Anyone who has studied linear algebra knows that a clear difference between square and non-square matrices is that we can calculate the determinant for a square matrix, but for non-square matrices, the concept of a determinant does not exist. In a mathematical system that pursues unity and harmony, why do non-square matrices lack a determinant? Perhaps the most appropriate answer to this question is—because it is not "beautiful" enough.
In fact, one could define a determinant for non-square matrices in a similar fashion. Such a definition could retain the same properties as the determinant of a square matrix; for example, if a row is multiplied by a constant, the value of the determinant is also multiplied by that constant, and so on. Furthermore, its geometric meaning could be preserved. However, the determinant of a non-square matrix is not elegant enough. For a general square matrix with integer elements, the determinant is an integer; however, for a general non-square matrix with integer elements, the definition leads to a determinant value that is an irrational number. Additionally, another important reason is that the determinant of square matrices alone is sufficient for most applications. Based on these two reasons, the concept of a determinant for non-square matrices has been discarded.
The Determinant of a Non-Square Matrix is Not Elegant
The determinant of an n \times n square matrix is a linear function of each vector, representing the linear dependence between vectors. Geometrically, it represents the (signed) volume of the n-dimensional parallelepiped formed by the vectors. We naturally expect the determinant of a non-square matrix to retain these properties, as only then would the operational properties of square matrix determinants be preserved—such as the aforementioned property where multiplying a row by a constant multiplies the determinant by the same constant.
Let us consider an m \times n matrix where m < n. We view this as a combination of m vectors in n-dimensional space. For the simplest case, let us consider the determinant of a 1 \times 2 matrix, which is the determinant of a single 2D vector (a, b).
We already know that the absolute value of the determinant of a 2 \times 2 matrix is the area of the parallelogram formed by the two vectors. Similarly, the absolute value of the determinant of a 2D vector (a, b) should be equal to the length of the vector itself, which is: \det(a, b) = \sqrt{a^2 + b^2} For rational numbers a and b, in most cases, this results in an irrational determinant value. This is inconsistent. Since the determinant is a linear function, it feels "uncomfortable" for linear operations on rational numbers to result in irrational numbers.
Similarly, we can consider the determinant of a 2 \times 3 matrix: \boldsymbol{A} = \begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix} According to its geometric meaning, the magnitude of its determinant should be the area of the parallelogram formed by these two vectors. We could use the cross product to calculate this area, but to be more general, we will use the method of orthogonalization.
Starting with the vector \boldsymbol{x}_1 = (a, b, c), we choose: \boldsymbol{e}_1 = \frac{(a, b, c)}{\sqrt{a^2 + b^2 + c^2}} Then \boldsymbol{x}_1 = |\boldsymbol{x}_1|\boldsymbol{e}_1. Orthogonalizing \boldsymbol{x}_2 = (d, e, f) against \boldsymbol{e}_1, we get: \boldsymbol{x}_2 - \langle \boldsymbol{e}_1, \boldsymbol{x}_2 \rangle \boldsymbol{e}_1 Taking: \boldsymbol{e}_2 = \frac{\boldsymbol{x}_2 - \langle \boldsymbol{e}_1, \boldsymbol{x}_2 \rangle \boldsymbol{e}_1}{|\boldsymbol{x}_2 - \langle \boldsymbol{e}_1, \boldsymbol{x}_2 \rangle \boldsymbol{e}_1|} We obtain: \boldsymbol{x}_2 = |\boldsymbol{x}_2 - \langle \boldsymbol{e}_1, \boldsymbol{x}_2 \rangle \boldsymbol{e}_1|\boldsymbol{e}_2 + \langle \boldsymbol{e}_1, \boldsymbol{x}_2 \rangle \boldsymbol{e}_1 Therefore, the magnitude of the determinant, which is the area of the parallelogram formed by the two vectors, equals: |\det \boldsymbol{A}| = |\boldsymbol{x}_1| \cdot |\boldsymbol{x}_2 - \langle \boldsymbol{e}_1, \boldsymbol{x}_2 \rangle \boldsymbol{e}_1|
The final expression involves two norms, which means two square roots. By picking any specific example, one can verify that even if all elements of \boldsymbol{A} are integers, the result is generally not a rational number. This reflects the lack of elegance in the determinant of non-square matrices.
The Determinant of a Square Matrix is Sufficient
Since the determinant of a non-square matrix is not elegant, we simply discard it. However, does this cause any "side effects"? That is to say, are there places where the determinant of a non-square matrix is absolutely necessary? In fact, at least to my current knowledge, the answer is no.
For example, to determine the linear independence of m row vectors in n-dimensional space, we can do the following: The first method is to use elementary transformations to see if the rank of the transformed matrix is m. If the first method involves too many steps, the second method is relatively more direct: check if the determinant of any m \times m submatrix (formed by deleting n-m columns) is non-zero. If at least one such determinant is non-zero, the vectors are linearly independent. Thus, the determinant of square matrices is sufficient.
What if we need to calculate area or volume? In fact, we use formulas similar to |\boldsymbol{x}_1| \cdot |\boldsymbol{x}_2 - \langle \boldsymbol{e}_1, \boldsymbol{x}_2 \rangle \boldsymbol{e}_1|. This formula is not difficult to understand or remember. Since it contains irreducible square roots, there is no simpler calculation formula anyway.
Finally, a core reason that limits the use of non-square determinants is that they have few applications. Square determinants can be used to find various factors, such as the Jacobian determinant in coordinate transformations for multiple integrals, etc. Due to the invertibility of square matrices, these determinants have direct practical significance. In contrast, there is no concept of invertibility for non-square matrices, so the applications for non-square determinants are very narrow.
When reprinting, please include the original address of this article: https://kexue.fm/archives/2757
For more detailed information regarding reproduction, please refer to: Scientific Space FAQ