Three determinants evaluated in circular products

Three determinants for symmetric and skew-symmetric matrices are explicitly evaluated, in closed form, as circular products. One of them gives a solution to a problem proposed by Dzhumadil’daev [ Amer. Math. Monthly 129 (2022) 486].


Introduction and the main results
Evaluating determinants is one of the important topics in mathematics and physics. There are two classical results (cf. [2] and [3, §4.3]) about a skew-symmetric matrix Ξ. When the order of Ξ is odd, then det Ξ = 0. When the order of Ξ is even, the determinant det Ξ results in a square of a polynomial in the entries of Ξ. In this paper, we explicitly evaluate three determinants, in closed form, as circular products. The main results are announced in advance as follows, whose proofs are given in the next section. The first one is about the symmetric matrix U n = u i,j 1≤i,j≤n : u i,j =    x i − x j , i ≤ j; x j − x i , i > j. The next two results are concerned with the following skew-symmetric matrices A n = a i,j 1≤i,j≤n : where λ and {x k } 1≤k≤n are real numbers. Theorem 1.2 (Determinant identity for skew-symmetric matrices).
It is remarked here that Theorem 1.2 resolves a monthly problem proposed recently by Dzhumadil'daev [1].

Theorem 1.3 (Determinant identity for skew-symmetric matrices).
det This section is divided into three subsections, dedicating to proofs of the three corresponding theorems anticipated in the previous section.

Proof of Theorem 1.1
By examining the difference between the ith row and the (i + 1)th row, we see that the resulting entry u i,j in the (i, j) position equals Iterating this operation downwards from the first row to the penultimate row, and then extracting the common row factors, we get the following equality where the matrix V n is given by Next, for the matrix V n , we make the same row operations. Considering the difference between the ith row and the (i + 1)th row, where 1 ≤ i ≤ n − 2, we can check without difficulty that the resulting entry v i,j in the (i, j) position becomes Repeating this operation for i from 1 to n − 2, we derive another equality where the matrix W n is given by Write this matrix explicitly Expanding the determinant of W n with respect to the first and the last columns, we find that which confirms the circular product formula stated in Theorem 1.1

Proof of Theorem 1.2
Firstly, we reduce the matrix A 2n by row and column operations. If we subtract the (i + 1)th row from the ith row, then the resulting entry a i,j in the (i, j) position becomes Repeating this operation downwards for all the rows except for the last one and then pulling out the common factors from the first row to the (2n − 1)th row, we get the equality where the square matrix B 2n is given by Analogously, for the matrix B 2n , we make the corresponding column operations. By subtracting the (j + 1)th column from the jth column, the resulting entry b i,j in the (i, j) position becomes Iterating this operation rightwards for all the columns except for the last one and then extracting the common column factors from the first column to the (2n − 1)th column, we derive another equality where the matrix C 2n returns to skew-symmetric one: Finally, for each k with 1 ≤ k < 2n, performing simultaneously the operations on the matrix C 2n by subtracting the (k+1)th row and column from the kth row and column, respectively, we find the following reduced expression where D 2n is the double bordered skew-symmetric matrix Now, we write the matrix D 2n in blocks and y = (y 1 , y 2 , · · · , y 2n−1 ) : By means of the Laplace formula, expanding the determinant of D 2n along the last row and then the last column, we have the double sum expression det D 2n = 1≤i,j<2n where E 2n−1 (i, j) stands for the minor of E 2n−1 after the ith row and the jth column having been removed. In general, we have the following remarkable formula According to this formula, we can rewrite, under the replacements i → 2ı − 1 and j → 2 − 1, the former double sum for det D 2n as where the last step is justified by applying twice the telescoping method. Summing up, we have proved that Induction principle Now, we return to present an inductive proof for . It is routine to verify that the formula is true for E 3 . Suppose that the same formula is valid for E 2n−1 . Then we have to validate it for E 2n+1 . In order to facilitate the intuitive reasoning below, we write explicitly the corresponding matrix Now, we can prove the formula in for E 2n+1 case by case as follows: • i, j < 2n Expanding the determinant along the last row and then the last column, we see that E 2n+1 (i, j) = E 2n−1 (i, j).
• i = 2n When j = 2n + 1, we have directly E 2n+1 (2n, j) = 0 since all the entries in the last column result in zero. Instead, for j = 2n + 1, we have E 2n+1 (2n, 2n + 1) = 0 because we can reduce the related matrix, by simple row and column operations, to a lower triangular matrix containing zero diagonal entries.
In conclusion, we have verified that is true for all the E 2n+1 with n > 1.

Proof of Theorem 1.3
For the matrix Ω 2n , by examining the difference of the ith row minus (i−1)th row and then iterating this operation upwards for all the rows except for the first one, we get the resulting matrix Similarly, by making the column operations leftwards, we transform the matrix Ω 2n into another one Now, extracting the common factor x λ 1 from both the first row and the first column, and then x k − x k−1 from the kth row and the kth column for k from 2 to 2n, we find the following determinant equality The above skew-symmetric matrix Φ 2n can be expressed in blocks as where the submatrix Ψ 2n−1 is defined by Now, by subtracting y 2 times the first row and the first column, respectively, from the second row and the second column, we can further reduce the matrix Φ 2n to the following skew-symmetric matrix: −y i , 2 < i < j ≤ 2n; where Ψ 2n−2 is the skew-symmetric matrix explicitly given by Expanding the determinant of Φ 2n along the second row and then the second column, we find the determinant equality det Φ 2n (y 2 , y 3 , · · · , y 2n−1 ) = det Φ 2n = det Ψ 2n−2 .