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.