1
$\begingroup$

Suppose you have an irreducible, stochastic matrix $A$ with left Perron-Frobenius eigenvector $v$ (corresponding to the eigenvalue $1$), and suppose the next largest eigenvalue for $A$ is $\lambda$. Let $V = \text{diag}(v)$ and define $B = V^{-1} A' V$. Then $B$ and $A$ have the same eigenvalues, so let $E = \ker(\lambda I - B)$. Let $Q = V^{-1} P V$ where $P$ is the projection onto $E$. I'm curious about the terms $$ R_1(x) %= \frac{((\sqrt{V} AQB \sqrt{V}^{-1}) \sqrt{V} x)' (\sqrt{V} \overline{x})} % {(\sqrt{V}x)'(\sqrt{V}\overline{x})} = \frac{(AQB x)' V \overline{x}} {x'V\overline{x}} $$ and $$ R_2(x) = \frac{(A(I-Q)B x)' V \overline{x}} {x'V\overline{x}}; $$ these aren't quite Rayleigh quotients, but my intution is that they should behave something like them. In particular, I expect an upper bound for the first term to be something like $|\lambda|^2$. For the second term, if we constrain $x$ to be orthogonal to $[1,1,\ldots,1]$ I expect it to be something like the square of the modulus of the next eigenvalue down after $\lambda$. Is this the case? Also, what happens if we replace $A$ by $A^n$?

Thanks!

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.