I have a special case where $X=\left(\begin{array}{cc} A & B\\ C & 0 \end{array}\right)$ and:

  1. $X$ is non-singular

  2. $A$ is singular

  3. $B$ is full column rank

  4. $C$ is full row rank

How do you calculate $X^{-1}$ in this case?

$A\in R^{n\times n}$ , $B\in R^{n\times m}$ , $C\in R^{m\times n}$ and $D\in 0^{m\times m}$

For example: $$X=\begin{pmatrix}0&1\\1&0\end{pmatrix}$$

  • $\begingroup$ It may not change the analysis, but what are the dimensions of the submatrices? $\endgroup$ – Daryl Jun 4 '13 at 23:11
  • $\begingroup$ In general this matrix is not invertible - take $A$ and $D$ sto be $m\times m$ and $n\times n$ respectively. If $m>n$ and $A=0$ then the rank of $X$ is $2n$, which is less than $m+n$. $\endgroup$ – Chris Godsil Jun 5 '13 at 1:31
  • $\begingroup$ An example of X is $$\begin{pmatrix}0&1\\1&0\end{pmatrix}$$ $\endgroup$ – Shyam Jun 5 '13 at 1:38

If you are looking for a blockwise closed-form formula in terms of $A,B$ and $C$, I am very skeptical about its usefulness. Yet that doesn't mean there isn't one: since $X$ is invertible, $$ X^{-1} = (X^TX)^{-1}X^T ={\underbrace{\begin{bmatrix}A^TA+C^TC & A^TB\\ B^TA & B^TB\end{bmatrix}}_M}^{-1} \begin{bmatrix}A^T & C^T\\ B^T & 0\end{bmatrix}. $$ As $X$ is invertible, $M=X^TX$ is positive definite. Hence the diagonal sub-block $B^TB$ is invertible (and in fact, positive definite). In turn, its Schur complement in $M$ is also invertible. By applying the matrix inversion formula in terms of Schur complement, we get $$ X^{-1}= \begin{bmatrix} S^{-1} & -S^{-1} A^TB (B^TB)^{-1} \\ -(B^TB)^{-1} B^TA S^{-1} & (B^TB)^{-1} + (B^TB)^{-1} B^TA S^{-1} A^TB (B^TB)^{-1} \end{bmatrix} \begin{bmatrix}A^T & C^T\\ B^T & 0\end{bmatrix}, $$ where $S=A^TA+C^TC-A^TB(B^TB)^{-1}B^TA$ is the Schur complement of $B^TB$ in $M$.

Edit. The above idea can be generalised to any invertible complex matrix $X=\begin{bmatrix}A&B\\ C&D\end{bmatrix}$ with square diagonal sub-blocks $A$ and $D$ of possibly different sizes. We have \begin{aligned} X^{-1} = (X^\ast X)^{-1}X^\ast &={\begin{bmatrix}A^\ast A+C^\ast C & A^\ast B+C^\ast D\\ B^\ast A+D^\ast C & B^\ast B+D^\ast D\end{bmatrix}}^{-1} \begin{bmatrix}A^\ast & C^\ast\\ B^\ast & D^\ast\end{bmatrix}\\ &=:{\underbrace{\begin{bmatrix}P&Q\\ Q^\ast&R\end{bmatrix}}_M}^{-1} \begin{bmatrix}A^\ast & C^\ast\\ B^\ast & D^\ast\end{bmatrix} \end{aligned} where $P= A^\ast A+C^\ast C,\ Q=A^\ast B+C^\ast D$ and $R=B^\ast B+D^\ast D$.

Since $X$ is invertible, $M=X^\ast X$ is positive definite. Thus the principal submatrices $R$ and its Schur complement $S=P-QR^{-1}Q^\ast$ in $M$ are also positive definite and we may apply the usual block matrix inverse formula to $M$ to obtain $$ X^{-1}= \begin{bmatrix} S^{-1} & -S^{-1} QR^{-1} \\ -R^{-1}Q^\ast S^{-1} & R^{-1} + R^{-1}Q^\ast S^{-1}QR^{-1} \end{bmatrix} \begin{bmatrix}A^\ast & C^\ast\\ B^\ast & D^\ast\end{bmatrix}. $$

  • $\begingroup$ Thank you, this answers my question! $\endgroup$ – Shyam Jun 5 '13 at 18:02
  • $\begingroup$ user1551, can I get your contact information? I want to acknowledge you in my dissertation and in any publications if I get them. $\endgroup$ – Shyam Jun 5 '13 at 20:24
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    $\begingroup$ @Shyam Thanks, but I would like to remain anonymous. See this question and footnote 7 of this paper for an example of citing $\mathtt{mathoverflow.net}$. In our case, you may write something like "this fact was explained to us by an anonymous user ($\mathtt{user1551}$) of $\mathtt{math.stackexchange.com}$ in answer no. 412136 to question no. 411492." $\endgroup$ – user1551 Jun 5 '13 at 20:40

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