Exercise 3.1.4

Answers

We have $Ax = b$ , $Az = u$ and $D = 1 - v^{T} z$ , set $y = x + \frac{v^{T} x}{D} z$ , then we have

\begin{array}{l} My & = (A - u v^{T}) y \\ = (A - u v^{T}) (x + \frac{v^{T} x}{D} z) \\ = Ax + A \frac{v^{T} x}{D} z - u v^{T} x - u v^{T} \frac{v^{T} x}{D} z \\ = b + \frac{v^{T} xAz}{D} - u v^{T} x - u v^{T} \frac{v^{T} x}{D} z \\ = b + \frac{v^{T} xu}{D} - u v^{T} x - u v^{T} \frac{v^{T} x}{D} z \\ = b + u v^{T} x (\frac{1}{D} - 1 - v^{T} \frac{1}{D} z) \\ = b \end{array}

niuers

2020-03-20 00:00