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Suppose I have some block diagonal matrix $A$, defined as:

$A = \begin{bmatrix} A_1 & 0 & ... & 0 \\ 0 & A_2 & ... & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & ... & A_n \end{bmatrix}$

Where $\{A_i\}_{i \in \{1,...n\}}$ are the blocks of $A$. Is it true that the Pseudo-inverse of $A$, $A^+$, is given by:

$A^+ = \begin{bmatrix} A_1^+ & 0 & ... & 0 \\ 0 & A_2^+ & ... & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & ... & A_n^+ \end{bmatrix}$

If so, why?/why not?

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Yes it is. Just check the Moore-Penrose conditions by doing matrix multiplication.

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