13
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
10
$\begingroup$

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 $$ X^{-1}= \begin{bmatrix} S^{-1} & -S^{-1} QR^{-1} \\ -R^{-1}Q^\ast S^{-1} & R^{-1} + R^{-1}Q^\ast S^{-1}QR^{-1} \end{bmatrix} \begin{bmatrix}A^\ast & C^\ast\\ B^\ast & D^\ast\end{bmatrix}. $$

$\endgroup$
  • $\begingroup$ 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
  • 8
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.