理顺 QR 分解算法

咱们网站的这个公式编辑器，估计是后台生成图片后贴回来的，固定分辨率而且分辨率不高。

还不如先离线 latex 生成 pdf 后再截图上来

1. Why QR

When A and b are known, to solver the minimization of $|| Ax - b ||_{2}$ , where $A \in \mathbb{R}^{m*n}, b \in \mathbb{R}^{m}, and\,\,m \ge n$ .

The reduction of A to various canonical form via orthogonal transformations should use Householder reflections and Givens rotations.

2. preview on orthogonal matrix

2.1 Orthogonal matrix

$Q \in \mathbb{R}^{mxm} , \, Q \, is \, orthogonal \, matrix\, \, if:\\Q^{T}Q=QQ^{T}=I_{m}$

2.2 rotation matrix is orthogonal matrix

$If\\\mathbf{Q} = \left( \begin{array}{cc} \,\,\,\,\,\cos(\theta) & \sin(\theta)\\ -\sin(\theta)& \cos(\theta)\\ \end{array} \right)\\$

$\mathbf{Q}$ is orthogonal and $\mathbf{Q}$ is a rotation matrix.

If $y = Q^Tx$ , then $y$ is obtained by rotating $x$ counterclockwise through an angle $\theta$ .

2.3 reflection matrix is orthogonal matrix

$If\\ \mathbf{Q} = \left( \begin{array}{cc} \cos(\theta) & \sin(\theta)\\ \sin(\theta) & -\cos(\theta)\\ \end{array} \right)$

$\mathbf{Q}$ is orthogonal and $\mathbf{Q}$ is a reflection matrix.

If $y = Q^Tx = Qx$ , then $y$ is obtained by reflecting the vector $x$ across the line defined by

$\mathbf{S}=\mathbf{span}\left\{\left[ \begin{array}{c} \cos(\theta/2)\\ \sin(\theta/2)\\ \end{array}\right]\right\}$

That means $\overrightarrow{x}$ and $\overrightarrow{y}$ are axial symmetry, looks like :

3. Householder transformation

未完待续 ... ...

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