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Let $A$ be a matrix with $A^+$ Moore-Penrose inverse. Let also $Det()$ denote the pseudo-determinant of a matrix.

Does the formula (which assumes the existence of $A^{-1}$)

$$ det\left( \begin{array}{cc} A & B \\ C & D \end{array} \right) = det(A)det(D-CA^{-1}B), $$

where $det()$ denotes the usual determinant, applies to the use of pseudo determinants and Moore-Penrose inverse? This is

$$ Det\left( \begin{array}{cc} A & B \\ C & D \end{array} \right) = Det(A)Det(D-CA^{+}B)\,? $$

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The classical formula ${\rm Det}(R) = {\rm Det}( \left[ \begin{array}{cc} A & B \\ C & D \end{array} \right] ) = {\rm Det}(D) {\rm Det}(A-BD^+C)$ for block matrices does not hold for pseudo determinants ${\rm Det}$ and Moore pseudo inverse $D^+$, as already $$ \left[ \begin{array}{cc} A & B \\ C & D \end{array} \right] \left[ \begin{array}{cc} I & 0 \\ -D^+ C & I \end{array} \right] = \left[ \begin{array}{cc} A-B D^+ C & B \\ 0 & D \end{array} \right] $$ does not holds with the pseudo inverse $D^+$. But here is an example: $$ A=\left[ \begin{array}{cc} -1 & 0 \\ -2 & -1 \\ \end{array} \right], B=\left[ \begin{array}{cc} 2 & 1 \\ 0 & 2 \\ \end{array} \right], C=\left[ \begin{array}{cc} 2 & -2 \\ 2 & -1 \\ \end{array} \right], D=\left[ \begin{array}{cc} 1 & 1 \\ 0 & 0 \\ \end{array} \right] $$ so that the block matrix is $$ R=\left[ \begin{array}{cccc} -1 & 0 & 2 & 1 \\ -2 & -1 & 0 & 2 \\ 2 & -2 & 1 & 1 \\ 2 & -1 & 0 & 0 \\ \end{array} \right] $$ is even invertible with $\det(R)={\rm Det}(R)=2$. The pseudo inverse of $D$ is $D^+ = \left[ \begin{array}{cc} \frac{1}{2} & 0 \\ \frac{1}{2} & 0 \\ \end{array} \right]$. Also the matrix $A - B D^+ C= \left[ \begin{array}{cc} -4 & 3 \\ -4 & 1 \\ \end{array} \right]$ is invertible with determinant $8$. We have ${\rm Det}(D)=1$. The left hand side is $2$, the right hand side is $8$. It is a nice question as we can answer also the question whether the block diagonal formula for diagonal block matrices generalizes to upper triangular block matrices. There are a few surprises with pseudo determinants: see my (paper) about it and the (ArXiv pre-print).

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  • $\begingroup$ Hi Oliver, thank you for your answer. I have a follow-up question. If in the original question, $A$ is always invertible. Then does the formula for the determinant of block matrix holds for pseudo-determinant? $\endgroup$
    – Shi James
    Mar 3 at 15:44
  • $\begingroup$ Yes, sure, your identity then holds. It does not involve pseudo inverses any more. I actually think it is a cute identity for pseudo determinants and I'm not aware that it has appeared already, but the literature on matrix stuff is huge. The identity is a consequence of the known identity $$ \left[ \begin{array}{cc} 1 & 0 \\ -C A^{-1} & 1 \\ \end{array} \right] \left[ \begin{array}{cc} A & B \\ C & D \\ \end{array} \right] = \left[ \begin{array}{cc} A & B \\ 0 & D-CA^{-1} B \\ \end{array} \right] $$ Then use that for block triangular matrices the peudo determinant is the product of blocks. $\endgroup$ Mar 4 at 22:00

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