*New Understandings of Matrices 6*: Why Do Only Square Matrices Have Determinants?
Anyone who has studied linear algebra knows that one obvious difference between square and non-square matrices is that for a square matrix we can compute its determinant, whereas for a non-square matrix the notion of a determinant simply doesn't exist. In a mathematical system that strives for unity and harmony, why should non-square matrices be denied a determinant? Perhaps the most fitting answer to this question is—because it just isn't beautiful enough. For non-square matrices, one can in fact define something analogous to a determinant, and the resulting object shares many of the same properties as the determinant of a square matrix: for instance, multiplying one row by a constant also multiplies the "determinant" value by that constant, and so on; its geometric meaning can likewise be preserved. But the determinant of a non-square matrix is not elegant enough, because for a general square matrix with integer entries, the determinant is an integer, whereas for a general non-square matrix with integer entries, the resulting "determinant" value turns out to be irrational. Another fairly important reason is that the determinant of square matrices alone already suffices for our needs. Taking both of these reasons together, the determinant of non-square matrices was simply abandoned.
The determinant of a non-square matrix isn't beautiful enough
The determinant of an $n$-order square matrix is a linear function of each vector, and it represents the linear dependence among the vectors; geometrically, it is the (signed) volume of the n-dimensional parallelepiped spanned by those vectors. We would naturally hope that the determinant of a non-square matrix could preserve these properties too, because only then would the familiar algebraic properties of determinants — such as the one mentioned above, where multiplying a row by a constant multiplies the determinant value by that constant — survive. Consider a matrix of shape $m\times n$, where $ m < n $，我们将它看成是$m$个$n$维向量的组合。最简单的，我们先考虑$1\times 2$矩阵的行列式，也就是二维向量$(a,b)$'s determinant. more
We already know that the absolute value of the determinant of a $2\times 2$ matrix equals the area of the parallelogram spanned by the two vectors. Similarly, the absolute value of the "determinant" of a two-dimensional vector $(a,b)$ should equal the length of the vector itself, that is, $\det (a,b)=\sqrt{a^2+b^2}$. For rational $a,b$, this will, in most cases, give an irrational determinant value — which feels inconsistent. Since the determinant is supposed to be a linear function, it is quite uncomfortable that a linear operation on rational numbers should produce an irrational result.
Similarly, consider the $2\times 3$ matrix
$$\boldsymbol{A}=\left(\begin{array}{\cdot {20}{c}} a&b&c\\ d&e&f \end{array}\right)$$
and its determinant. Geometrically, the modulus of its determinant should be the area of the parallelogram spanned by the two vectors. We could use the cross product to compute this area, but to keep things more general, let's compute it via orthogonalization.
Starting from the vector $\boldsymbol{x}_1=(a,b,c)$, choose
$$\boldsymbol{e}_1=\frac{(a,b,c)}{\sqrt{a^2+b^2+c^2}}$$
so that $\boldsymbol{x}_1=\left|\boldsymbol{x}_1\right|\boldsymbol{e}_1$. Orthogonalizing $\boldsymbol{x}_2=(d,e,f)$, we get
$$\boldsymbol{x}_2 - \left\langle {\boldsymbol{e}_1,\boldsymbol{x}_2} \right\rangle\boldsymbol{e}_1$$
Taking
$$\boldsymbol{e}_2=\frac{\boldsymbol{x}_2 - \left\langle {\boldsymbol{e}_1,\boldsymbol{x}_2} \right\rangle\boldsymbol{e}_1}{\left|\boldsymbol{x}_2 - \left\langle {\boldsymbol{e}_1,\boldsymbol{x}_2} \right\rangle\boldsymbol{e}_1\right|}$$
we obtain
$$\boldsymbol{x}_2=\left|\boldsymbol{x}_2 - \left\langle {\boldsymbol{e}_1,\boldsymbol{x}_2} \right\rangle\boldsymbol{e}_1\right|\boldsymbol{e}_2+\left\langle {\boldsymbol{e}_1,\boldsymbol{x}_2} \right\rangle\boldsymbol{e}_1$$
Hence the modulus of the determinant, i.e., the area of the parallelogram formed by the two vectors, equals
$$\left|\det\boldsymbol{A}\right|=\left|\boldsymbol{x}_1\right|\cdot \left|\boldsymbol{x}_2 - \left\langle {\boldsymbol{e}_1,\boldsymbol{x}_2} \right\rangle\boldsymbol{e}_1\right|$$
The final expression contains two moduli, meaning two square roots — and picking almost any concrete example is enough to verify that even when the entries of $\boldsymbol{A}$ are all integers, we still don't get a rational number. This illustrates the inelegance of the non-square-matrix determinant.
The determinant of square matrices is already enough
Since the determinant of non-square matrices isn't beautiful enough, we might as well discard it altogether. But does this cause any "side effects"? That is, are there situations where a non-square-matrix determinant is truly indispensable? In fact, as far as I currently understand, the answer is no.
For example, to determine whether $m$ row vectors of dimension $n$ are linearly independent, we can proceed as follows. The first method uses elementary transformations and checks whether the rank of the resulting matrix equals $m$. If that feels like too many steps, a second, more direct method is to check whether, after deleting any $n-m$ columns, the determinant of the remaining square matrix is nonzero; if there exists such a nonzero determinant, the vectors are linearly independent. So the determinant of square matrices already suffices.
What if we need to compute an area or a volume? In fact, we simply use a formula analogous to $\left|\boldsymbol{x}_1\right|\cdot \left|\boldsymbol{x}_2 - \left\langle {\boldsymbol{e}_1,\boldsymbol{x}_2} \right\rangle\boldsymbol{e}_1\right|$ — not hard to understand, and not hard to remember either. Since it involves an irreducible square root, there is no simple closed-form computational formula to be had.
Finally, one fairly central reason limiting the usefulness of non-square-matrix determinants is that they have very few applications. The determinant of a square matrix can be used to compute all sorts of factors — for example, the Jacobian determinant in coordinate transformations for multiple integrals, and so on — and because square matrices can be invertible, these determinants carry direct practical significance. Non-square matrices, by contrast, have no notion of invertibility, so the applications of their "determinants" remain very narrow indeed.
Translated automatically with claude-sonnet-5; all equations are reproduced verbatim from the source. Copyright remains with the original author.