Inverse Matrix of a Special Matrix for Optimization

Question

For some portfolio optimization problems, it finally comes to an inverse matrix of a special matrix $$B=\begin{bmatrix} A & \mathbf{1} \\ \mathbf{1}' & 0 \end{bmatrix}$$ where $A$ is a symmetric matrix and $\mathbf{1}=(1,\cdots,1)'\in\mathbb{R}^n$. Can anyone help to find the inverse matrix of matrix $B$?

Thanks a lot!

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Below is the background of the above problem:

Assume $\omega$ is the weight vector of the portfolio, $\mu$ is the return vector and $\Sigma$ is the covariance matrix. The constraint is $\sum\omega=1$. For different optimization problems, we will have different targets:

1. For the optimization problem of minimum variance: $$\min_\omega\omega'\Sigma\omega$$ applying the Lagrange multiplier, we have the matrix equation $$\begin{bmatrix} 2\Sigma & \mathbf{1} \\ \mathbf{1}' & 0 \end{bmatrix} \begin{bmatrix} \omega \\ \lambda \end{bmatrix}= \begin{bmatrix} \mathbf{0} \\ 1 \end{bmatrix}$$

2. For the optimization problem of mean-variance optimization: $$\max_\omega\omega'\mu-\frac{1}{2}\delta\omega'\Sigma\omega$$ applying the Lagrange multiplier, we have the matrix equation $$\begin{bmatrix} -\delta\Sigma & \mathbf{1} \\ \mathbf{1}' & 0 \end{bmatrix} \begin{bmatrix} \omega \\ \lambda \end{bmatrix}= \begin{bmatrix} -\mu \\ 1 \end{bmatrix}$$

Anyway, it comes to the problem of inverse matrix of $$B=\begin{bmatrix} A & \mathbf{1} \\ \mathbf{1}' & 0 \end{bmatrix}$$

Answer 1

We can use the Schur complement of $A$ to get the formula $$ B^{-1} = \pmatrix{A^{-1} + A^{-1}\mathbf 1 (B/A)^{-1}\mathbf 1' A^{-1} & -A^{-1}\mathbf 1(B/A)^{-1}\\ -(B/A)^{-1}\mathbf 1' A^{-1} & (B/A)^{-1}}, $$ where $$ B/A = 0 - \mathbf 1' A^{-1} \mathbf 1 = -\mathbf 1' A^{-1} \mathbf 1 \in \Bbb R. $$ We can rewrite the above in the following arguably simpler form: $$ B^{-1} = \pmatrix{A^{-1} & 0\\0 & 0} + \frac 1{\mathbf 1' A^{-1} \mathbf 1}\pmatrix{\mathbf{vv}' & - \mathbf v\\ -\mathbf v' & 1} \\ = \pmatrix{A^{-1} & 0\\0 & 0} + \frac 1{\mathbf 1' A^{-1} \mathbf 1} \pmatrix{-\mathbf v\\ 1} \pmatrix{-\mathbf v \\ 1}', $$ where $\mathbf v = A^{-1} \mathbf 1$.