## 每週問題 June 23, 2014

The general solution to a linear system of equations is described by

$\displaystyle \begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 1\\ 2\\ 4 \end{bmatrix}+\alpha\left[\!\!\begin{array}{r} 1\\ 0\\ -1 \end{array}\!\!\right]+\beta\begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix}$,

where $\alpha$ and $\beta$ are arbitrary parameters. Determine the vector $(x,y,z)$ that has minimum length.

$\displaystyle L=x^2+y^2+z^2=(1+\alpha)^2+(2+\beta)^2+(4-\alpha+2\beta)^2$

\displaystyle\begin{aligned} \frac{\partial L}{\partial\alpha}&=2(1+\alpha)-2(4-\alpha+2\beta)=2(2\alpha-2\beta-3)=0\\ \frac{\partial L}{\partial\beta}&=2(2+\beta)+4(4-\alpha+2\beta)=2(-2\alpha+5\beta+10)=0. \end{aligned}

$\displaystyle \begin{vmatrix} x-1&y-2&z-4\\ 1&0&-1\\ 0&1&2 \end{vmatrix}=x-2y+z-1=0$

$\displaystyle B=\left[\!\!\begin{array}{rc} 1&0\\ 0&1\\ -1&2 \end{array}\!\!\right]$

$\displaystyle P=B(B^TB)^{-1}B^T=\frac{1}{6}\left[\!\!\begin{array}{rcr} 5&2&-1\\ 2&2&2\\ -1&2&5 \end{array}\!\!\right]$

$\displaystyle \begin{bmatrix} 1\\ 2\\ 4 \end{bmatrix}-P\begin{bmatrix} 1\\ 2\\ 4 \end{bmatrix}=\begin{bmatrix} 1\\ 2\\ 4 \end{bmatrix}-\frac{1}{6}\left[\!\!\begin{array}{r} 5\\ 14\\ 23 \end{array}\!\!\right]=\frac{1}{6}\left[\!\!\begin{array}{r} 1\\ -2\\ 1 \end{array}\!\!\right]$

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