Exercise 2.1.3

Problem 1 says that $A=D{D}^T+e{e}^T$, where $e=(1,0,\dots ,0)$. From secition III.1 we have $A^{-1}={(D{D}^T)}^{-1}-z{z}^T$.

We see that $A{A}^{-1}=(D{D}^T+e{e}^T)({(D{D}^T)}^{-1}-z{z}^T)=I-D{D}^{T}z{z}^T+e{e}^T{(D{D}^T)}^{-1}-e{e}^{T}z{z}^T$ For $n=3$, we have $D{D}^T=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$, and ${(D{D}^T)}^{-1}=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$, $e{e}^T=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$, let $z=(z_1,z_2,z_3)$, we have

$D{D}^{T}z{z}^T=\left[\begin{array}{ccc}\hfill z_1(z_1-z_2)\hfill & \hfill z_2(z_1-z_2)\hfill & \hfill z_3(z_1-z_2)\hfill \\ \hfill z_1(2z_2-z_1-z_3)\hfill & \hfill z_2(2z_2-z_1-z_3)\hfill & \hfill z_3(2z_2-z_1-z_3)\hfill \\ \hfill z_1(2z_3-z_2)\hfill & \hfill z_2(2z_3-z_2)\hfill & \hfill z_3(2z_3-z_2)\hfill \end{array}\right]$

$e{e}^T{(D{D}^T)}^{-1}=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

$e{e}^{T}z{z}^T=\left[\begin{array}{ccc}\hfill z_1^2\hfill & \hfill z_1{z}_2\hfill & \hfill z_1{z}_3\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

So we have $z_3(2z_3-z_2)=0$, $z_3(2z_2-z_1-z_3)=0$, and $-z_1(z_1-z_2)+3-z_1^2=0$ we see that $z=(\frac{3}{2},1,\frac{1}{2})$ can be a solution here.

So rank-1 change in $D{D}^T$ produces rank-1 change in its inverse.