \det(A)\det(B)$, and you actually get some really nice geometry out of it. – Trey Reynolds Mar 18 '24 at 18:33

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