%! TEX root = GT.tex % vim: tw=50 % 25/11/2023 09AM \begin{flashcard}[eigenvalue-facts-prop] \begin{proposition}[Eigenvalue facts] \label{eval_facts} Let $G$ be a \gls{graph}. Then: \begin{enumerate}[(i)] \item \cloze{$\lambda$ is an \gls{eval} $\implies$ $|\lambda| \le \maxdeg$.} \item \cloze{For $G$ \gls{connected}: $\maxdeg$ is an \gls{eval} $\iff$ $G$ \gls{regular}.} \item \cloze{For $G$ \gls{connected}: $-\maxdeg$ is an \gls{eval} $\iff$ $G$ \gls{regular} and \gls{bipartite}.} \item \cloze{$\lambda_{\max} \ge \mindeg$.} \end{enumerate} \end{proposition} \begin{proof} \phantom{} \begin{enumerate}[(i)] \item \cloze{Let $\lambda$ be an \gls{eval} with \gls{evec} $x$. Choose $i$ with $|x_i|$ maximal -- without loss of generality $x_i = 1$. Then $(Ax)_i = \sum_{j \in \nbd(i)} x_j$, so \[ |(Ax)_i| \le \sum_{j \in \nbd(i)} |x_j|\ le \maxdeg \cdot 1 .\] Then $|\lambda| \le \maxdeg$.} \item \cloze{\begin{enumerate}[$\Leftarrow$] \item[$\Leftarrow$] Let $x = (1, \ldots, 1)$ -- then $Ax = (\maxdeg, \maxdeg, \ldots, \maxdeg)$. \item[$\Rightarrow$] Choose $i$ with $|x_i|$ maximal. Without loss of generality $x_i = 1$. Then as we have equality in the inequality of (i) above, we must have $\vdeg(i) = \maxdeg$ and $x_j = 1$ for all $j \in \nbd(i)$. We can repeat at each $j \in \nbd(i)$ (as $x_j = 1$) to obtain $\vdeg(j) = \maxdeg$ and $x_{j' = 1}$ for all $j' \in \nbd(j)$. Continue: we obtain $\vdeg(j) = 1$ for all $j$ (as $G$ \gls{connected}). \end{enumerate}} \item \cloze{\begin{enumerate}[$\Leftarrow$] \item[$\Leftarrow$] Let $x = 1$ on $X$, $-1$ on $Y$ ($X, Y$ our bipartition). Then $Ax = (-\maxdeg) x$. \item[$\Rightarrow$] Choose $i$ with $|x_i|$ maximal. Without loss of generality $x_i = 1$. Then since we have equality in the inequality of (i), we must have $\vdeg(i)= \maxdeg$ and $x_j = -1$ for all $j \in \nbd(i)$. So can repeat at each $j \in \nbd(i)$ to obtain $\vdeg(j) = \maxdeg$ and $x_{j'} = 1$ for all $j' \in \nbd(j)$. Continue: we get that $x_j = \pm 1$ for all $j$, and $jj' \in E$ implies $x_j = 1$, $x_{j'} = -1$ or vice versa (as $G$ \gls{connected}). So $G$ has no odd cycle, hence is bipartite (and we've already seen that it is \kreg{\maxdeg}). \end{enumerate}} \item \cloze{By \eqref{lambda_minmax_eq}, enough to find $x$, $\|x\| = 1$, with $\langle Ax, x \rangle \ge \delta$. Let $x = (1, 1, \ldots, 1)$. Then $(Ax)_i \ge \delta$ for all $i$. So $\langle Ax, x \rangle \ge \delta n$, with $\langle x, x \rangle = n$.\qedhere} \end{enumerate} \end{proof} \end{flashcard} \begin{remark*} Proof of (ii) actually gives that if $G$ is \gls{connected} then \gls{eval} $k$ has multiplicity $1$. \end{remark*} \vspace{-1em} \glsref[eval]{Eigenvalues} can relate to other graph parameters. For example, we know $\chromnum(G) \le \maxdeg + 1$. Can strengthen this to: \begin{flashcard}[chromnum-le-lambdamax-plus-1-prop] \begin{proposition} For any \gls{graph} $G$, have $\chromnum(G) \le \lambda_{\max} + 1$. \end{proposition} \begin{proof} \cloze{Without loss of generality $|G| \ge 2$ ($|G| = 1$ trivial). Choose $v \in \sq[n] = \vset(G)$ with $\vdeg(v) = \mindeg$, and let $G' = G - v$. \textbf{Claim:} $\lambda_{\max}(G') \le \lambda_{\max}$. Once we prove this we're done, because we can colour $G'$ in $\lambda_{\max} + 1$ colours using induction, and now $\vdeg(v) = \mindeg \le \lambda_{\max}$ (by \cref{eval_facts}(iv)), so can colour $v$ as well. \textit{Proof of claim:} Adjacency matrix of $G'$, $B$ say, is formed from $A$ by deleting $v$-th row and column: say $n$-th row and column. By \eqref{lambda_minmax_eq}, enough to show that \[ \max_{\substack{x \in \RR^{n - 1} \\ \|x\| = 1}} \langle Bx, x \rangle \le \max_{\substack{x \in \RR^n \\ \|x\| = 1}} \langle Ax \rangle x .\] But if $x \in \RR^{n - 1}$, say $x = (x_1, \ldots, x_{n - 1})$, then $y = (x_i, \ldots, x_{n - 1}, 0) \in \RR^n$ with $\|y\| = \|x\|$ and $\langle B x, x \rangle = \langle A y, y\rangle$.} \end{proof} \end{flashcard} \subsection{Towards Moore Graphs} \begin{flashcard}[strongly-regular-defn] \begin{definition*}[Strongly regular] \glsadjdefn{streg}{strongly regular}{\gls{graph}} \glssymboldefn{streg}{$(k, a, b)$}{$(k, a, b)$} \cloze{Say a \gls{graph} $G$ is \emph{strongly regular} with \emph{parameters} $(k, a, b)$ if $G$ is \kreg{k}, with any two \gls{adj_v} \glspl{vertex} having $a$ common \glspl{neighbour} and any two non-\gls{adj_v} \glspl{vertex} having $b$ common \glspl{neighbours}.} \end{definition*} \end{flashcard} \vspace{-1em} ``One step up from being \gls{regular}''. \begin{example*} \phantom{} \begin{enumerate} \item $C_4$: \begin{tsqx} ! size(0.5cm); A .= dir 45 B .= dir 135 C .= dir 225 D .= dir 315 A--B--C--D--A \end{tsqx} has $\strp(2, 0, -2)$. \item $C_5$: \begin{tsqx} ! size(0.7cm); A .= dir 0 B .= dir 72 C .= dir 144 D .= dir 216 E .= dir 288 A--B--C--D--E--A \end{tsqx} has $\strp(2, 0, 1)$. \item And in general, if $G$ is a \gls{moore_graph} of degree $k$, then $G$ is \gls{streg}, with parameters $\strp(k, 0, 1)$. \item Triangle $\times$ triangle: \begin{center} \includegraphics[width=0.4\linewidth]{images/b1a9ed33e1de4f35.png} \end{center} has $\strp(4, 1, 2)$. \end{enumerate} \end{example*} \vspace{-1em} Seems that `\gls{streg}' is a bit more than \gls{regular}, but\ldots