Inversion of the Hessian

Using the preceding results, I can invert for the Hessian in equation (9), either with or without regularization.

• Without regularization

The fitting goal is

$${\bf 0} \approx {\bf Lm}-{\bf d},$$ (28)

with ${\bf L}=({\bf H} \;\; {\bf B})$ and ${\bf m'}=({\bf m_s} \;\; {\bf m_n})$.The matrix equation we want to solve is

$$\left( \begin{array} {cc} {\bf H'H} & {\bf H'B} \\ {\bf B... ... \begin{array} {c} {\bf H'd} \\ {\bf B'd}\end{array}\right),$$ (29)

where ${\bf m_s}$ and ${\bf m_n}$ are the unknowns. For ${\bf m_s}$, I use the bottom row of equation (25). For ${\bf m_n}$, I use the top row of equation (26). We have, then,

$$\begin{eqnarray} \hat{{\bf m_s}}&=&({\bf H'H}-{\bf H'B}({\bf B'B})^{-1}{\bf B'H... ...\bf H'H})^{-1} {\bf H'B})^{-1}{\bf B'H}({\bf H'H})^{-1}{\bf H'd},\end{eqnarray}$$ (30) (31)

which can be simplified as follows:

$$\begin{eqnarray} \hat{{\bf m_s}}&=&({\bf H'}({\bf I}-{\bf B}({\bf B'B})^{-1}{\b... ...B})^{-1}{\bf B'}({\bf I}-{\bf H}({\bf H'H})^{-1}{\bf H'}){\bf d}.\end{eqnarray}$$ (32) (33)

${\bf B}({\bf B'B})^{-1}{\bf B'}$ is the coherent noise resolution matrix, whereas ${\bf H}({\bf H'H})^{-1}{\bf H'}$ is the signal resolution matrix. Denoting $\overline{{\bf R_s}} = {\bf I}-{\bf H}({\bf H'H})^{-1} {\bf H'}$ and $\overline{{\bf R_n}} = {\bf I}-{\bf B}({\bf B'B})^{-1} {\bf B'}$ yields the following simplified expression for $\hat{{\bf m_s}}$ and $\hat{{\bf m_n}}$: 

$$\left( \begin{array} {c} \hat{{\bf m_s}} \\ \hat{{\bf m_n... ...ne{R_s}B})^{-1}{\bf B'\overline{R_s}}\end{array}\right){\bf d}.$$ (34)

• With model space regularization

The fitting goal becomes

$$\begin{eqnarray} {\bf 0} &\approx& {\bf Lm}-{\bf d}, \\ {\bf 0} &\approx& \epsilon {\bf Cm},\end{eqnarray}$$ (35) (36)

with ${\bf L}=({\bf H} \;\; {\bf B})$, ${\bf m'}=({\bf m_s} \;\; {\bf m_n})$ and

$${\bf C}= \left(\begin{array} {cc} {\bf C_s} & {\bf 0} \\ {\bf 0} & {\bf C_n} \end{array}\right).$$ (37)

The matrix equation we want to solve is

$$\left( \begin{array} {cc} {\bf H'H} + \epsilon^2 {\bf C'_s... ... \begin{array} {c} {\bf H'd} \\ {\bf B'd}\end{array}\right).$$ (38)

Using equations (25) and (26) we obtain  

$$\left( \begin{array} {c} \hat{{\bf m_s}} \\ \hat{{\bf m_n... ... C_n'C_n})^{-1}{\bf B'\overline{R_s}}\end{array}\right){\bf d},$$ (39)

where $\overline{{\bf R_s}} = {\bf I}-{\bf H}({\bf H'H})^{-1} {\bf H'}$ and $\overline{{\bf R_n}} = {\bf I}-{\bf B}({\bf B'B})^{-1} {\bf B'}$.