On the Determinants of the Square-Type Stirling Matrix and Bell Matrix
We study determinants of the square-type Stirling matrix and the square-type Bell matrix . For this purpose, we prove that and have LU factorizations and where the diagonal entries of are , while those of are ().
The Stirling numbers () of second kind count the number of ways to partition an element set into subsets. The Stirling matrix satisfies a recurrence rule . The sum of the row of is called the Bell number , so . A triangular matrix having Bell numbers on both border and holding is called the Bell matrix . Since every entries in the first row and column of are zeros except , we denote by the Stirling matrix deleted in the first row and column from .
Letbe a square-type Stirling matrix and a square-type Bell matrix. In the work, we study the determinants of and by finding their LU factorizations. In fact, we prove that has an LU factorization , where and has diagonal entries (Theorem 2), and has an LU factorization where along with the Pascal matrix and has diagonal entries (Theorem 12). This consideration is motivated by the square-type Pascal matrixwhere its LU factorization is  so . Note that was discussed in  by means of Hankel transformation. Our feature in the work is to study by recurrence rules of Bell numbers over an LU factorization of . For our notations, with a matrix , denotes the size , and let and be the row and column of . Write and simply by and with the copies of . Therefore, means a row matrix having ’s followed by a row matrix , and similarly, means a column matrix . Let be a diagonal matrix having diagonal entries .
2. Square-Type Stirling Matrix
For , let and (resp., and ) be the row and column of (resp., ). Since and are lower triangular matrices, and can be considered as of size , while and are of size . But, if necessary, like the case of multiplication , we may regard filled with infinitely many zeros after the first entries.
Lemma 1. Let , , and for . Let and be the row and column of .(1)In , and .(2)In , . Thus, and for .(3)In , and . And, .
Proof. The recurrence in (1) is well known . The column isThe inverse is the signed Stirling matrix of first kind  satisfying the recurrence . From , we have for , since is composed of all 1s. Moreover, simple computation of shows () equals , respectively. Hence, if we assume for some , then .
Comparingwith , it is easy to see and . Now, for the column , we have
Theorem 1. for all .
Proof. Note that and . Hence, we have from . When , by Lemma 1 (3), we haveThus, by assuming for , we have .
Theorem 2. has an LU factorization , where is an upper triangular matrix having diagonal entries (). Therefore, .
Proof. Let . Then, Lemma 1 (2) shows for all , since all entries in are 1. And,by Theorem 1. Thus, by assuming , we havewhich shows is an upper triangular matrix. Now, for , we have
3. Bell Matrix with the Pascal Matrix
Let and be the row and column of the Pascal matrix (). Well-known recurrence rules of and are
Theorem 3. For any , and .
Proof. Due to the binomial identity for , we haveClearly, . Assume for some . Note that is the set of coefficients of and equals which is the set of coefficients of expanded in descending order. Thus, (12) with implies
A matrix is called a flipped matrix of if it is horizontally flipped sideways of . Hence,
Theorem 4. Let be the row of for . Then,(1) and (2)
Proof. Clearly, and . And, for , we haveThus, it follows immediately that
We now develop some interrelations of the Bell matrix , square-type Bell matrix , and Pascal matrix .
Theorem 5. Let be the column of for . Then,
Proof. So if we assumefor some , thenby recurrence (12). Comparing with , we haveThus, with , (1) impliesMoreover, (1) gives rise to (3) such that
Proof. Clearly, by Theorem 5 (2). And, observe () from such that and .
From , () satisfy , , and .
Now, for some , we assume for all . Then, equalsBut, sinceby Theorem 5, we have
4. LU Factorization of the Square-Type Bell Matrix
We are ready to have an LU factorization of with diagonal entries.
Theorem 7. () where the lower triangular matrix and the upper triangular matrix has diagonal entries .
Proof. Let () be withThen, yields . And, is a lower triangular matrix; denote it by .
From , clearly and by (28). LetThen, the identity in Theorem 5 (2) implies , so because . Similarly, , , sois a lower triangular matrix and denote it by .
From , (28), and (29), we have , , and . Now, letSince and , we have , , , and . Therefore,Denote it by .
From(28), (29), and (31), give , , and . Now, letand let for .
Since , , and , we have . Also, and imply , , and . Hence,Denote it by .
Note for . From (34), we let