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Let

$$S := \pmatrix{A&B\\C&D}$$

If $A^{-1}$ or $D^{-1}$ exist, we know that matrix $S$ can be inverted.

$$S^{-1} = \pmatrix{A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\\-(D-CA^{-1}B)^{-1}CA^{-1}&(D-CA^{-1}B)^{-1}}$$

But, what if $A^{-1}$ and $D^{-1}$ do not exist? Can we invert matrix $S$?

For example,

$$S = \pmatrix{0&1\\1&0}$$

or

$$S = \pmatrix{2&3&1&1\\4&6&1&2\\1&1&3&1\\4&1&12&4}$$

both their $A^{-1}$ and $D^{-1}$ don't exist, but $S^{-1}$ exists.

Rodrigo de Azevedo
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Claire
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    You're right: the existence of $A^{-1}$ or $D^{-1}$ is sufficient, but not necessary, for $S$ to be invertible. If neither $A$ nor $D$ is invertible, then sometimes $S$ is invertible (as your examples show), and sometimes it isn't. I don't know of any general criterion, other than the usual algorithms for finding matrix inverses. – Greg Martin Sep 29 '16 at 16:45
  • @GregMartin Thanks! Can we prove if $S_{n\times n}^{-1}$ exists, there exists m<n such that $A_{m\times m}^{-1}$ or $D_{n-m\times n-m}^{-1}$ exists? – Claire Sep 30 '16 at 09:35
  • Take my second example, if m=1, then A=[2] and its inverse exists. – Claire Sep 30 '16 at 09:41
  • By $A_{m\times m}$ do you mean the upper-left $m\times m$ block (and similarly for $D$)? If you allow arbitrary blocks (equivalently, exchanging rows or columns) then it becomes trivial (take $m=1$). – Greg Martin Sep 30 '16 at 16:04
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    @GregMartin I found the statement is false. If S=$\pmatrix{0&0&1\1&1&1\1&0&0}$ , then $S^{-1}$ exists. But for all $A_{m\times m}$ or $D_{n-m\times n-m}$ , their inverses are don't exist for all m. – Claire Oct 01 '16 at 16:27
  • Now I wonder if $S^{-1}$ exists, can we prove at least one of its block matrices(A, B, C, or D) can be inverted? – Claire Oct 01 '16 at 16:41
  • No, indeed the permutation matrix $\displaystyle S = \begin{pmatrix} 0&1&0&0 \ 0&0&0&1 \ 1&0&0&0 \ 0&0&1&0 \end{pmatrix}$ is a counterexample. – Greg Martin Oct 01 '16 at 19:09
  • I don't get it: What if $A$ is invertible and $B$, $C$ and $D$ are all zero matrices of appropriate (non-zero) dimensions. Your matrix $S$ looks non-invertible then. The formula that you give for $S^{-1}$ also runs into troubles if you consider that case. – Hecatonchires Jul 24 '22 at 15:34

4 Answers4

8

I saw a paper on this topic by Lu and Shiou some years ago. Here is the link. They first introduced the formula you mentioned and then investigated other special cases.

polfosol
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8

One good reference is the Handbook Matrix Mathematics Theory, Facts and Formulas by Dennis S. Bernstein. Let $\mathbb{F}$ equal to $\mathbb{R}$ or $\mathbb{C}$.

Proposition 3.9.7. Let $A \in \mathbb{F}^{n \times n}, B \in \mathbb{F}^{n \times m}, C \in \mathbb{F}^{m \times n}$, and $D \in \mathbb{P}^{m \times m}$. If $A$ and $D-C A^{-1} B$ are nonsingular, then $$ \left[\begin{array}{ll} A & B \\ C & D \end{array}\right]^{-1}=\left[\begin{array}{rr} A^{-1}+A^{-1} B\left(D-C A^{-1} B\right)^{-1} C A^{-1} & -A^{-1} B\left(D-C A^{-1} B\right)^{-1} \\ -\left(D-C A^{-1} B\right)^{-1} C A^{-1} & \left(D-C A^{-1} B\right)^{-1} \end{array}\right] $$

There are three more cases to consider in addition to the above proposition.

  • If $D$ and $A-B D^{-1} C$ are nonsingular;
  • If $B$ and $C-D B^{-1} A$ are nonsingular;
  • If $C$ and $B-A C^{-1} D$ are nonsingular;

For the sake of illustration we will prove only the case where $C$ and $B-A C^{-1} D$ are nonsingular;

To prove these cases it is enough to use the matrix $J=\left[\begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]$ to reduce them to the same case as the proposition above. Observe that $ \left[\begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]^{-1} = \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right] \mbox{ once } \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right] \cdot \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right] = \left[ \begin{array}{cc} I & 0 \\ 0 & I \end{array} \right]. $ Another important thing to note is that $$ \;(\ast)\;\;\qquad \left[ \begin{array}{cc} U & V \\ X & Y \end{array} \right] = \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right] \cdot \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right] \cdot \left[ \begin{array}{cc} U & V \\ X & Y \end{array} \right] = \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right] \cdot \left[ \begin{array}{cc} X & Y \\ U & V \end{array} \right]. $$ and $$ (\ast\ast)\qquad \left[ \begin{array}{cc} U & V \\ X & Y \end{array} \right] = \left[ \begin{array}{cc} U & V \\ X & Y \end{array} \right] \cdot \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right] \cdot \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right] = \left[ \begin{array}{cc} V & U \\ Y & X \end{array} \right] \cdot \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right] $$

Suppose there exists the inverse $C^{-1}$ of $C$ and there exists the inverse $(B -AC^{-1}D)^{-1}$ of $(B -AC^{-1}D)$. What can we say about the inverse of $\left[ \begin{array}{cc} C & D \\ A & B \end{array} \right]$? By proposition above we have $$ \left[ \begin{array}{cc} C & D \\ A & B \end{array} \right]^{-1} =\left[\begin{array}{rr} C^{-1}+C^{-1} D\left(B-A C^{-1} D\right)^{-1} A C^{-1} & -C^{-1} D\left(B-A C^{-1} D\right)^{-1} \\ -\left(B-A C^{-1} D\right)^{-1} A C^{-1} & \left(B-A C^{-1} D\right)^{-1} \end{array}\right] $$ And more, by $(\ast)$ and $(\ast\ast)$ we have \begin{align} \left[ \begin{array}{cc} A & B \\ C & D \end{array} \right]^{-1} &= \left( \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right] \cdot \left[ \begin{array}{cc} C & D \\ A & B \end{array} \right] \right)^{-1} \\ &= \left[ \begin{array}{cc} C & D \\ A & B \end{array} \right]^{-1} \cdot \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]^{-1} \\ &= \left[ \begin{array}{cc} C & D \\ A & B \end{array} \right]^{-1} \cdot \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right] \\ &= \left[\begin{array}{rr} C^{-1}+C^{-1} D\left(B-A C^{-1} D\right)^{-1} A C^{-1} & -C^{-1} D\left(B-A C^{-1} D\right)^{-1} \\ -\left(B-A C^{-1} D\right)^{-1} A C^{-1} & \left(B-A C^{-1} D\right)^{-1} \end{array}\right] \cdot \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right] \\ &= \left[\begin{array}{rr} -C^{-1} D\left(B-A C^{-1} D\right)^{-1} & C^{-1}+C^{-1} D\left(B-A C^{-1} D\right)^{-1} A C^{-1} \\ \left(B-A C^{-1} D\right)^{-1} & -\left(B-A C^{-1} D\right)^{-1} A C^{-1} \end{array}\right] \end{align}

Elias Costa
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3

If your block matrix is real or complex and known to be invertible, you may apply the usual block matrix inverse formula to find $(S^\ast S)^{-1}$ and thus to calculate $S^{-1}=(S^\ast S)^{-1}S^\ast$. See my answer to another question for more details.

user1551
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3

As the comments discuss, the claim that $S$ is invertible if $A$ is invertible as claimed in the question is false. What is true is that:

Theorem: If $A$ is invertible, then $S$ is invertible if and only if the Schur complement $D-CA^{-1}B$ is invertible. In which case, the stated formula for the inverse of $S$ in $2\times 2$ block form holds.

So $A$ does not need to be invertible for $S$ to be invertible, but if $A$ is, then $D-CA^{-1}B$ must be invertible as well for $S$ to be invertible.

As the OP notes with their example $\begin{bmatrix}0&1\\1&0\end{bmatrix}$, the top-left matrix need not be invertible for $S$ to be invertible. Instead, for $S$ to be invertible, it just has to be possible for one to reorder the rows and columns of the matrix $S$ (left- and right-multiplying by permutation matrices) to bring a nonsingular matrix $A$ into the block $(1,1)$-position and the resulting reordered Schur complement $D-CA^{-1}B$ to be nonsingular. Indeed, the additional cases in Elias’ answer all boil down to reordering the rows and columns of $S$ to put either $B$, $C$, or $D$ into the top-left spot and then applying the theorem.

An important observation is that one only needs to reorder the rows of $S$ to check invertibility of $S$, the core observation behind LU factorization with partial pivoting:

Theorem: Fix $k$ between $1$ and the size of $S$, $S$ is invertible if and only if it is possible to find a permutation matrix $P$ where $PS$ has block form $PS=\begin{bmatrix} A&B\\C&D\end{bmatrix}$, with $A$ being $k\times k$ and $A$ and $D-CA^{-1}B$ are invertible.

eepperly16
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