0
$\begingroup$

The eigenvalue is for the adjacency matrix. I think there must be some clever inequality chain using the product of the matrix $A^2$ by an eigenvector, but i couldn't get it to work.

$\endgroup$
2
$\begingroup$

Hint: Let $v$ be a vertex of degree $\Delta$, and $\chi$ be the vector in $\mathbb{R}^{V(G)}$ such that $\chi(v) = \sqrt{\Delta}$; $\chi(u) = 1$ for all $u \in N_G(v)$; and $\chi(w) = 0$ for every other vertex $w$. Then from a fact in linear algebra, the largest eigenvalue $\lambda$ satisfies the following:

$$\lambda \ge \frac{\chi^{T}{\bf{A}}\chi}{\chi^T\chi}.$$

So what is left for you is to evaluate $\chi^T{\bf{A}}\chi$ and $\chi^T\chi$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.