Linear Algebra of Pascal Matrices

LINEAR ALGEBRA OF PASCAL MATRICES

LINDSAY YATES

Abstract. The famous Pascal’s triangle appears in many areas of mathematics, such as number theory, combinatorics and algebra. Pascal matrices are derived from this triangle of binomial coeﬃ- cients, which create simplistic matrices with interesting properties. We explore properties of these matrices and the inverse of the Pas- cal matrix plus the identity matrix times any positive integer. We further consider a unique matrix called the Stirling matrix, which can be factorized in terms of the Pascal matrix.

1. Introduction The ancient arithmetic triangle, today known as Pascal’s triangle, is an inﬁnite numerical table represented in triangular form. The num- n bers displayed in the triangle are called binomial coeﬃcients, k , which represent the number of ways of picking k unordered outcomes from n possibilities. Each entry in the triangle is obtained by adding together two entries from the row above: the one directly to the left and the one directly to the right; this pattern can be seen in the image below.

The Pascal’s triangle has been known for over ten centuries. The set of numbers that form the Pascal’s triangle were known before Blaise

Date: December 5, 2014. 1 LINEAR ALGEBRA OF PASCAL MATRICES 2

Pascal, although he is attributed with being the first one to publish the information known about the triangle in his treatise, Traitédu triangle arithmétique. The numbers originally arose from Indian studies of combinatorics and the Greeks interest in figurate numbers. These numbers were continually discussed by Islamic mathematicians during the 10th century and in the 11th century by a Persian poet named Omar Khayyam. They were also seen in China during the 13th century. The Pascal’s triangle was officially published in Pascal’s treatise soon after his death in 1665.

This triangle arises in many areas of mathematics such as algebra, probability, and combinatorics. We were motivated by the Pascal’s triangle prominence in the ﬁeld of mathematics and its many applications, in particular Pascal matrices. We wanted to further our studies to consider the various properties and unique connections that Pascal matrices has to other functions and number sequences.

2. Pascal Matrices The Pascal’s triangle can be transcribed into a matrix containing the binomial coeﬃcients as its elements. We can form three types of matrices: symmetric, lower triangular, and upper triangular, for any integer n > 0.

The symmetric Pascal matrix of order n is deﬁned by Sn = (sij), where

i + j − 2 s = for i, j = 1, 2, ...., n (1) ij j − 1

We can deﬁne the lower triangular Pascal matrix of order n by Ln = (lij), where

(i−1 j−1 if i ≥ j lij = (2) 0 otherwise

The upper triangular Pascal matrix of order n is deﬁned by Un = (uij), where

( j−1 if j ≥ i u = i−1 (3) ij 0 otherwise LINEAR ALGEBRA OF PASCAL MATRICES 3

T We notice that Un = (Ln) , for any positive integer n.

For example, for n = 5 we have:

 1 1 1 1 1   1 0 0 0 0   1 1 1 1 1   1 2 3 4 5   1 1 0 0 0   0 1 2 3 4        S5 =  1 3 6 10 15  L5 =  1 2 1 0 0  U5 =  0 0 1 3 6         1 4 10 20 35   1 3 3 1 0   0 0 0 1 4  1 5 15 35 70 1 4 6 4 1 0 0 0 0 1

These Pascal matrices have some interesting properties, which we present next.

Theorem 2.1. [1] Let Sn be the symmetric Pascal matrix of order n defined by (1), Ln be the lower triangular Pascal matrix of order n defined by (2), and Un be the upper triangular Pascal matrix of order n defined by (3), then Sn = LnUn.

Proof. Let Ln be the lower triangular Pascal matrix of order n deﬁned by (2) and Un be the upper triangular Pascal matrix of order n deﬁned by (3). By direct multiplication of matrices Ln and Un we obtain the ij-th element of the product LnUn: n n X X T likukj = liklkj, since Un = (Ln) . k=1 k=1 n n j X Xi − 1j − 1 Xi − 1j − 1 Then, l l = = = ik jk k − 1 k − 1 k − 1 k − 1 k=1 k=1 k=1 j Xi − 1j − 1 = , since l = 0 for k>j. k − 1 j − k ik k=1 The Vandermonde identity says that:

n X m n m + n = , for any m,n,t ∈ (4) t n − t n N t=0 Let m = i − 1, n = j − 1, and t = k − 1 in (4). j Xi − 1j − 1 i + j − 2 Then, = = s , the entries of the sym- k − 1 j − k j − 1 ij k=1 metric Pascal matrix Sn. Hence, Sn = LnUn. LINEAR ALGEBRA OF PASCAL MATRICES 4

This result can be used to determine the determinant of the symmetric Pascal matrix, Sn.

Corollary 2.2. If Sn is the symmetric Pascal matrix of order n deﬁned by (1), then det(Sn) = 1, for any positive integer n.

Proof. Let Sn be the symmetric Pascal matrix of order n defined by (1). By Theorem 2.1, we know that Sn = LnUn, where Ln is the lower triangular Pascal matrix of order n defined by (2) and Un is the upper triangular Pascal matrix of order n defined by (3). Since Ln and Un are triangular matrices, then det(Ln) = 1 and det(Un) = 1. It follows that det(Sn) = det(LnUn) = det(Ln)det(Un) = 1. Definition 2.3. [5] Let A and B be n × n matrices. We say that A is similar to B if there is an invertible n × n matrix P such that P −1AP = B.

Theorem 2.4. [1] Let Sn be the symmetric Pascal matrix of order n −1 deﬁned by (1), then Sn is similar to its inverse Sn .

This result shows the following property of the eigenvalues of Sn.

Corollary 2.5. [1] Let Sn be the symmetric Pascal matrix of order n deﬁned by (1). Then the eigenvalues of Sn are pairs of reciprocal numbers.

Proof. Let Sn be the symmetric Pascal matrix of order n deﬁned by (1) and λ be an eigenvalue of Sn. Since the det(Sn) = 1, we know Sn −1 −1 is invertible. It follows that λ 6= 0 and, λ is an eigenvalue of Sn . −1 −1 Since Sn and Sn are similar by Theorem 2.4, then Sn and Sn have −1 the same eigenvalues. Hence, λ and λ are eigenvalues of Sn, and the eigenvalues of Sn are pairs of reciprocal numbers. Remark 1. If n is odd, since the eigenvalues must come in pairs, one of the eigenvalues must be equal to 1.

Example 2.6.√ The eigenvalues√ of the symmetric Pascal matrix, S2, 3 + 5 3 − 5 are λ = and λ = , where λ λ = 1 gives a reciprocal 1 2 2 2 1 2 pair.

Example 2.7. For n odd, let n = 3. Then√ the eigenvalues√ of the symmetric Pascal matrix, S3, are λ1 = 4+ 15, λ2 = 4− 15, and λ3 = 1. We note that λ1λ2 = 1 gives a reciprocal pair and λ3 = 1 is a self- reciprocal. LINEAR ALGEBRA OF PASCAL MATRICES 5

In their paper, A Note on Pascal’s Matrix, Cheon, Kim, and Yoon found an interesting factorization of the lower triangular Pascal matrix, Ln. T In−k O Theorem 2.8. [4] Let Gk = OSk be a matrix of order n, where Sk is the matrix of order k deﬁned by: ( 1 if i ≥ j s = ij 0 j>i for every k = 1, 2, ..., n. Then the lower triangular Pascal matrix of order n can be written as: Ln = GnGn−1 ··· G1.

For example,  1 0 0 0 0   1 0 0 0 0   1 0 0 0 0   1 0 0 0 0   1 1 0 0 0   0 1 0 0 0   0 1 0 0 0   0 1 0 0 0          L4 =  1 1 1 0 0   0 1 1 0 0   0 0 1 0 0   0 0 1 0 0           1 1 1 1 0   0 1 1 1 0   0 0 1 1 0   0 0 0 1 0  1 1 1 1 1 0 1 1 1 1 0 0 1 1 1 0 0 0 1 1

1 0 0 0 0 1 1 0 0 0   = 1 2 1 0 0   1 3 3 1 0 1 4 6 4 1 To further our studies of the lower triangular Pascal matrix, we are interested in studying the inverse of this matrix.

Theorem 2.9. [6] Let Ln be the lower triangular Pascal matrix of order n deﬁned by (2), then

−1 i−j Ln = ((−1) lij). −1 Proof. We will show that Ln · Ln = In. −1 By direct multiplication of Ln and Ln we get the ij-th element of the product:

n X k−j (−1) liklkj. (5) k=1 LINEAR ALGEBRA OF PASCAL MATRICES 6

If i < j then the element (5) is zero and if i = j, then the element (5) is 1. We will show that for i > j, the element (5) is zero. If i > j, the element is: i − 1j − 1 i − 1 j i − 1i − 1 − + ... + ( − 1)i − j = j − 1 j − 1 j j − 1 i − 1 j − 1 i − 1 i − j i − j i − j = − + ... + ( − 1)i − j = 0. j − 1 i − j i − j − 1 0 −1 −1 i−j Hence, Ln · Ln = In and Ln = ((−1) lij) is the inverse of Ln. There is another unique way in which the inverse of the lower triangular Pascal matrix can be written, using the Hadamard product of matrices.

Deﬁnition 2.10. [8] Let A, B be m × n matrices. The Hadamard product of A and B is deﬁned by:

[A ◦ B]ij = [A]ij[B]ij, for 1 ≤ i ≤ m, 1 ≤ j ≤ n.

Theorem 2.11. [8] Let τn be a n × n lower triangular matrix deﬁned below as: ( (−1)i−j if i ≥ j ≥ 1, τ = ij 0 otherwise The inverse of the lower triangular matrix can be found using the Hadamard product:

−1 Ln = Ln ◦ τn For example, if n = 4, then:

 1 0 0 0   1 0 0 0   1 0 0 0  −1  −1 1 0 0   1 1 0 0   −1 1 0 0  L =   =   ◦   4  1 −2 1 0   1 2 1 0   1 −1 1 0  −1 3 −3 1 1 3 3 1 −1 1 −1 1

3. Inverse of the Pascal Matrix Plus An Integer

In this section we are going to describe the inverse of Ln +kIn where Ln is the lower triangular matrix of order n deﬁned by (2), In is the identity matrix and k is a positive integer. We call Ln + kIn the Pascal matrix plus an integer. First, we are considering the case for k = 1. By direct computation of the inverse of Ln+In, we can observe that there is LINEAR ALGEBRA OF PASCAL MATRICES 7 a close relation between the inverse of Ln +In and the Pascal matrix Ln.

For example, for n = 4,

2 0 0 0 1 2 0 0 L4 + I4 =  , 1 2 2 0 1 3 3 2

 1     1  2 0 0 0 1 0 0 0 2 0 0 0 −1 1 −1 1 −1  4 2 0 0   1 1 0 0   4 2 0 0  (L4 + I4) =  −1 1  =   ◦  −1 1   0 2 2 0   1 2 1 0   0 2 2 0  1 −3 1 1 −3 1 8 0 4 2 1 3 3 1 8 0 4 2 In their paper, Explicit Inverse of the Pascal Matrix Plus One, S.L. Yang and Z.K. Liu showed that the inverse of Ln + In is the Hadamard product between Ln and a lower triangular matrix. We are going to describe this unique lower triangular matrix next. For this we need to deﬁne the Euler polynomials.

Euler polynomials, En(x), can be deﬁned by the following generating function:

∞ 2etx X tn = E (x) et + 1 n n! n=0 The ﬁrst few Euler polynomials are:

E0(x) = 1 1 E1(x) = x − 2 2 E2(x) = x − x 3 3 2 1 E3(x) = x − 2 x + 4

−1 Theorem 3.1. [8] For n ≥ 1, the n×n inverse matrix Qn = (Pn+In) is given by:

( 1 i−1 2 j−1 Ei−j(0) if i ≥ j ≥ 1, qij = 0 if i < j

Proof. From the deﬁnition of Euler polynomials, En, we have: LINEAR ALGEBRA OF PASCAL MATRICES 8

∞ 2etx X tn = E (x) . et + 1 n n! n=0 ∞ X tn Then, (E (x + 1) + E (x)) n n n! n=0 ∞ X tn ∞ tn = E (x + 1) + P E (x) n n! n n! n=0 n=0 2et(x+1) 2etx = + et + 1 et + 1 ∞ X tn = 2etx = 2xn . n! n=0 tn Comparing the coeﬃcients of in the above equation, we obtain: n!

n En(x + 1) + En(x) = 2x , for n ≥ 0. (6) In [8] it was proved that for all n ≥ 0:

n X n E (x + y) = E (x)yn−k (7) n k k k=0 n X n E (x + 1) = E (x) (8) n k k k=0 From (6) and (8), we get:

n 1 X n 1 E (x) + E (x) = xn, for n ≥ 0. (9) 2 k k 2 n k=0 Setting x = 0 in (7), we obtain:

n X n E (y) = E (0)yk (10) n k n−k k=0 In other terms:

n X n E (x) = E (0)xk, for n ≥ 0 (11) n k n−k k=0 LINEAR ALGEBRA OF PASCAL MATRICES 9

Let E(x) and X(x) be column matrices deﬁned by: T n−1 T E(x) = [E0(x),E1(x), ..., En−1(x)] and X(x) = [1, x, ..., x ] . ¯ Let En be the lower triangular n × n matrix deﬁned by:

(i−1 j−1 Ei−j(0) if i ≥ j ≥ 1, eij = 0 otherwise

From (9) and (11), we get:

1 2 (Pn + In)E(x) = X(x), ¯ E(x) =EnX(x). Then we obtain,

−1 1 ¯ (Pn + In) = 2 En.

A connection between Euler polynomials and Bernoulli numbers can be found. Bernoulli numbers, Bk, can be deﬁned by:

∞ n t P t t = Bn e − 1 n=0 n! Euler polynomials can be expressed in terms of Bernoulli numbers by the following expression:

n+1 1 X E (x) = (2 − 2k+1)n+1B xn+1−k n n + 1 k k k=1 This connection between Euler polynomials and Bernoulli numbers gives another expression of the inverse of Ln + In.

−1 Theorem 3.2. [8] For n ≥ 1, the n×n inverse matrix Qn = (Pn+In) is given by:

 (1 − 2i−j+1)B i−1 i−j+1 j−1 if i ≥ j ≥ 1, qij = i − j + 1 0 if i < j LINEAR ALGEBRA OF PASCAL MATRICES 10

We also note that the inverse of the Pascal matrix plus one can be written in terms of a Hadamard product.

Theorem 3.3. [8] Let ∆n be a matrix of order n deﬁned as: ( 1 E (0) if i ≥ j ≥ 1, ∆ = 2 i−j ij 0 if i

 1 − 2i−j+1B i−1 i−j+1 j−1 if i ≥ j ≥ 1, ∆ij = i − j + 1 0 if i

We observe that the inverses of the matrices Ln and Ln +In both can be written as a Hadamard product between Ln and a lower triangular matrix. We now want to extend our scope of study to consider the inverse of Ln +kIn, where k is a positive integer, to see if these inverses can also be expressed as a Hadamard product of Ln and a lower triangular matrix.

By direct computation of the inverse of (Ln +kIn), for k = 2 and k = 3, we obtain for n = 4:

 1     1  3 0 0 0 1 0 0 0 3 0 0 0 −1 1 −1 1 −1  9 3 0 0   1 1 0 0   9 3 0 0  (L4 + 2I4) =  −1 −2 1  =   ◦  −1 −1 1  .  27 9 3 0   1 2 1 0   27 9 3 0  1 −3 −3 1 1 −1 −1 1 27 27 9 3 1 3 3 1 27 27 9 3

 1     1  4 0 0 0 1 0 0 0 4 0 0 0 −1 1 −1 1 −1  16 4 0 0   1 1 0 0   16 4 0 0  (L4 + 3I4) =  −1 −2 1  =   ◦  −1 −1 1  .  32 16 4 0   1 2 1 0   32 16 4 0  1 −3 −3 1 1 −1 −1 1 128 32 16 4 1 3 3 1 128 32 16 4 We note from these examples that the pattern is preserved. The inverse of Ln + kIn can be written as Ln ◦ ∆n. The question now is what are the elements of the matrix ∆n. In their paper, Inverting the Pascal Matrix Plus One, Aggarwala and Lamoureux found that these numbers are values of polylogarthmic functions. LINEAR ALGEBRA OF PASCAL MATRICES 11

Deﬁnition 3.4. The polylogarithm function, denoted Lim(λ), is the function deﬁned as:

∞ X λk λ2 λ3 Li (λ) = = λ + + + ... m km 2m 3m k=1 where m is an integer and λ is a complex parameter.

For integer values of the polylogarithm order, we have:

Li1(λ) = −ln(1 − λ) λ Li (λ) = 0 1 − λ λ Li (λ) = −1 (1 − λ)2 λ(1 + λ) Li (λ) = −2 (1 − λ)3 The polylogarithmic functions satisfy the following recurrence relation:

∂Li Li = λ m m−1 ∂λ

Using the result found by Aggarwala and Lamoureux in [2], we can describe the inverse of the Pascal matrix plus an integer in terms of the Hadamard product.

Theorem 3.5. Let Ln be the lower triangular matrix of order n deﬁned by (2), and ∆n be the n × n lower triangular matrix deﬁned by:

 1 (−1)i−j+1 (Li (−k)) if i>j  k j−i  1 ∆ij = if i = j 1 + k 0 if i

4. Stirling Matrix via Pascal Matrix Very similar to the Pascal matrix, the Stirling matrix can be deﬁned using the Stirling numbers of the second kind. In this section we describe LINEAR ALGEBRA OF PASCAL MATRICES 12 a unique connection between Stirling matrices and Pascal matrices.

Deﬁnition 4.1. [4] The Stirling numbers of the second kind denoted S(n, k) for integers n and k, where 1 ≤ k ≤ n, count the ways to divide a set of n objects into k nonempty subsets. Example 4.2. Let X = {a, b, c, d}, then the partitions for each k = 1, 2, 3, 4: k = 1 : X; k = 2 : [{a}, {b, c, d}], [{b}, {a, c, d}], [{c}, {a, b, d}], [{d}, {a, b, c}], [{a, b}, {c, d, }], [{a, c}, {b, d}], [{a, d}, {b, c}]; k = 3 : [{a}, {b}, {c, d}], [{a}, {c}, {b, d}], [{a}, {d}, {b, c}], [{c}, {d}, {a, b}], [{b}, {d}, {a, c}], [{b}, {c}, {a, d}]; k = 4 : [{a}, {b}, {c}, {d}].

Thus,

1 if k = 1  7 if k = 2 S(4, k) = 6 if k = 3  1 if k = 4 We can take these Stirling numbers of the second kind and transcribe them into a matrix, just as the binomial coeﬃcients of the Pascal’s triangle can be written as a Pascal matrix.

Deﬁnition 4.3. Let Sn be the Stirling matrix of order n, deﬁned by: ( S(i, j) if i ≥ j s = ij 0 otherwise For example, when n = 4:

1 0 0 0 1 1 0 0 S4 =  , 1 3 1 0 1 7 6 1 We note that the elements of the Stirling matrix satisfy the following recurrence relation:

sij = si−1,j−1 + jsi−1,j LINEAR ALGEBRA OF PASCAL MATRICES 13

The Stirling matrix can be factorized via the Pascal matrix as the following:

T In−k 0 Theorem 4.4. [4] Let Ak = 0 Pk be a matrix of order n, where Pk is the lower triangular Pascal matrix of order k deﬁned by (2). Then the Stirling matrix of order n can be written as: Sn = AnAn−1 ··· A1.

Remark 2. When k = n in Theorem 4.4, then An is the lower triangular Pascal matrix. Example 4.5. The Stirling matrix of order 5 is:

1 0 0 0 0 1 1 0 0 0   S5 = 1 3 1 0 0 =   1 7 6 1 0 1 15 25 10 1

1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0         = 1 2 1 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0         1 3 3 1 0 0 1 2 1 0 0 0 1 1 0 0 0 0 1 0 1 4 6 4 1 0 1 3 3 1 0 0 1 2 1 0 0 0 1 1

5. Conclusion We see from our studies that the Pascal’s triangle is very prominent in many areas of mathematics. The ability to transcribe the binomial coeﬃcients into matrices that have many unique properties and connections is what has so many mathematicians interested and excited to study Pascal matrices.

References [1] A. Edelman and G. Strang, Pascal Matrices, The Mathematical Association of America, Monthly 111 (2004) 189-197. [2] R. Aggarwala and M. Lamoureux, Inverting the Pascal Matrix Plus One, The Mathematical Association of America, Monthly 109, No. 4 (April 2002) 371- 377. LINEAR ALGEBRA OF PASCAL MATRICES 14

[3] G. Call and D. Velleman, Pascal’s Matrices, The Mathematical Association of America, Monthly 100 (1993), 372-376. [4] G. Cheon, J. Kim and H.Yoon, A Note on Pascal’s Matrix, Mathematics Sub- ject Classiﬁcation (1999) 121. [5] Lay, David C. (2012). Linear Algebra and Its Applications: 4th Ed. Boston: Pearson, Addison-Wesley. [6] Lawden, G.H. Pascal matrices, The Mathematical Gazette, 325-327. [7] R. Brawer and M. Pirovino, The Linear Algebra of the Pascal Matrix, Elsevier Science Publlishing Co., Inc (1992), 13-23. [8] S. Yang and Z. Liu, Explicit Inverse of the Pascal Matrix Plus One, Interna- tional Journal of Mathematics and Mathematical Sciences (2006) 1-7.

Lindsay Yates, Mathematics Department, Georgia College, Milledgeville, GA 31061, U.S.A. E-mail address: [email protected]