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

$Ax=f-By$

or

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

On the above solution the result x is related to the unknown vector y. Since y can be any values, this way the result x has very strong artifacts which is $-A^{-1}By$.

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

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

Here $Q^{+}$ is generalized inverse of the matrix $Q$. x can be solved from the above formula. 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 $QQ=Q$

$QQ=[I-BB^{-1}][I-BB^{-1}]=I-2BB^{-1}+BB^{-1}BB^{-1}=I-2BB^{-1}+BB^{-1}=I-BB^{-1}=Q$

and hence

$QQQ=Q$

Hence Q is also the generalized inverse of Q

That means

$QQ^{+}=QQ=Q$

$x=A^{+}[Q]^{+}Qf=A^{+}Qf$

or

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

The matrix $[A]^{+}[I-BB^{+}]$

is defined as the local inverse of Matrix A. Using local inverse instead of generalized inverse or inverse can avoid the artifacts from unknown input data.