Local inverse

Local inverse

${\begin{bmatrix}f&g\\\end{bmatrix}}={\begin{bmatrix}A&B\\C&D\\\end{bmatrix}}{\begin{bmatrix}x\\y\\\end{bmatrix}}$

Assume A, B, C, D are known matrices; x and y are unknown vectors; f is known vector; g is unknown vector. We are interested to know the vector x. What is the best solution? Shuang-ren Zhao defined a Local inverse to solve the above problem. First we consider the simplest solution.

$f=Ax+By$

or

$Ax=f-By$

We write the solution of the above equation as $x_{0}$ instead to distinguish it with the original vector $x$

$x_{0}=A^{-1}(f-By)$

On the above solution the result $x_{0}$ is related to the unknown vector $y$. Since $y$ can be any values, this way the result $x_{0}$ has very strong artifacts which is

$error_{0}=|-A^{-1}By|$.

In order to minimize the above artifacts of the solution, we considered a special matrix $Q$ which satisfy

$QB=0$

$QAx=Qf-QBy=Qf$

or

$x_{1}=[QA]^{+}Qf=[A]^{+}Q^{+}Qf$

Here $Q^{+}$ is generalized inverse of the matrix $Q$. $x_{1}$ is a solution for $x$. It is easy to see that Q can be written as following:

$Q=I-BB^{+}$

Here $B^{+}$ is the generalized inverse of B which satisfy

$BB^{+}B=B$

hence

$QB=[I-BB^{+}]B=B-BB^{+}B=B-B=0$

It is easy to prove that $QQ=Q$

$QQ=[I-BB^{+}][I-BB^{+}]=I-2BB^{+}+BB^{+}BB^{+}=I-2BB^{+}+BB^{+}=I-BB^{+}=Q$

and hence

$QQQ=(QQ)Q=QQ=Q$

Hence Q is also the generalized inverse of Q

That means

$Q^{+}Q=QQ=Q$

$x_{1}=A^{+}[Q]^{+}Qf=A^{+}Qf$

or

$x_{1}=[A]^{+}[I-BB^{+}]f$

The matrix

$A^{L}=[A]^{+}[I-BB^{+}]$

$A^{L}$ is referred as the local inverse of Matrix A. Using local inverse instead of generalized inverse or inverse can avoid the artifacts from unknown input data.

It is clear that if $y=0$, We can find a better solution than the above local inverse solution which is:

$x_{0}=[A]^{+}f_{0}$

where$f_{0}$ is where $f$ in the case $y=0$.

In this situation we can found that the difference of above solutions are

$error_{1}=|x_{0}-x_{1}|=|[A]^{+}BB^{+}f_{0}|$

In case the contribution of $[A]^{+}BB^{+}$ to $x$ are smaller than that of $[A]^{+}$, or

$error_{1}<<error_{0}$

the local inverse solution $x_{1}$ is a much better solution compared to $x_{0}$ to this problem.