# The Largest Entry in the Inverse of a Vandermonde Matrix

We investigate the size of the largest entry (in absolute value) in the inverse of certain Vandermonde matrices. More precisely, for every real b > 1, let M_b(n) be the maximum of the absolute values of the entries of the inverse of the n × n matrix [b^i j]_0 ≤ i, j < n. We prove that lim_n → +∞ M_b(n) exists, and we provide some formulas for it.

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09/02/2021

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## 1 Introduction

Let be a list of real numbers. The classical Vandermonde matrix is defined as follows:

 V(a):=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1a0a20⋯an−101a1a21⋯an−11⋮⋮⋮⋱⋮1an−1a2n−1⋯an−1n−1⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦.

As is well-known, the Vandermonde matrix is invertible if and only if the are pairwise distinct. See, for example, [3].

In what follows, is a positive integer and is a fixed real number. Let us define the entries by

 [ci,j,n]0≤i,j

and let , the maximum of the absolute values of the entries of . The size of the entries of inverses of Vandermonde matrices have been studied for a long time (e.g., [1]). Recently, in a paper by the first two authors and Daniel Kane [2]

, we needed to estimate

, and we proved that . In fact, even more is true: the limit exists and equals . In this paper, we generalize this result, replacing with any real number greater than .

Our main results are as follows:

###### Theorem 1.

Let and . Then for . Hence .

Let and . Then .

###### Theorem 3.

For all real the limit exists.

## 2 Preliminaries

For every real number , and for all integers , let us define the power sum

 σi,j,n(x):=∑0≤h1<⋯

The following lemma will be useful in later arguments.

###### Lemma 4.

Let be integers with , , and let be a positive real number.

1. [(a)]

2. If , then .

3. If , then .

###### Proof.

We have

 σi,j+1,n(x)−σi,j,n(x)=∑(h1,…,hi)∈Si,j,nxh1+⋯+hi−∑(h1,…,hi)∈Ti,j,nxh1+⋯+hi,

where

 Si,j,n:={0≤h1<⋯

and

 Ti,j,n:={0≤h1<⋯

Now there is a bijection given by

 (h1,…,hi)↦(h1,…,hi0−1,hi0+1,hi0+1,…,hi),

where is the unique integer such that . Hence, it follows easily that for , and for . ∎

Recall the following formula for the entries of the inverse of a Vandermonde matrix (see, e.g., [4, §1.2.3, Exercise 40]).

###### Lemma 5.

Let be pairwise distinct real numbers. If then

 cn−1,jXn−1+cn−2,jXn−2+⋯+c0,jX0=∏0≤i

For define

 πj,n:=∏0≤h

We now obtain a relationship between the entries of and and .

###### Lemma 6.

Let . Then

 |ci,j,n|=σn−i−1,j,nπj,n (1)

for .

###### Proof.

By Lemma 5, we have

 ∏0≤h

which in turn, by Vieta’s formulas, gives

 cn−i−1,j,n=(−1)i⎛⎜ ⎜ ⎜⎝∏0≤h

for . The result now follows by the definitions of and . ∎

Next, we obtain some inequalities for .

###### Lemma 7.

Define . Then

 πj,n≤πj+1,nfor n0≤j
###### Proof.

For , we have

 πj+1,n:=∏0≤h

A quick computation shows that the inequality

 bn+j−1−bn−2bn−1−bj≥1

is equivalent to

 bj≥bn−1+bn−2bn−1+1.

Let be the minimum positive integer such that . Then . Hence, for , we have

 bj≥1+1b>bn−1+bn−2bn−1+1,

so that

 πj,n≤πj+1,nfor n0≤j

Finally, we have the easy

For we have .

###### Proof.

is a symmetric matrix, so its inverse is also. ∎

## 3 Proof of Theorem 1

###### Proof.

Suppose . Then

 |ci,j,n| =σn−i−1,j,nπj,n(by (???)) ≤σn−i−1,n0,nπj,n(by Lemma~% {}??? (a)) ≤σn−i−1,n0,nπn0,n(by % Lemma~{}???) =|ci,n0,n|(by (???)),

and so we get

 |ci,j,n|≤|ci,n0,n|. (4)

But

 ci,n0,n=cn0,i,n (5)

by Lemma 8. Make the substitutions for and for in (4) to get

 |cn0,i,n|≤|cn0,n0,n|. (6)

The result now follows by combining Eqs. (4), (5), and (6). ∎

## 4 Proof of Theorem 2

###### Proof.

Since , it follows that . Hence in Theorem 1 we can take , and this gives . However, by explicit calculation, we have

 σn−1,1,n =bn(n−1)/2−1 σn−2,1,n =bn(n−1)/2−1+∑(n−1)(n−2)/2−1≤i≤n(n−1)/2−3bi,

so that

 σn−1,1,n≤σn−2,1,n. (7)

Hence

 |c1,0,n| =|c0,1,n|(by Lemma~{}???) =σn−1,1,nπ1,n(by (???)) ≤σn−2,1,nπ1,n(by (???)) =|c1,1,n|(by (???)),

and the result follows. ∎

## 5 Proof of Theorem 3

###### Proof.

We have

 |ci,j,n| =σn−i−1,j,nπj,n =σn−i−1,j,n(b)∏0≤h

where the equality

 σn−i−1,j,n(b)bn(n−1)/2−j=σi,j,n(b−1)

arises from the one-to-one correspondence between the subsets of of cardinality and those of cardinality .

For define

 σi,j,∞(x)=∑0≤h1<⋯

Hence the limits

 ℓi,j :=limn→+∞|ci,j,n| =limn→+∞σi,j,n(b−1)1∏0≤h

exist and are finite.

From Theorem 1 we see that

 limn→+∞Mb(n)=max0≤i≤j

and the proof is complete. ∎

From this theorem we can explicitly compute for .

###### Corollary 9.

Let be the real zero of the polynomial .

1. [(a)]

2. If , then .

3. If , then .

###### Proof.

From Theorem 2 we know that for we have . Now an easy calculation based on (8) shows that

 ℓ0,0 =∏t≥1(1−b−t)−1 ℓ1,1 =b2−b+1b(b−1)2 ∏t≥1(1−b−t)−1.

By solving the equation , we see that for we have , while if we have . This proves both parts of the claim. ∎

###### Remark 10.

The quantity converges rather slowly to its limit when is close to . The following table gives some numerical estimates for .

## 6 Final remarks

We close with a conjecture we have been unable to prove.

###### Conjecture 11.

Let and . Then, for all sufficiently large , we have for some , .

## References

• [1] W. Gautschi. On inverses of Vandermonde and confluent Vandermonde matrix. Numer. Mathematik 4 (1962), 117–123.
• [2] D. M. Kane, C. Sanna, and J. Shallit. Waring’s theorem for binary powers. Combinatorica 39 (2019), 1335–1350.
• [3] A. Klinger. The Vandermonde matrix. Amer. Math. Monthly 74 (1967), 571–574.
• [4] D. E. Knuth. The Art of Computer Programming, Vol. 1, Fundamental Algorithms. Addison-Wesley, third edition, 1997.