SOLUTION TECHNIQUES FOR TRUST REGION PROBLEMS

easy case: no eigenvector of $B$ is orthogonal to $\eta$; $B-\lambda I$ is
invertible for all $\lambda \in \Lambda$ and 
  $x^{\lambda} = (B-\lambda I)^{-1} \eta.$

implicit secular equation:
\[ 1 - \eta^t (B-\lambda I)^{-2} \eta = 0  \]

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denoting the spectrum of $B$ by
$\gamma_1 \geq \gamma_2 \geq \ldots \geq \gamma_{n-1}$, let $P$ be an
orthogonal matrix such that
\[
P^{t}BP = D = \rm{diag}(\gamma_1, \gamma_2, \ldots ,\gamma_{n-1}),
\]
the diagonal matrix with diagonal elements $\gamma_1,\gamma_2,\ldots,
\gamma_{n-1}$.
Then the Lagrange stationarity condition becomes
\[
(D-\lambda I)\hat{x} = \hat{\eta},
\]
where
\[ \hat{x} = P^{t}x,~~~\hat{\eta} = P^{t} \eta.\]
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The set of Lagrange multipliers $\Lambda$ is not changed by this
transformation of the Lagrange equation, and for every $x \in R^{n-1}$ we
have
\[
\hat{\mu}(\hat{x}) := \hat{x}^tD\hat{x} -2\hat{\eta}^{t}\hat{x} = \mu(x).
\]
 
 
{\em explicit secular equation}
\[
f_{\mu}(\lambda) :=1 - \sum_{i=1}^{n-1}\left(\frac{\hat{\eta}_i}
{\gamma_i-\lambda}\right)^2=0.
\]