%&latex \documentclass[11pt]{article} \parskip=0.3cm \parindent=0.5cm \renewcommand{\baselinestretch}{1.4} \pagestyle{plain} \setlength{\textwidth}{6.0in} \setlength{\textheight}{8.5in} \setlength{\topmargin}{0cm} \setlength{\oddsidemargin}{0cm} \setlength{\evensidemargin}{0cm} \def\j{\bold{j}} \def\a{\alpha} \def\la{\lambda} \def\bu{\mathbf{u}} \usepackage{amsmath} \begin{document} \begin{center} An Involutory Matrix of Eigenvectors \end{center} \begin{center} David Callan,\footnote{corresponding author} \\ Department of Statistics, University of Wisconsin-Madison, \\ 1210 W. Dayton Street, Madison, WI 53706-1693, U.S.A. \end{center} \begin{center} Helmut Prodinger, \\ The John Knopfmacher Centre for Applicable Analysis and Number Theory, \\ School of Mathematics, University of the Witwatersrand, \\ P.~O. Wits, 2050 Johannesburg, South Africa. \end{center} %\today A matrix with a full set of linearly independent eigenvectors is diagonalizable: if the $n$ by $n$ matrix $A$ has eigenvalues $\la_{j}$ with corresponding eigenvectors $u_{j}\ (1 \leq j \leq n$), if $U = (u_{1}|u_{2}|\ldots|u_{n})$ and $ D=\textrm{diag}(\la_{1},\la_{2},\ldots,\la_{n})$, then $U$ is a diagonalizing matrix for $A$: $U^{-1}AU=D$. Taking transposes shows that $(U^{-1})^{t}$ is a diagonalizing matrix for $A^{t}$. Hence $U^{t}$ itself is a diagonalizing matrix for $A^{t}$ if $U^{2}$ is the identity matrix, or more generally, due to the scalability of eigenvectors, if $U^{2}$ is a scalar matrix. The purpose of this note is to point out that the right-justified Pascal-triangle matrix $R = \left( \binom{i-1}{n-j} \right)_{1 \leq i,j \leq n}$ is an example of this phenomenon. Let $a$ denote the golden ratio $(1+\sqrt{5}\,)/2$. The eigenvalues of $R^{t}$ (which of course are the same as the eigenvalues of $R$) were found in \cite{cooper}: $\la_{i} = (-1)^{n-i}a^{2i-n-1},\ 1 \leq i \le n$. The corresponding eigenvectors $u_{i}$ of $R^{t}$ were also found in \cite{cooper} (here suitably scaled for our purposes): $u_{i} = (u_{ij})_{1 \leq j \leq n}$ where \[ u_{ij} = (-a)^{n-j}\sum_{k=1}^{j} (-1)^{i-k}\binom{i-1}{k-1}\binom{n-i}{j-k}a^{2k-i-1}. \] Let $U=(u_{ij})_{1 \le i,j \le n}$. For example, when $n=5$, \[ R = \left( \begin{matrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 2 & 1 \\ 0 & 1 & 3 & 3 & 1 \\ 1 & 4 & 6 & 4 & 1 \end{matrix} \right) \quad \mathrm{and} \quad U = \left( \begin{matrix} a^4 & -4\, a^3 & 6\, a^2 & -4\,a & 1 \\ - a^3 & 3\, a^2 - a^4 & -3\,a + 3\, a^3 & 1 - 3\, a^2 & a \\ a^2 & -2\,a + 2\, a^3 & 1 - 4\, a^2 + a^4 & 2\,a - 2\, a^3 & a^2 \\ -a & 1 - 3\, a^2 & 3\,a - 3\, a^3 & 3\, a^2 - a^4 & a^3 \\ 1 & 4\,a & 6\, a^2 & 4\, a^3 & a^4 \end{matrix}\right). \] Since the rows of $U$ are eigenvectors of $R^{t}$, $U^{t}$ is a diagonalizing matrix for $R^{t}$. By the first paragraph applied to $A=R^{t}$, $U$ will be a diagonalizing matrix for $R$ if we can show that $(U^{t})^{2}$ (equivalently $U^{2}$) is a scalar matrix. We now proceed to show that $U^{2}=(1+a^{2})^{n-1}I_{n}$ and in fact this holds for arbitrary $a$. We use the notation $[x^{k}]p(x)$ to denote the coefficient of $x^{k}$ in the polynomial $p(x)$. Consider the generating function $U_i(z)=z(z-a)^{n-i}(az+1)^{i-1}$. Using the binomial theorem to expand $U_{i}(z)$, it is immediate that \begin{equation*} U_i(z)=\sum_{j=1}^{n} u_{ij}z^j. \end{equation*} Now the $(i,k)$ entry of $U^{2}$ is \begin{align*} (U^2)_{ik}&=\sum_{j=1}^{n}[x^{k-1}](x-a)^{n- j}(ax+1)^{j- 1} \cdot[z^{j-1}](z-a)^{n-i}(az+1)^{i-1}\\ &=[x^{k-1}]\sum_{j=1}^{n}[z^{j-1}](x-a)^{n-j}(ax+1)^{j- 1} (z-a)^{n-i}(az+1)^{i-1}\\ &=[x^{k-1}](x-a)^{n-1}\sum_{j=1}^{n}[z^{j- 1}]\left(\frac{ax+1}{x-a}\right)^{j-1} (z-a)^{n-i}(az+1)^{i-1}\\ &=[x^{k-1}](x-a)^{n-1}\sum_{j=1}^{n}[z^{j-1}] \Big(z\frac{ax+1}{x-a}-a\Big)^{n-i}\Big(az\frac{ax+1}{x- a}+1\Big)^{i- 1}\\ &=[x^{k-1}](x-a)^{n-1} \Big(\frac{ax+1}{x-a}-a\Big)^{n-i}\Big(a\frac{ax+1}{x- a}+1\Big)^{i-1}\\ &=[x^{k-1}]({ax+1}-ax+a^2)^{n-i}(a^2x+a+x-a)^{i-1}\\ &=[x^{k-1}](1+a^2)^{n-i}x^{i-1}(1+a^2)^{i-1}\\ &=[x^{k-i}](1+a^2)^{n-1}\\ &=(1+a^2)^{n-1}\delta_{ki},\\ \end{align*} as desired. \bibliographystyle{amsplain} \begin{thebibliography}{1} \bibitem{cooper} R.~S.~Melham and Curtis Cooper. ``The Eigenvectors of a Certain Matrix of binomial Coefficients.'' \emph{The Fibonacci Quarterly} \textbf{38.2} (2000):123-26. \end{thebibliography} AMS Classification Numbers: 11B65, 15A36, 15A18 \end{document} Professor Helmut Prodinger Mathematics Department University of the Witwatersrand P.O. Wits 2050 Johannesburg South Africa Tel. +27-11-717 6224 Fax. +27-11-4032017 Email. helmut@gauss.cam.wits.ac.za Homepage. http://www.wits.ac.za/helmut/index.htm