In remote sensing we often have a linear problem that can be written in a matrix form as $aAx+bAy=B$, where $a$ and $b$ are scalar constants, $A$ is a matrix; and $x$, $y$, and $B$ are vectors (i.e. 1-column matrix). This can be further re-written as $\hat{A} \hat{x} = B$, where $\hat{A}$ is a matrix with 2 times more columns than $A$ (i.e. $\{aAbA\}$) and $\hat{x}$ is a vector with two times more rows than $x$ and $y$. Then we solve this problem by inverting the matrix $\hat{A}$ using the pseudoinverse (i.e. SVD): $\hat{x}=\hat{A}^{-1}B$.
The problem is that the computation of $\hat{A}^{-1}$ for each pixel is expensive. Knowing that the matrix $\hat{A}$ has a particular column structure, can the problem be reduced to inverting the matrix $A$ instead, which has half the amount of columns. In general, the matrix $A$ is rank deficient and overdetermined (i.e. more rows than columns, but rank is less or equal than the number of columns).
I think that yes due to the following equation (this equation should be true for block-diagonal matrices) \begin{equation} \big(\begin{smallmatrix} aA & 0\\ 0 & bA \end{smallmatrix}\big)\big( \begin{smallmatrix} a^{-1}A^{+} & 0\\ 0 & b^{-1}A^{+} \end{smallmatrix}\big) = \big(\begin{smallmatrix} aA a^{-1}A^{+} & 0\\ 0 & bA b^{-1}A^{+} \end{smallmatrix}\big) = \big(\begin{smallmatrix} I & 0\\ 0 & I \end{smallmatrix}\big), \end{equation} where $A^{+}$ is the pseudoinverse of $A$ and $I$ is the identity matrix. So, If I'm right, the pseudoinverse of $\hat{A}$ can be calculated using the following formula \begin{equation} \hat{A}^{+} = \big(\begin{smallmatrix} a^{-1}A^{+} & 0\\ 0 & b^{-1}A^{+} \end{smallmatrix}\big). \end{equation}