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In Paul.A Clement's (1959) paper:

A Class of Triple-Diagonal Matrices for Test Purposes

SIAM Review, Vol. 1, No. 1 (Jan., 1959), pp. 50-52

He makes the claim that the eigenvalues of :

$ \begin{pmatrix} 0 & y_{1} & 0 & ... & 0 \\\ x_{1} & 0 & y_{2} & & ... \\\ 0 & x_{2} & 0 & ... & 0 \\\ ... & & ... & & y_{n-1} \\\ 0 & ... & 0 & x_{n-1} & 0 \end{pmatrix} $

are $\pm (n), \pm (n-2), ..., (\pm1 \; or \; 0)$ for $x_{k} = k$ and $y_{k} = n-k+1$.

Specifically, and I quote, "then a theorem of Sylvester establishes that the eigenvalues of this An+, are the numbers".

I can't for the life of me figure out what theorem and/or how it follows from them. I am familiar with Sylvester's formula for matrices in terms of their eigenvalues, but to get Frobenius covariants of a matrix A one needs to know the eigenvalues to start with.

Am I overlooking something trivial here?

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    $\begingroup$ This website is for questions of research interest. Your question might get a better reception at math.stackexchange.com $\endgroup$
    – Gerry Myerson
    Commented May 29, 2013 at 5:44
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    $\begingroup$ This is a legit question and I don't see why it will get a better response on M.SE. $\endgroup$
    – darij grinberg
    Commented May 29, 2013 at 14:18
  • $\begingroup$ @darij, it certainly got a quicker response at m.se; I won't be drawn on whether it got a better response there. $\endgroup$
    – Gerry Myerson
    Commented May 29, 2013 at 23:49
  • $\begingroup$ (of course, since all I put in my answer there were links to related papers, it can't really be said to be "better"...) $\endgroup$
    – J. M. isn't a mathematician
    Commented May 30, 2013 at 19:04

1 Answer 1

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I think the claim in this form is wrong. The eigenvalues are not integral. For example, with $n=4$ the matrix is $$ A=\begin{pmatrix} 0 & 4 & 0 & 0\cr 1 & 0 & 3 & 0\cr 0 & 2 & 0 & 2 \cr 0 & 0 & 3 & 0 \end{pmatrix}. $$

The characteristic polynomial of this matrix is $\chi (t)=t^4-16 t^2 +24$, which has no integral roots. Am I overlooking something ?

Edit: just visited the site http://math.nist.gov/MatrixMarket/deli/Clement/ Here I saw that the upper diagonal must be $3,2,1$, not $4,3,2$. Then everything is OK. So $y_k=n-k$ rather than $y_k=n-k+1$, what you wrote. We have a recursion for $\chi(t)=\det(t\cdot id-A)$, see Eigenvalues of Symmetric Tridiagonal Matrices.

Edit: A proof for the correct claim can be found in the paper of Taussky and Todd, "Another look at a matrix of Mark Kac", Linear Algebra Appl. 150 (1991), 341-360. More details are also at https://math.stackexchange.com/questions/405670/regarding-a-paper-by-paul-a-clement-on-tridiagonal-matrices, which was remarked by Gerry (thank you ). Also, Darij's remark is correct, that the proof does not follow easily, so I have edited this here.

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  • $\begingroup$ Why easily? Where is the symmetry? $\endgroup$
    – darij grinberg
    Commented May 29, 2013 at 20:46
  • $\begingroup$ Yes, OP posted to m.se, I posted a $3\times3$ counterexample, then the correction to the problem was noted, and a link was given to a previous m.se question about the same matrix. $\endgroup$
    – Gerry Myerson
    Commented May 29, 2013 at 23:47
  • $\begingroup$ @Darij: you are right, the proof is not evident. My answer was (is) a comment about the mistake of the matrix, not more. @Gerry: I have found your link and the answers, thank you. The correction to the problem got lost, as we see. $\endgroup$
    – Dietrich Burde
    Commented May 30, 2013 at 11:58
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    $\begingroup$ As you have linked to the paper by Taussky and Todd, you might want to mention www-math.mit.edu/~edelman/homepage/papers/kac.pdf as well... $\endgroup$
    – J. M. isn't a mathematician
    Commented May 30, 2013 at 19:03

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