# Inverse of a block matrix

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

• It may not change the analysis, but what are the dimensions of the submatrices? – Daryl Jun 4 '13 at 23:11
• 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 $m+n$. – Chris Godsil Jun 5 '13 at 1:31
• An example of X is $$\begin{pmatrix}0&1\\1&0\end{pmatrix}$$ – 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 $$X^{-1}= \begin{bmatrix} S^{-1} & -S^{-1} QR^{-1} \\ -R^{-1}Q^\ast S^{-1} & R^{-1} + R^{-1}Q^\ast S^{-1}QR^{-1} \end{bmatrix} \begin{bmatrix}A^\ast & C^\ast\\ B^\ast & D^\ast\end{bmatrix}.$$
• @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." – user1551 Jun 5 '13 at 20:40