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Before we begin, we remark that (unsurprisingly) diagonalizing the usual “Cesaró matrix” [2]

1 1 1  2 2  C = . . ,  .. .     1 ··· 1   n n  1 n turns out to be much easier. It can be easily verified that V CV − = Diag([1/i]i=1) for V given by (2.2); the anti-triangular form of (1.1) makes it harder to diagonalize explicitly.

2 Explicit Diagonalization

Let J denote the reverse , i.e., the matrix with ones on its anti-diagonal. Then JP is upper-triangular, so that eigenvalues of JP can be read off of the diagonal. These eigenvalues are λi(JP )=1/i. It turns out that the eigenvalues of P are also 1/i multiplied with alternating signs. Let us prove this observation.

i+1 Proposition 2.1. Let P be given by (1.1). Then, λi(P ) = (−1) /i for i =1,...,n.

1 1 Proof. It proves more convenient to analyze P − . An easy verification shows that P − is the anti- bidiagonal matrix 1 − n n 2 − n n − 1 1  . .  P − = .. .. . (2.1) −   1 2   1    1 To obtain eigenvalues of P − it suffices to find a similar to it. To that end, we use 1 the following matrix (curiously, this matrix is the strict lower-triangular part of JP − ):

0 −1 0  0 −2 0  L = . . .  .. ..    ... 1 − n 0   Setting V = exp(L), we obtain

(i j) i − 1 V = [Vij ]= (−1) − for i ≥ j. (2.2)  j − 1

1 Explicitly carrying out the multiplication, we obtain the triangular matrix (using V − = exp(−L))

1 ∗ ··· ∗ −2 ∗ ∗ 1 1 V P − V − =  .  , (2.3) .. ∗  − n+1n  ( 1)  1 i+1 where ∗ represents unspecified entries. From (2.3) it is clear that λi(P − ) = (−1) i.

2 Obtaining eigenvectors is harder and requires more work. A quick computation shows that P − 2 1 is tridiagonal but asymmetric, which rules out an easy solution. However, V P − V − turns out to

2 2 EXPLICIT DIAGONALIZATION be a highly structured bidiagonal matrix:

1 −2(n − 1) 4 −3(n − 2)  9 −4(n − 3)  2 1 .. .. B := V P − V − =  . .  .      2   (n − 1) −n(1)  n2    1 2 n Suppose now that there is a matrix S that diagonalizes B, that is SBS− = Λ with Λ = [i ]i=1 diagonal. Then,

1 2 1 2 1 1 2 n S− ΛS = V P − V − =⇒ SV P − V − S− =Λ= Diag([i ]i=1), (2.4)

2 which shows that SV diagonalizes P − , completing the answer.

Theorem 2.2. Let (x)k := x(x + 1) ··· (x + k − 1) be the Pochammer symbol, and let M be lower- triangular with nonzero entries in column j given by

(j + 1)k j (n − k + 1)k j mkj := − − , k ≥ j (1 ≤ j ≤ n). (2.5) (2j + 1)k j (k − j)! − Then, S = M T diagonalizes B. Proof. To find S we need to solve the system of equations:

SB =ΛS ⇐⇒ BT M = MΛ.

2 Consider the jth eigenvalue λj = j ; denote the corresponding column of M by m and its kth entry by mk. To obtain m we must solve the linear system

T 2 2 B m = j m, =⇒ m1 = j m1 2 2 −(k + 1)(n − k)mk + (k + 1) mk+1 = j mk+1, 1 ≤ k ≤ n. Since B is bidiagonal, a brief reflection shows that M is lower-triangular with 1s on its diagonal—the 1s come from the equation corresponding to the index k +1= j. The subsequent entries of m are nonzero. Symbolic computation with a few different values of j suggests the general solution (which can be formally proved using an easy but tedious induction):

(j + 1)k j (n − k + 1)k j mk = − − , k ≥ j. (2j + 1)k j (k − j)! −

Remark 2.3 (Added Oct 31, 2014). Similarly we can diagonalize the transition matrix

1 1 1 n n ··· n 1 1 1  n 1 n ··· n 1 0 − . . − . Z =  . .. .  . (2.6)    1 1   2 2   1    1 Clearly, from a diagonalization of P − we can recover a diagonalization of Z above. To see why, 1 T 1 T observe that P − = (J Z− J) where J is the “reverse identity” (anti-diagonal) matrix.

3 3 CESARÓ KERNELS

3 Cesaró Kernels

In this section we mention kernel matrices closely related to the Cesraró matrix discussed above.

3.1 Brownian bridge kernel

Interestingly, matrix P is closely related to the Brownian bridge kernel K = [kij ] = min(i, j) [3]. Proposition 3.1 makes this connection precise and highlights a well-known property of this kernel. T r Proposition 3.1. The kernel matrix K = P P is infinitely divisible (i.e., [kij ]  0 for r ≥ 0). r r Proof. We show that the Schur power K◦ = [kij ], for r > 0 is positive definite. This claim follows after we realize that 1 K = [kij ]= 1 ≤ i, j ≤ n, max(i, j) which is well-known to be infinitely divisible [1, Ch. 5]. We include a short proof below. Let 1 n D = Diag([i− ]i=1); also let M := [min(i, j)]. As K = DMD, it suffices to establish infinite divisibility of M. We prove a more general statement. Let f be a positive monotonic function, and for any set C ⊂ R, define 1C(x)=1 if x ∈ C and 0 otherwise. Then,

∞ 1 1 f(mij ) = min(f(i),f(j)) = [0,f(i)](x) [0,f(j)](x)dx, Z0 which is nothing but an inner-product; thus [f(mij )] is a and hence positive definite. Setting f(t)= tr,r> 0, infinite divisibility of M (and hence of K) follows.

3.2 Operator Norm bounds When using the kernels min(i, j) or 1/ max(i, j) in an application, one may need to bound their operator norms (e.g., for approximation or optimization). Let us first illustrate a bound on kKk that one may obtain naively. Since K = DMD, a naive bound is kKk≤kDk2kMk = kMk (as kDk =1). Proposition 3.2 bounds kMk. 2kπ 1 Proposition 3.2. Let M = [mij ] = [min(i, j)]; define θk := 2n+1 . Then, M = V ΛV − with 1 λk = (2+2cos θk)− , k =1, 2,...,n 2 1 (3.1) vjk = √2n+1 sin(k − 2 )θj , θj 6= π, j, k =1, 2, . . . , n. Consequently, kMk≈ 4n2/π2. Proof. The key is to observe that M has the Cholesky factorization M = LLT , where L is the all ones lower-triangular matrix 1 1 1 L = . .  . . ..  ···  1 1 1 We wish to explicitly diagonalize LLT , a task that becomes much easier if we consider the inverse 1 T 1 M − = L− L− , as this is a perturbed Toeplitz- 2 −1 −1 2 −1 T 1  . . .  L− L− = ...... (3.2)  − −   1 2 1  −1 1   

4 3.2 Operator Norm bounds 3 CESARÓ KERNELS

1 1 Applying the derivation of [8, Thm. 3.2-(ix)] , we find that the eigenvalues of (3.2) are λk− = 2kπ 1 2+2cos θk where θk = 2n+1 . In fact M − can also be diagonalized explicitly: its eigenvectors are given by (again by resorting to arguments of [8, Thm. 3.3]):

√2n+1 1 2 vjk := sin(k − 2 )θj , j, k =1, 2, . . . , n. (3.3)

1 T T T 1 1 With V = [vjk] and noting V − = V , we obtain V (LL )− V = Diag(λk− ). T 1 2 2 Hence, it immediately follows that kMk = λmax(LL )= (2+2cos θn)− ≈ 4n /π . The eigendecomposition of M derived above is no surprise; M = [min(i, j)] is essentially the Brownian bridge covariance function whose spectrum is well-studied [3]. But it is worth noting that our derivation uses elementary , compared with a more advanced Fourier-analytic derivation typically employed when studying eigenfunctions of kernels.

Remark 3.3. Using kKk = kDMDk≤kDk2kMk≤kMk ∼ 4n2/π2, we get a very pessimistic bound on kKk. Remark 3.4 analyzes kKk directly, yielding a much better dependence on n. Finally, Proposition 3.5 actually provides a dimension independent bound on kKk. Remark 3.4. The bound from Remark 3.3 can be greatly improved rather easily. Indeed,

T T T 1 kKk = kP P k = λmax(P P ) ≤ tr(P P )= = Hn, (3.4) i i X where Hn denotes the n-th Harmonic number. However, even bound (3.4) is suboptimal as it depends on the dimension of the matrix P .

1 2 Proposition 3.5. Let K = [1/ max(i, j)] for 1 ≤ i, j ≤ n. Then, kKk≤ 4, while kK− k≤ 4n .

T T Proof. Since K = P P , we analyze kKk = λmax(P P ), where P is defined by (1.1). T 1 T 2 Let Z = JP . Then, kKk = λmax(JP P J − ) = λmax(JP (JP ) ) = kZk . A quick calculation T n show that (I − Z)(I − Z ) = Diag[(n − j)/(n − j + 1)]j=1. Thus, n − 1 kI − Zk2 = k(I − Z)(I − ZT )k = . n

Thus, kI − Zk≤ (n − 1)/n, so by the triangle-inequality we obtain kZk≤ 1+ (n − 1)/n. But Z = JP and thep operator norm is unitarily invariant, hence kZk = kP k≤ 1+ p(n − 1)/n. Thus, kKk = kP k2 ≤ (1 + 1 − 1/n)2 ≤ 4. p 1 To compute kK−pk, consider

2(n − 1)n + 1 −(n − 1)2 −(n − 1)2 2(n − 2)(n − 1) + 1 −(n − 2)2  .  1 T .. W := P − P − = ,      −4 5 −1  −   1 1  which is a highly structured tridiagonal matrix. Thus, using Gershgorin’s theorem, we obtain

2 λ(W ) ≤ |λ(W ) − wii| + wii ≤ 2(n − i) + 2(n − i)(n − i +1)+1, 1 ≤ i ≤ n.

2 2 1 Thus, λmax(W ) ≤ 4n − 6n +3 ≤ 4n . Since kK− k = kW k, the proof is complete.

1There seems to be a typo in the cited theorem; the cases (viii) and (ix) stated in that paper seem to be switched.

5 REFERENCES REFERENCES

Acknowledgments I would like to thank David Speyer, whose comment to an initial answer of mine on MathOverflow prodded me to think more carefully about the problem, which ultimately led to this paper.

References

[1] R. Bhatia. Positive Definite Matrices. Princeton University Press, 2007. [2] M.-D. Choi. Tricks or treats with the . American Mathematical Monthly, pages 301–312, 1983. [3] T. Hofmann, B. Schölkopf, and A. J. Smola. A review of kernel methods in machine learning. Technical Report 156, Max Planck Institute for Biological Cybernetics, 2006. [4] O. Holtz. The inverse eigenvalue problem for symmetric anti-bidiagonal matrices. Linear algebra and its applications, 408:268–274, 2005. [5] P. Koev and F. Dopico. Accurate eigenvalues of certain sign regular matrices. Linear Algebra and its Applications, 424:435–447, 2007. [6] C. Mehl. Anti-triangular and anti-m-Hessenberg forms for Hermitian matrices and pencils. Linear Algebra and its Applications, 317(1):143–176, 2000. [7] H. Ochiai, M. Sasada, T. Shirai, and T. Tsuboi. Eigenvalue problem for some special class of anti-triangular matrices. arXiv:1403.6797, 2014. [8] W.-C. Yueh and S. S. Cheng. Explicit Eigenvalues and Inverses of Tridiagonal Toeplitz Matrices With Four Perturbed Corners. The ANZIAM Journal, 49(03):361, Aug. 2008.

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