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Let $c$ be a complex number. Consider the family of $n\times n$ matrices $M_n$ which have $c$'s on one off-diagonal, $\bar{c}$'s on the other off-diagonal, and zero everywhere else. So $M_4$ looks like this:$$\left(\begin{array}{cccc}0&c&0&0\\\bar{c}&0&c&0\\0&\bar{c}&0&c\\0&0&\bar{c}&0\end{array}\right)$$How does one find a general formula for the operator norm of $M_n$?

Martin Argerami
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David Hillman
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1 Answers1

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Write $c=\alpha|c|$, with $|\alpha|=1$. Then $$ M_4=|c|\,\begin{bmatrix} 1&0&0&0\\ 0&\alpha&0&0\\ 0&0&\alpha^2&0\\ 0&0&0&\alpha^3\end{bmatrix} \begin{bmatrix} 0&1&0&0\\ 1&0&1&0\\ 0&1&0&1\\ 0&0&1&0 \end{bmatrix} \begin{bmatrix} 1&0&0&0\\ 0&\alpha&0&0\\ 0&0&\alpha^2&0\\ 0&0&0&\alpha^3\end{bmatrix}^*, $$ and so $M_4(c)$ is unitarily equivalent to $|c|\,M_4(1)$. Thus we only need to determine the spectrum of $M_4(1)$. The same argument applies for $M_n$ for any $n$.

The eigenvalues of $M_n(1)$ are $\lambda_k=2\cos\frac{k\pi}{n+1}$, where an eigenvector for $\lambda_k$ is $\sum_{s=1}^n\sin\frac{ks\pi}{n+1}e_s$, and $e_1,\ldots,e_n$ is the canonical basis.

As $M_n(1)$ is selfadjoint, its norm is the greatest eigenvalue in absolute value, so $$ \|M_n(c)\|=|c|\,\|M_n(1)\|=2|c|\,\cos\frac{\pi}{n+1}. $$

$\newcommand\abajo{\\ \ \\}$

Edit: The Eigenvalues of $M_n(1)$.

Let $e_1,\ldots,e_n$ be the canonical basis of $\mathbb C^n$. Let $\eta_k=\sum_{\ell=1}^n\sin\frac{k\ell\pi}{n+1}\,e_\ell$, $k=1,\ldots,n$. Then

\begin{align*} M_n(1)\eta_k &=\sum_{s=1}^{n-1}\sin\frac{sk\pi}{n+1}\,e_{s+1}+\sum_{s=2}^{n}\sin\frac{sk\pi}{n+1}\,e_{s-1} \abajo &=\sin\frac{2k\pi}{n+1}\,e_1+\sum_{s=2}^{n-1}\left(\sin\frac{(s-1)k\pi}{n+1}+\sin\frac{(s+1)k\pi}{n+1}\right)\,e_{s} +\sin\frac{(n-1)k\pi}{n+1}\,e_{n} \abajo &=2\cos\frac{k\pi}{n+1}\sin\frac{k\pi}{n+1}\,e_1+\sum_{s=2}^{n-1}2\cos\frac{k\pi}{n+1}\sin\frac{sk\pi}{n+1}\,e_s+2\cos\frac{k\pi}{n+1}\sin\frac{nk\pi}{n+1}\abajo &=2\cos\frac{k\pi}{n+1}\,\eta_k. \end{align*} All the above requires is the sine of a sum formula.

Martin Argerami
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  • Thanks!...but you have left for me the larger mystery of how one arrives at the solution for $M_n(1)$ in terms of trig functions. Any hints there? – David Hillman Oct 17 '18 at 14:39
  • It's fairly basic, just a matter of patience. I've included a few lines. – Martin Argerami Oct 17 '18 at 16:28
  • That proves it's correct, thanks. But how does one find those eigenvalues and eigenvectors to begin with? I'm trying to improve my skill set. When I try to solve this (not knowing the solution in advance), either by hand or using Mathematica, I get various roots of polynomials, with sin and cos nowhere to be found. Do you know of a general approach that leads to this solution? – David Hillman Oct 17 '18 at 22:14
  • I don't have a general answer for that. In this case, these numbers are fairly well known, but I don't think I would have come up with the eigenvector and the eigenvalue on my own. The numbers $\cos k\pi/(n+1)$ are for instance the roots of the Chebyshev polynomials of the second kind. The numbers $2\cos k\pi/(n+1)$ are the norms of matrices of zeroes and ones, provided that the norm is less than 2. – Martin Argerami Oct 18 '18 at 02:52
  • Ah: the characteristic polynomial of $M_n(1)$ is the $n$th Chebyshev polynomial of the second kind with $x$ replaced by $-x/2$! – David Hillman Oct 18 '18 at 12:55
  • ...and a way I could perhaps have figured this out on my own would have been to search for the polynomial coefficients in the encyclopedia of integer sequences. That thing you say about norms of matrices of zeroes and ones is wild. Thanks for your help! – David Hillman Oct 18 '18 at 13:30
  • You are welcome. The thing about norms is indeed wild. I'll write a proof here one of these days. – Martin Argerami Oct 18 '18 at 15:43