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

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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 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, $$X^{-1} = (X^TX)^{-1}X^T =\begin{bmatrix}A^TA+C^TC & A^TB\\ B^TA & B^TB\end{bmatrix}^{-1} \begin{bmatrix}A^T & C^T\\ B^T & 0\end{bmatrix}.$$ As $B$ has full column rank, $B^TB$ is invertible. Therefore you can use the formula for Schur complement to calculate the inverse of the block matrix on the RHS above, and the result is $$X^{-1}= \begin{bmatrix} S^{-1} & -S^{-1} A^TB (B^TB)^{-1} \\ -(B^TB)^{-1} B^TA S^{-1} & (B^TB)^{-1} + (B^TB)^{-1} B^TA S^{-1} A^TB (B^TB)^{-1} \end{bmatrix} \begin{bmatrix}A^T & C^T\\ B^T & 0\end{bmatrix},$$ where $S=A^TA+C^TC-A^TB(B^TB)^{-1}B^TA$.
@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