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I came across the following argument in a lecture about algebraic methods in combinatorics:

Suppose we have $L$, the Laplacian of some graph, and $\mu_1 \leq \cdots \leq \mu_n$ are its eigenvalues. Then, for every $x \in {\Bbb R}^n$, $x \cdot \bar 1=0$, we have $$ \frac{x^tLx}{x^tx} \geq \mu_2 $$

I think that the bound was reasoned by Courant-Fisher and that actually $L$ could be any symmetric real matrix. A similiar bound, only in the other direction and regarding the 2nd largest eigenvalue can be found in this paper. Can you please explain how exactly this bound was derived from Courant-Fisher?

(I'm aware of the similarity to 2nd largest eigenvalue and rayleigh quotient, but this argument might be weaker).

Rodrigo de Azevedo
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Guy
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  • If I add that $\bar 1$ is an eigenvector of $L$ corresponding to $\mu_1$, will that make it easier? – Guy Jul 29 '13 at 20:43

1 Answers1

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Referencing Courant-Fisher was apparently misleading. The explaination is actually pretty basic:

Suppose $(\bar 1, x_2, \ldots, x_n)$ is an orthogonal basis of $\mathbb{R}^n$ using eigenvectors corresponding to $\mu_1,\ldots,\mu_n$ (here comes the assumption that $\bar 1$ is an eigenvector belonging to $\mu_1$). Then every $x\in\mathbb{R}^n$, $x\perp\bar 1$ can be represented as $\alpha_2x_2+\cdots+\alpha_nx_n$. Substituting in the Rayleigh quotient: $$ \frac{x^tLx}{x^tx} = \frac{(\alpha_2x_2+\cdots+\alpha_nx_n)^tL(\alpha_2x_2+\cdots+\alpha_nx_n)}{(\alpha_2x_2+\cdots+\alpha_nx_n)^t(\alpha_2x_2+\cdots+\alpha_nx_n)} = \frac{\mu_2\alpha_2\|x_2\|^2+\cdots\mu_n\alpha_n\|x_n\|^2}{\alpha_2\|x_2\|^2+\cdots\alpha_n\|x_n\|^2} \geq \frac{\mu_2 (\alpha_2\|x_2\|^2+\cdots\alpha_n\|x_n\|^2)}{\alpha_2\|x_2\|^2+\cdots\alpha_n\|x_n\|^2}=\mu_2 $$

Guy
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