A few years ago, based on my superficial understanding of matrices, I wrote a series titled “Understanding Matrices.” One of the articles, “Why Do Only Square Matrices Have Determinants?”, discussed the issue of determinants for non-square matrices. In that post, I presented the viewpoints that “determinants of non-square matrices are not elegant” and that “determinants of square matrices are sufficient.” In this article, I will revisit this question.
First, let us review the determinant of a square matrix. In fact, the most important value of a determinant lies in its geometric meaning:
The absolute value of the determinant of an n-dimensional square matrix is equal to the hypervolume of the n-dimensional solid spanned by its row (or column) vectors.
This geometric meaning is the source of all the importance of the determinant. For related discussions, one can refer to “Bits of the Determinant”; it also serves as the foundation for our discussion on the determinants of non-square matrices.
Analysis
For a square matrix \boldsymbol{A}_{n\times n}, it can be viewed as a combination of n row vectors or a combination of n column vectors. In either case, the absolute value of the determinant equals the hypervolume of the n-dimensional solid spanned by these n vectors. In other words, for square matrices, the distinction between row and column vectors does not change the determinant.
For a non-square matrix \boldsymbol{B}_{n \times k}, the situation is different. Without loss of generality, assume n > k. We can view it as a combination of n row vectors of dimension k, or as a combination of k column vectors of dimension n. The determinant of a non-square matrix should carry the same meaning: the hypervolume of the solid they span.
Let’s look at the first case. If we view it as n row vectors of dimension k, we must consider the hypervolume of the n-dimensional solid spanned by these n vectors. However, note that n > k, so these n vectors must be linearly dependent. Consequently, they cannot span an n-dimensional solid; they might span an (n-1)-dimensional solid or even lower. Thus, their n-dimensional hypervolume is naturally 0.
However, the second case is not so trivial. If we view it as k column vectors of dimension n, although these k vectors are n-dimensional, they span a k-dimensional solid. The hypervolume of this k-dimensional solid is not necessarily 0. We shall take this non-trivial volume as the definition of the determinant for a non-square matrix.
Definition
For the second case, there is a very clever definition that can be borrowed from the determinant of square matrices: |\det \boldsymbol{B}| = \sqrt{\det (\boldsymbol{B}^{\top}\boldsymbol{B})} \label{eq:dingyi} Of course, this only defines the absolute value of the determinant, but it is usually sufficient. In most cases, we only utilize the absolute value of the determinant.
It can be observed that this definition is compatible with the results for square matrices. Furthermore, we will prove later that this definition indeed preserves the geometric meaning of the determinant.
Let us now calculate two examples. For the first example, consider an n\times 1 matrix: \boldsymbol{X} = \begin{pmatrix}x_1 \\ x_2 \\ \vdots \\ x_n\end{pmatrix} According to definition [eq:dingyi], we calculate: |\det \boldsymbol{X}| = \sqrt{x_1^2 + x_2^2 + \dots + x_n^2} According to our definition, this should represent the “1-dimensional volume” of a single n-dimensional column vector. By analogy, the so-called “1-dimensional volume” is simply the length, and the above formula is exactly the formula for the magnitude (norm) of a vector. That is to say, in the n\times 1 case, definition [eq:dingyi] is compatible with our hypothesis.
The second example is an n\times 2 matrix: \boldsymbol{Z} = \begin{pmatrix}x_1 & y_1 \\ x_2 & y_2 \\ \vdots & \vdots \\ x_n & y_n\end{pmatrix}=(\boldsymbol{x}, \boldsymbol{y}) Calculating according to definition [eq:dingyi], the final result is: |\det \boldsymbol{Z}| = \sqrt{\boldsymbol{x}^{\top}\boldsymbol{x}\boldsymbol{y}^{\top}\boldsymbol{y} - (\boldsymbol{x}^{\top}\boldsymbol{y})^2} It is not difficult to find that this result is exactly the square of the area of the parallelogram spanned by \boldsymbol{x} and \boldsymbol{y}. This is because the area of the parallelogram calculated by definition should be: \begin{aligned} |\boldsymbol{x}|\cdot|\boldsymbol{y}|\cdot\sin\theta &= |\boldsymbol{x}|\cdot|\boldsymbol{y}|\cdot\sqrt{1-\cos^2\theta}\\ &=|\boldsymbol{x}|\cdot|\boldsymbol{y}|\cdot\sqrt{1-\left(\frac{\boldsymbol{x}^{\top}\boldsymbol{y}}{|\boldsymbol{x}|\cdot|\boldsymbol{y}|}\right)^2} \end{aligned} In other words, for an n\times 2 matrix, definition [eq:dingyi] is also consistent with our expectations.
Proof
Now consider the general proof for an n \times k matrix \boldsymbol{B}: \boldsymbol{B}_{n \times k} = \begin{pmatrix}b_{11} & \dots & b_{1k}\\ b_{21} & \dots & b_{2k}\\ \vdots & \ddots & \vdots\\ b_{n1} & \dots & b_{nk}\end{pmatrix} = (\boldsymbol{b}_1,\dots,\boldsymbol{b}_k) where n > k. First, from the familiar Gram–Schmidt orthogonalization process, we know there exists an n\times k orthogonal matrix \boldsymbol{U}_{n\times k} (k mutually orthogonal n-dimensional unit column vectors) and a k\times k upper triangular matrix \boldsymbol{C}_{k\times k} such that: \boldsymbol{B}_{n \times k}=\boldsymbol{U}_{n\times k}\boldsymbol{C}_{k\times k} This is known in mathematics as the “QR decomposition.” We know that orthogonal transformations do not change any geometric properties, so the determinant of \boldsymbol{B}_{n \times k} should be equal to the (absolute value of the) determinant of \boldsymbol{C}_{k\times k}, i.e., |\det \boldsymbol{C}_{k\times k}|.
Thus, we have: \begin{aligned} |\det \boldsymbol{B}_{n\times k}| &= |\det \boldsymbol{C}_{k\times k}|\\ &= \sqrt{\det\left(\boldsymbol{C}_{k\times k}^{\top}\boldsymbol{C}_{k\times k}\right)}\\ &= \sqrt{\det\left[\left(\boldsymbol{U}_{n\times k}^{\top}\boldsymbol{B}_{n \times k}\right)^{\top}\left(\boldsymbol{U}_{n\times k}^{\top}\boldsymbol{B}_{n \times k}\right)\right]}\\ &= \sqrt{\det\left(\boldsymbol{B}_{n \times k}^{\top}\boldsymbol{B}_{n \times k}\right)} \end{aligned} Therefore, for an n\times k matrix \boldsymbol{B} where n > k, a non-trivial and reasonable definition for the matrix determinant is \sqrt{\det (\boldsymbol{B}^{\top}\boldsymbol{B})}. Obviously, if n < k, the definition would be \sqrt{\det (\boldsymbol{B}\boldsymbol{B}^{\top})}.
Conclusion
Starting from geometric significance, we discussed the problem of determinants for non-square matrices. The results indicate that Equation [eq:dingyi] can serve as a reasonable definition for the determinant of a non-square matrix. Although theoretically [eq:dingyi] only defines the absolute value of the determinant, it is sufficient for most cases.
As for the applications of non-square determinants, we know that when performing integral transformations, we use the Jacobian determinant to ensure the non-triviality of the transformation. Similarly, perhaps non-square determinants can be used to ensure the non-triviality of dimensionality increasing or decreasing transformations. Of course, this is a conceptual idea; I am currently still reflecting on such problems and welcome interested readers to join the discussion.
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