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

Thank you for your help


1 Answer 1


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.


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