Derivation: Wilk’s $\Lambda$ as $\prod \frac{1}{1+\lambda_i}$.

Show that Wilk’s $\Lambda$ can be represented as $\Lambda = \frac{|E|}{|E+H|} = \prod_{1=1}^s\frac{1}{1+\lambda_i}$, where $\lambda_i$ is the eigenvalues of $E^{-1}H$.

Proof:

$\frac{|E|}{|E+H|} = \frac{|E|/|E|}{|E+H|/|E|} = \frac{1}{|E^{-1}E+E^{-1}H|} = \frac{1}{|I+E^{-1}H|}$

Now suppose $\mu$ as an eigenvalue of $|I+E^{-1}H|$, so that:
$|I+E^{-1}H – \mu I|=0$
$\implies |E^{-1}H – (\mu-1) I|=0$

This implies
$\mu – 1 = \lambda \implies \mu = \lambda + 1$.

Thus
$\Lambda = \prod \frac{1}{\mu_i} = \prod \frac{1}{1+\lambda_i}$

QED.