Find inverse of a block $2\times 2$ array

Question

I got an array $D = \begin{bmatrix}A & b\\ b^T & c\end{bmatrix}$ where $A$ is a real $N\times N$ array, $b$ is a column vector of length $N$, and $c$ is a real scalar. I am confused how to find the inverse of matrix $D$, with respect to $A$.

One idea is to say that: $D D^{-1} = I$ ,and continue ,but cant go any further.

My main problem is how to handle the fact that i got an array with block structure. That's the confusing part. Any idea?

Answer 1

Let us look for the inverse under the very same form:

$$\Delta = \begin{bmatrix}U & v\\ v^T & w\end{bmatrix}$$

Let us express that (as you intended to begin):

$$\tag{1}D\Delta=I_{N+1} \ \ \Longleftrightarrow \ \ \begin{bmatrix}A & b\\ b^T & c\end{bmatrix}\begin{bmatrix}U & v\\ v^T & w\end{bmatrix}=\begin{bmatrix}I_N & 0\\ 0 & 1\end{bmatrix}$$

$$\Longleftrightarrow \ \ \begin{cases}AU+bv^T=I_N&\ (i) \ (\text{matrices})\\Av+bw=0&\ (ii) \ (\text{column vectors})\\b^TU+cv^T=0 &\ (iii) \ (\text{row vectors})\\b^Tv+cw=1&\ (iv) \ (\text{real numbers})\end{cases}$$

Now, from (ii), one deduces

$\tag{2}v=-wA^{-1}b$.

Plugging expression (2) in (iv) gives:

$$-wb^TA^{-1}b+wc=1$$

Thus, $w$ can be expressed as an expression using in an exclusive way known quantities:

$$\tag{3}w:=\dfrac{1}{c-b^TA^{-1}b}$$

(inverse of the so called "Schur's complement"). Relationship (3) allows to place now $w$ in the category of known quantities, and consequently the same for $v$ (due to (2)).

It remains to obtain an expression of $U$ as a function of known quantities ($A,b,c$). It is evident that the key relationship for that is (i).

I let you do this last step.

Remark: we have been obliged to assume that $A$ is invertible and $c \neq b^TA^{-1}b$.

Answer 2

There are generic formulae for blockwise inversion of a block matrix $\begin{bmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D}\end{bmatrix}$ available at Wikipedia here. In the present case, we can use these to write the inverse of $D$ in two forms:

\begin{align} \begin{bmatrix} A & b \\ b^T & c\end{bmatrix}^{-1} &= \begin{bmatrix} A^{-1} +A^{-1}b(D/A)^{-1}b^T A^{-1} & -A^{-1} b(D/A)^{-1} \\ -(D/A)^{-1}b^T A^{-1} & (D/A)^{-1}\end{bmatrix}\\ &= \begin{bmatrix} (D/c)^{-1} & -(D/c)^{-1} bc^{-1} \\ -c^{-1}b^T(D/c)^{-1} & c^{-1}+c^{-1}b^T (D/c)^{-1}bc^{-1}\end{bmatrix} \end{align}

where $D/A=c-b^T A^{-1}b$ and $D/c=A-b c^{-1} b^T$ are the so-called Schur complements of $A$ and $c$ respectively. The first formula requires that $A,D/A$ be invertible; for the latter, $c$ must be nonzero and $D/c$ must be invertible. Which is more useful will likely depend on whether $A$ or $D/c=A-c^{-1}bb^T$ is more readily inverted.