Q:

An expression subtly different from the one in Sherman-Morrison formula

Let $0\ne x\in\mathbb{R},\,\mathbf{y} \in \mathbb{R}^{n},\,\mathbf{1}=(1,1, \cdots, 1)^{T} \in \mathbb{R}^{n}$ and assume that $\Sigma=x I_{n}+\mathbf{y} \mathbf{1}^{T}+\mathbf{1} \mathbf{y}^{T}$ is positive definite. Prove that
$$
\Sigma^{-1}-\frac{\Sigma^{-1} \mathbf{1 1}^{T} \Sigma^{-1}}{\mathbf{1}^{T} \Sigma^{-1} \mathbf{1}}=\frac{1}{x}\left(I_{n}-\frac{1}{n} \mathbf{1 1}^{T}\right).
$$

A:

With the Sherman Morrison identity,
$$
\Sigma^{-1} - \frac{\Sigma^{-1}11^T\Sigma^{-1}}{1^T\Sigma 1} = 
\lim_{m \to \infty}
\Sigma^{-1} - \frac{\Sigma^{-1}(m1)(m1)^T\Sigma^{-1}}{1 + (m1)^T\Sigma (m1)} 
\\ = 
\lim_{m \to \infty}(\Sigma + (m \mathbf 1)(m \mathbf 1)^T)^{-1}.
$$
On the other hand, we can write
$$
\Sigma(m) := \Sigma + (m \mathbf 1)(m \mathbf 1)^T = x I_n + \pmatrix{\mathbf 1 & \mathbf y}
\pmatrix{m^2 & 1\\1 & 0}
\pmatrix{\mathbf 1 & \mathbf y}^T.
$$
To show that $\Sigma(m)^{-1} \to \frac 1x (I - \frac{11^T}{n})$, it suffices to block-diagonalize $\Sigma$. We see that $\Sigma \mathbf z = x\mathbf z$ for all $\mathbf z$ perpendicular to both $\mathbf 1,\mathbf y$. On the other hand, let $v_1 = \mathbf 1/\sqrt{n}$, let $v_2$ be the component of $\mathbf y$ orthogonal to $\mathbf 1$. We have
$$
v_1^T\pmatrix{\mathbf 1 & \mathbf y}
\pmatrix{m^2 & 1\\1 & 0}
\pmatrix{\mathbf 1 & \mathbf y}^Tv_1 = m^2n + 2(v_1^Ty)^2\\
$$
$$
v_1^T\pmatrix{\mathbf 1 & \mathbf y}
\pmatrix{m^2 & 1\\1 & 0}
\pmatrix{\mathbf 1 & \mathbf y}^Tv_2 = \sqrt{n}\, v_2^Ty.
$$
$$
v_2^T\pmatrix{\mathbf 1 & \mathbf y}
\pmatrix{m^2 & 1\\1 & 0}
\pmatrix{\mathbf 1 & \mathbf y}^Tv_2 = 
\pmatrix{0 & v_2^Ty} \pmatrix{m^2 & 1\\1 & 0}\pmatrix{0 & v_2^Ty} = 0
$$
So, relative to an orthogonal basis that extends $v_1,v_2$, the matrix of $\Sigma$ is
$$
x I_n + \pmatrix{m^2n + 2(v_1^Ty)^2 & \sqrt{n}\, v_2^Ty\\
\sqrt{n}\, v_2^Ty & 0} \oplus 0_{(n-2) \times (n-2)}.
$$
The inverse of this matrix is
$$
\pmatrix{x + m^2n + 2(v_1^Ty)^2 & \sqrt{n}\, v_2^Ty\\
\sqrt{n}\, v_2^Ty & x}^{-1} \oplus \frac 1x I = \\
\frac 1{xm^2 n + o(m)}\pmatrix{x & -\sqrt{n}\,v_2^Ty\\
-\sqrt{n}\, v_2^Ty & x + m^2n + 2(v_1^Ty)^2} \oplus \frac 1x I.
$$
As $m \to \infty$, this approaches
$$
\pmatrix{0 & 0\\
0 & 1/x} \oplus \frac 1x I,
$$
which is the matrix of $(I - \frac 1n 11^T)$ relative to the new basis. Thus, we have the desired conclusion.

Alternatively it suffices to show that $\Sigma(m)^{-1} \mathbf z \to (I - \frac{11^T}{n}) \mathbf z$ for all $\mathbf z$. In particular,

If $\mathbf z \perp \mathbf 1$, then $\Sigma(m)^{-1} \mathbf z \to \frac 1x \mathbf z$, and
$\Sigma(m)^{-1} \mathbf 1 \to 0$.

Alternatively, with the Woodbury identity, we have
$$
(x I_n)^{-1} - \Sigma^{-1} = 
\frac 1{x^2} \pmatrix{y & 1}\left(\pmatrix{m^2&1\\1&0}^{-1} + \pmatrix{n & 1^Ty\\ 1^Ty & y^Ty}\right)^{-1}\pmatrix{y & 1}^T.
$$
It would therefore suffice to compute the limit of the middle term
$$
\left(\pmatrix{m^2&1\\1&0}^{-1} + \pmatrix{n & 1^Ty\\ 1^Ty & y^Ty}\right)^{-1},
$$
and show that $(x I_n)^{-1} - \Sigma^{-1} \to \frac 1{xn}11^T$.

