For the classical definition of matrix inversion by Schur complement, given by: \begin{aligned} M^{-1}=\left[\begin{array}{ll} A & B \\ C & D \end{array}\right]^{-1} &=\left(\left[\begin{array}{cc} I_{p} & B D^{-1} \\ 0 & I_{q} \end{array}\right]\left[\begin{array}{cc} A-B D^{-1} C & 0 \\ 0 & D \end{array}\right]\left[\begin{array}{cc} I_{p} & 0 \\ D^{-1} C & I_{q} \end{array}\right]\right)^{-1} \\ &=\left[\begin{array}{cc} I_{p} & 0 \\ -D^{-1} C & I_{q} \end{array}\right]\left[\begin{array}{cc} \left(A-B D^{-1} C\right)^{-1} & 0 \\ 0 & D^{-1} \end{array}\right]\left[\begin{array}{cc} I_{p} & -B D^{-1} \\ 0 & I_{q} \end{array}\right] \\ &=\left[\begin{array}{cc} \left(A-B D^{-1} C\right)^{-1} & -\left(A-B D^{-1} C\right)^{-1} B D^{-1} \\ -D^{-1} C\left(A-B D^{-1} C\right)^{-1} & D^{-1}+D^{-1} C\left(A-B D^{-1} C\right)^{-1} B D^{-1} \end{array}\right] \\ &=\left[\begin{array}{cc} (M / D)^{-1} & -(M / D)^{-1} B D^{-1} \\ -D^{-1} C(M / D)^{-1} & D^{-1}+D^{-1} C(M / D)^{-1} B D^{-1} \end{array}\right] \qquad \qquad \qquad \qquad \qquad \qquad \qquad (1) \end{aligned} the matrix $D$ is considered non-singular.

However, for generalised Schur complement, when $A$, $D$ and $M$ are singular, substituting the formulation for Schur complement using a Moore-Penrose inverse, i.e.,:

\begin{aligned} M/A := A - (BD^{\dagger}C) \qquad \qquad(2) \end{aligned}

and substituting the value in (1) results in a value of $M^{-1}$ (or should it be denoted by $M^{\dagger}$ ?) which gives $MM^{-1} \neq I$.

Granted M is ill-conditioned for this problem, but how do we go about inverting such matrices?

Any ideas/hints would be very helpful. Thanks in advance.


1 Answer 1


While the Moore-Penrose inverse does indeed mean that $MM^\dagger \neq I$, this is more a "this is not always the case" statement. See the actual statement is: $$MM^\dagger M = M, ~~~~ M^\dagger MM^\dagger = M^\dagger$$ Which includes many use cases. For instance, this could mean that $MM^\dagger \neq I$ but $M^\dagger M = I$ and vise versa. However, it can also mean that $M = M^\dagger$ and $MM^\dagger = M$ (take for instance an Identity matrix where some diagonal entries are 0).

With respect to the Schur complement, I think this boils down to 2 things: First, is $(A-BD^\dagger C)$ invertible? Second, is either $D^\dagger D = I$ or $DD^\dagger = I$, or the same but for $A$. But this is when we only look for a $M^\dagger$ such that either $MM^\dagger = I$ or $M^\dagger M = I$. This does still mean that $M$ is ill-conditioned, but for singular matrices the question is never "find an unique matrix such that if multiplied with $M$ yields the identity matrix", but "find any matrix such that if multiplied with $M$ minimizes the squared difference with the Identity matrix"

  • $\begingroup$ Very clear answer, thanks for your response. Indeed, $(A - BD^{\dagger}C)$ remains non-singular despite $A$, $D$ and $M$ being singular. The last sentence of your answer really clears this up for me. Cheers! $\endgroup$
    – Audrey
    Commented Sep 15, 2021 at 14:38
  • 1
    $\begingroup$ Thank you! However, I must point out that you cannot guarantee $(A-BD^\dagger C)$ to be non-singular. Take for instance $B$ or $C$ equal to zero. $\endgroup$
    – Petrus1904
    Commented Sep 15, 2021 at 18:42
  • $\begingroup$ I understand it is not guaranteed. However, in that case, using $(A - BD^{\dagger}C)^{\dagger}$ should produce equivalent results for the generalised Schur complement based inversion, right? $\endgroup$
    – Audrey
    Commented Sep 16, 2021 at 9:37

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