A Test Matrix for an Inverse Eigenvalue Problem

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2 Main Result

Let A(n) be a real, symmetric, tridiagonal, n n matrix with entries × a = n 1 i =1, 2,...,n ii − 1 a = i(2n i 1) i =1, 2,...,n 2 i,i+1 2 − − − pn(n 1) an−1,n = − r 2 and let Ao(n) be the principal submatrix of A(n), that is, the (n 1) (n 1) matrix obtained from A(n) by deleting the last row and column. − × −

Theorem 2.1. A(n) has eigenvalues 0, 2,..., 2n 2 and Ao(n) has eigenvalues 1, 3,..., 2n 3 . { − } { − } Proof: By induction. When n =2 1 1 A(2) = 1 1 has eigenvalues 0, 2, and Ao(2) has eigenvalue 1. Assume the result holds for n. So A(n) has eigenvalues 0, 2,..., 2n 2 . Let B = Ao(n + 1) nI and A = A(n) (n 1)I. Then B and A { − } − − −

2 are similar via BR = RA where R is upper triangular, with entries

k(j−1)!(2n−j−1)! (i−1)!(2n−i+1)! i, j have same parity and j i, rij = ≥ ( 0q otherwise, and 2 j = n, k = 6 1 j = n. Therefore Ao(n + 1) has eigenvalues 1, 3,..., 2n 1 . Now we show that A(n+1) has eigenvalues{ 2n− }eigenvalues of A(n) . Let C = A(n+1) 2nI. Factorize C = LLT , where L is lower bidiagonal.{ }∪{ We ﬁnd: } − − l = 2n−i+1 ; l = i , i =1, 2,...,n 1, ii 2 i+1,i − 2 − l =q n+1 ; l = q√n; l =0. nn 2 n+1,n − n+1,n+1 q Therefore C has eigenvalue 0 and thus A(n + 1) has eigenvalue 2n. Deﬁne D =2nI LT L, so − Do O D = O 2n with d = 2n−1 ; d = 1 i(2n i), i =1, 2,...,n 1, and d = n−1 . ii 2 i+1,i 2 − − nn 2 Now Do has the same eigenvalues aspA(n) since they are similar matrices via SDo = A(n)S where S is upper triangular with entries

s = √2n i; s = √i i =1, 2,...,n 1, ii − i,i+1 − − snn = √2n; sij = 0 otherwise.

Therefore A(n + 1) has eigenvalues 2n eigenvalues of A(n) . { }∪{ }

3 Discussion

A real, symmetric n n tridiagonal matrix B is called a Jacobian matrix when its oﬀ-diagonal elements are non-zero× ([6], p. 46). We write

a b 0 0 0 1 − 1 ··· b1 a2 b2 0 0  − − ···  .. . 0 b2 a3 b3 . . B =  − . −. .  (3.1)  ......   0 0 0   . . .   . . .. b − a − b −   − n 2 n 1 − n 1   0 0 ... 0 b − a   − n 1 n    The similarity transformation, Bˆ = S−1BS, where S = S−1 is the alternating sign matrix, S = diag(1, 1, 1, 1,..., ( 1)n−1), produces a Jacobian matrix Bˆ with entries the same as B except for the− sign of− the oﬀ-diagonal− elements, which are all reversed. If instead we use the self-inverse

3 n−m sign matrix, S(m) = diag(1, 1,..., 1, 1, 1,..., 1), to transform B, then Bˆ is a Jacobian matrix − − − m z }| { identical to B except for a switched sign on the mth off-digonal element. In regards to the spectrum of the matrix, there is therefore| {z no} loss of generality in accepting the convention that a Jacobian matrix be expressed with negative off-diagonal elements, that is, b > 0, i =1,...,n 1 in (3.1). i ∀ − While Cauchy’s interlace theorem [13] guarantees that the eigenvalues of any square, real, symmetric (or even Hermitian) matrix will interlace those of its leading (or trailing) principal submatrix, the interlacing cannot be strict, in general [5]. However, specializing to the case of Jacobian matrices restricts the interlacing to strict inequalities. That is, Jacobian matrices possess distinct eigenvalues, and the eigenvalues of the leading (or trailing) principal submatrix are also distinct, and strictly interlace those of the original matrix (see [6], Theorems 3.1.3 and 3.1.4. See also [9] exercise P8.4.1, p. 475: when a tridiagonal matrix has algebraically multiple eigenvalues, the matrix fails to be Jacobian). The inverse problem is also well-posed: there is a unique (up to the signs of the off-diagonal elements) Jacobian matrix B having given spectra specified as per (1.1) (see o n−1 [6], Theorem 4.2.1, noting that the interlaced spectrum of n 1 eigenvalues (λ )1 can be used to calculate the last components of each of the n ortho-normalized− eigenvectors of B via equation 4.3.31). Therefore, the matrix A(n) in theorem 2.1 is the unique Jacobian matrix with eigenvalues equally spaced by two, starting with smallest eigenvalue zero, whose leading principal submatrix has eigenvalues also equally spaced by two, starting with smallest eigenvalue one. As a consequence of the theorem, we now have

Corollary 3.1. The eigenvalues of the real, symmetric n n tridiagonal matrix × a c n−1 0 0 0 − 2 ···  c n−1 qa c 2n−3 0 0  − 2 − 2 ··· − − . .  q 2n 3 q 3n 6 .. .   0 c 2 a c 2  W =  − −  (3.2) n  q .. q.. ..   0 0 . . . 0     . . .. (n−2)(n+1) n(n−1)   . . . c a c   − 4 − 2   n(n−1)   0 0 ... q 0 c qa   − 2   q  form the arithmetic sequence, λ(W )= a +2c(i 1) n , (3.3) n { o − }i=1 o while the eigenvalues of its leading principal submatrix, Wn , form the uniformally interlaced sequence

λ(W o)= a + c +2c(i 1) n−1, where a = a + c(n 1). (3.4) n { o − }i=1 o −

The form and properties of Wn were first hypothesised by the third author while programming Fortran algorithms to reconstruct band matrices from spectral data [17]. Initial attempts to prove the spectral properties of Wn by both he and his graduate supervisor (the first author) failed. Later, the first author produced the short induction argument of Theorem (2.1), in July, 1996. Alas, the fax on which the argument was communicated to the third author was lost in a cross- border academic move, and so the matter languished until recently. In summer of 2013, the second and third authors, looking for a tractable problem for a summer undergraduate research project, assigned the problem of this paper: “hypothesize, and then verify, if possible, the explicit entries

4 of an n n symmetric, tridiagonal matrix with eigenvalues (3.3), such that the eigenvalues of its principal× submatirx are (3.4).” Meanwhile a building renovation in summer of 2013 required the third author to clean out his oﬃce, including boxes of old papers and notes. During this process the misplaced fax from the ﬁrst author was found. However, to our delight, the student, Mr. Alexander De Serre-Rothney, was also able to independently complete both parts of the problem. His proof is now found in [15]. Though longer than the one presented here, his proof utilizes the spectral properties of another tridiagonal (non-symmetric) matrix, the so-called Kac-Sylvester matrix, Kn, of size (n + 1) (n + 1) [2, 4, 14, 16]: × n n 1 00 0 1 − n n 2 0 ··· 0  − ···. .  0 2 n n 3 .. . Kn =  . .− .  , (3.5)  ......   0 0 0   . . .   . . .. n 2 n 1   −   0 0 ... 0 n 1 n   −    which has eigenvalues λ(K )= 2k n n . n { − }k=0 4 A Spring-Mass Model problem

k1 k2 k3 k4 kn (a)

m1 m2 m3 mn

k1 k2 k3 k4 kn (b)

m1 m2 m3

Figure 4.1: Spring-mass system: (a) right hand end free, (b) right hand end ﬁxed

Let

k1 + k2 k2 m1 k k −+ k k m  − 2 2 3 − 3   2  C = and M = ··· ·  k − k − + k k   m −   − n 1 n 1 n − n  n 1   k k   m   − n n   n     Then the natural frequencies of the systems in Figure 4.1 are solutions of (C λM)x = 0 − and (Co λoM o)xo = 0, where Co is obtained from C by deleting the last row and column. The − 0 o solutions can be ordered 0 < λ1 < λ1 < λ2 < < λn−1 < λn−1 < λn. We can also rewrite the ··· − − systems as (B λI)u = 0 and (Bo λoI)uo = 0 where B = M 1/2CM 1/2 and u = M 1/2x. Note the natural frequencies− of the systems− are the eigenvalues of B and Bo.

5 We will consider the system with natural frequencies 1, 3,..., 2n 1 for system (a), and 2, 4,..., 2n 2 for system (b). We can use the matrix A{ (n) to help− solve} this system for the {spring stiﬀnesses− } and the masses. Let B(n) be the symmetric tridiagonal matrix:

Bii = ai = n i =1,...,n 1 B = b = i(2n i 1) i =1,...n 2 i,i+1 − i −2 − − − p n(n 1) Bn−1,n = bn−1 = − − −r 2 n−1 o n−1 Note that B(n)= A(n)+ I and has eigenvalues 2k +1 k=0, while B (n) has eigenvalues 2k k=1. 1/2 1/2 T { } n T { } Let u = m , , mn with m > 0 for all i. Let m = m = u u. We wish to solve h 1 ··· i i i=1 i − B(n)u = m 1/2k , 0,..., 0 PT (4.1) h 1 i i n for (mi)i=1, and k1. The bottom, nth, equation is

1/2 1/2 1/2 nmn n mn−1 = − = √2 α, b − n 1 − n 1 − 1/2 1/2 where we choose mn = α. We will thus be able to express mi in terms of the scaling parameter α. The (n 1)th equation is − n(n 1) n 1/2 − n√2 1/2 1/2 αbn−1 nmn 2 − n 1 n(n + 1) m − = = αr r = α√2 . n 2 − 1 − bn−2 (n 2)(n + 1) (n 1)(n 2) − − 2 − − − The ith equation, for i =1, n 1, n, isp 6 − 1/2 1/2 1/2 b − m − + nm b m =0. − i 1 m i i − i i+1 Then

1/2 1/2 1/2 1/2 2nmi (i(2n i 1)) mi+1 m − = − − − (4.2) i 1 ((i 1)(2n i))1/2 − − Now suppose

1/2 1/2 n(n + 1) (n + i 1) m − = α√2 ··· − (4.3) n i (n 1)(n 2) (n i) − − ··· − for i = 1, 2,...,j. Then cases i = 1, 2 are already veriﬁed, and the strong inductive assumption

6 applied in (4.2) with i 1= n (j + 1) implies i = n j. So − − − 1/2 √ n(n+1)···(n+j−1) 1/2 1/2 2nα 2 (n−1)(n−2)···(n−j) ((n j)(n + j 1)) mn−j+1 m − − = − − − n j 1 ((n j 1)(n + j))1/2 − − − 1/2 1/2 n+j 1 1/2 n(n + 1) (n + j 1) 2n n−j ((n j)(n + j 1)) = α√2 ··· − − − − (n 1)(n 2) (n j)  ((n j 1)(n + j))1/2  − − ··· − − −   n(n + 1) (n + j 1) 1/2  2n (n j)  = α√2 ··· − − − (n 1)(n 2) (n j) ((n j 1)(n + j))1/2 − − ··· − − − n(n + 1) (n + j 1)(n + j) 1/2 = α√2 ··· − (n 1)(n 2) (n j)(n j 1 − − ··· − − − 1/2 which veriﬁes, by strong induction, the closed form for mn−i given by (4.3). Finally, the ﬁrst equation of (4.1) is

− nm1/2 b m1/2 = m 1/2k 1 − 1 2 1 1 and so k = nm b (m m )1/2. 1 1 − 1 1 2 We note that the values mn−i can be written as

2 (n + i 1)!(n i 1)! m − =2α − − − (4.4) n i ((n 1)!)2 − (n +0 1)!(n 0 1)! for i =1,...,n 1, and m = α2 − − − = α2. − n ((n 1)!)2 Since C = M 1/2B(n)M 1/2, then −

i!(2n i 1)! k = C = m1/2 B m1/2 = α2 − − (4.5) i+1 − i,i+1 − n−(n−i) i,i+1 n−(n−i−1) ((n 1)!)2 − (2n 1)! and k = α2 − . 1 ((n 1)!)2 − 5 Conclusion

A family of n n symmetric tri-diagonal matrices, Wn, whose eigenvalues are simple and uniformally spaced, and whose× leading principle submatrix has uniformly interlaced, simple eigenvalues, has been presented (3.2). Members of the family are characterized by a specified smallest eigenvalue, ao and gap size, c between eigenvalues. The matrices are termed Jacobian, since the off-diagonal entries are all non-zero. The matrix entries are explicit functions of the size n, ao and c, so the matrices can be used as a test matrices for eigenproblems, both forward and inverse. The matrix Wn for specified smallest eigenvalue ao and gap c is unique up to the signs of the off-diagonal elements. In Section 4, the form of W was used as an explicit solution of a spring-mass vibration model § n (Figure 4.1), and the inverse problem to determine the lumped masses and spring stiffnesses was solved explicitely. Both the lumped masses, mn−i given by equation (4.4), and spring stiffnesses,

7 kn−i from equation (4.5) show super-exponential growth. Consequently mn/m1,kn/k1 become van- ishingly small as n . As a result, the spring-mass system of Figure 4.1 cannot be used as a discretized model for→ a ∞ physical beam in longitudinal vibration, as the model becomes unrealistic in the limit as n . →∞ References

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