# How find this invertible matrix $C=\left[\begin{smallmatrix} A&B\\ B^T&0 \end{smallmatrix}\right]$

let matrix $A_{n\times n}$,and $\det(A)>0$, and the matrix $B_{n\times m}$,and such $rank(B)=m$,and let $$C=\begin{bmatrix} A&B\\ B^T&0 \end{bmatrix}$$

Find this Invertible matrix $C^{-1}$

my try: I found this matrix Invertible matrix $C$,it must find $B^TAB$ Invertible matrix.But I can't

Let $C^{-1} = \begin{bmatrix} X & Y \\ Z & W \end{bmatrix}$ with appropriate sizes (i.e. $X$ is $n \times n$, $Y$ is $n \times m$, $Z$ is $m \times n$ and $W$ is $m \times m$). Then,

$C C^{-1} = \begin{bmatrix} A & B \\ B^T & 0 \end{bmatrix} \begin{bmatrix} X & Y \\ Z & W \end{bmatrix} = \begin{bmatrix} I_n & 0 \\ 0 & I_m \end{bmatrix}$

Hence, we have the following equations:

\begin{align} AX + BZ &= I_n \\ AY + BW &= 0 \\ B^T X &= 0 \\ B^T Y &= I_m \end{align}

From the first and third equations:

\begin{align} X + A^{-1}BZ &= A^{-1}\\ B^T X + B^T A^{-1}BZ &= B^T A^{-1}\\ Z &= (B^T A^{-1}B)^{-1}B^T A^{-1} \end{align}

Now, if we put this in the second equation we get $X = A^{-1} - A^{-1}B (B^T A^{-1}B)^{-1}B^T A^{-1}$

From the second and forth equations:

\begin{align} Y + A^{-1}BW &= 0\\ B^T Y + B^T A^{-1}BW &= 0\\ B^T A^{-1}BW &= -I_m\\ W &= -(B^T A^{-1}B)^{-1} \end{align}

Now, if we put this in the second equation we get $Y = A^{-1}B(B^T A^{-1}B)^{-1}$ which is consistent with the forth equation.

Note that $A$ and $B^T A^{-1}B$ must be invertible.

• – copper.hat Nov 29 '13 at 16:01
• I think $X \neq 0$... – dazaga Dec 14 '14 at 3:06
• @dazaga I think you are right. I'm editing the response. – obareey Dec 17 '14 at 17:07