Given a $nm \times m$ matrix $A = \begin{pmatrix} A_1 \\ A_2 \\ \vdots \\ A_n\end{pmatrix}$ over $\mathbb{C}$, where $A_i$'s are $m \times m$ and $rank(A) = m$, is there an expression for the pseudo-inverse of $A$ in terms of the pseudo-inverses or SVDs of $A_i$'s? The $A_i$'s are not generally full-rank.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ I don't think there is much more to say than you can read on en.wikipedia.org/wiki/Block_matrix_pseudoinverse $\endgroup$– Carlo BeenakkerCommented Aug 28, 2019 at 10:32
-
$\begingroup$ Thank you for your reply but the partitioning seems to be different. In this case, I am partitioning over the larger dimension as opposed to the smaller dimension in the Wikipedia article. As such, I don't think the results transfer over. $\endgroup$– Jinhui WangCommented Aug 29, 2019 at 21:40
Add a comment
|