Background: I have been studying a proof offered by Driscoll (1999) that shows that if $x'Ax \sim \chi^2(rank(A), \frac12 \mu'A\mu)$ then $AV$ is idempotent of rank $r$. Within this proof (necessity), he points out that (equation 6b on page 274):
$\sum_{i=1}^P (\lambda_i^2 + nv_i^2)\lambda_i^{n-2} = d + n\gamma,\qquad n = 2,3,...$
In this case, $\lambda_i$ represents the eigenvalues $i=1...p$ of $AV$. Considering this equation only for even values of $n$, Driscoll then argues if $|\lambda_i| > 1$, for any $i$, then the left side grows geometrically in $n$ whereas the right side grows arithmetically. He therefore concludes that $-1 \leqslant \lambda_i \leqslant 1$ for all $i$.
Question: Following this argument, can we really conclude that $|\lambda_i|$ must be equal or smaller than $1$? I would argue that if we compare the nature of growth on both sides then we can only conclude that $|\lambda_i| =1 $.
Of course, if $|\lambda_i| > 1$ the left side grows exponentially. On the other hand, however, if $|\lambda_i| < 1$ then the left side should converge, shouldn't it? And in this case, growth on both sides is again 'not aligned' for $n$ increasing. In fact, it is only aligned (i.e., arithmetic) if $|\lambda_i| = 1$. Is there anything I misunderstand?
