Problems for Lecture 14
From textbook Section III.1

1. Another approach to $(I-u{v}^{\mathtt{T}}{)}^{-1}$ starts with the formula for a geometric series:
$(1-x{)}^{-1}=1+x+x^2+x^3+\cdots$   Apply that formula when $x=u{v}^{\mathtt{T}}=$ matrix:

$$\begin{array}{rcl}(I-u{v}^{\mathtt{T}}{)}^{-1}& =& I+u{v}^{\mathtt{T}}+u{v}^{\mathtt{T}}u{v}^{\mathtt{T}}+u{v}^{\mathtt{T}}u{v}^{\mathtt{T}}u{v}^{\mathtt{T}}+\cdots \\ & =& I+u[1+v^{\mathtt{T}}u+v^{\mathtt{T}}u{v}^{\mathtt{T}}u+\cdots ]v^{\mathtt{T}}\end{array}$$

Take $x=v^{\mathtt{T}}u$  to see $I+{\displaystyle \frac{u{v}^{\mathtt{T}}}{1-v^{\mathtt{T}}u}}$. This is exactly equation (1) for $(I-u{v}^{\mathtt{T}}{)}^{-1}$.

4. Problem 3 found the inverse matrix $M^{-1}=(A-u{v}^{\mathtt{T}}{)}^{-1}$. In solving the equation $My=b$, we compute only the solution $y$ and not the whole inverse matrix $M^{-1}$. You can find $y$ in two easy steps:

$$ Step 1:   Solve $Ax=b$ and $Az=u$. Compute $D=1-v^{\mathtt{T}}z$.

$$ Step 2:   Then $y=x+{\displaystyle \frac{v^{\mathtt{T}}x}{D}}z$ is the solution to $My=(A-u{v}^{\mathtt{T}})y=b$.

Verify $(A-u{v}^{\mathtt{T}})y=b$. We solved two equations using $A$, no equations using $M$.