QR分解

QR分解

QR分解基于施密特正交化，对向量${\mathbf{a}}_1,{\mathbf{a}}_2,...,{\mathbf{a}}_n$，其施密特正交化为
$${\mathbf{b}}_i={\mathbf{a}}_i-\sum _{k=1}^{i-1}{\mathbf{q}}_k{\mathbf{q}}_k^T{\mathbf{a}}_i$$
其中
$${\mathbf{q}}_i=\frac{{\mathbf{b}}_i}{\mid {\mathbf{b}}_i\mid }$$
为单位化的正交向量，将求和项移到另一边得
$${\mathbf{a}}_i={\mathbf{b}}_i+\sum _{k=1}^{i-1}{\mathbf{q}}_k{\mathbf{q}}_k^T{\mathbf{a}}_i$$
${\mathbf{b}}_i$是${\mathbf{a}}_i$在${\mathbf{q}}_i$的投影，即
$${\mathbf{b}}_i={\mathbf{q}}_i{\mathbf{q}}_i^T{\mathbf{a}}_i$$
综上所述
$${\mathbf{a}}_i=\sum _{k=1}^i{\mathbf{q}}_k{\mathbf{q}}_k^T{\mathbf{a}}_i$$

这显然是矩阵和列向量的乘积，不失一般性，假设矩阵$\mathbf{A}=\left[\begin{array}{ccc}{\mathbf{a}}_1& {\mathbf{a}}_2& {\mathbf{a}}_3\end{array}\right]$，其QR分解可表示为
$$\mathbf{A}=\left[\begin{array}{ccc}{\mathbf{q}}_1& {\mathbf{q}}_2& {\mathbf{q}}_3\end{array}\right]\left[\begin{array}{ccc}{\mathbf{q}}_1^T{\mathbf{a}}_1& {\mathbf{q}}_1^T{\mathbf{a}}_2& {\mathbf{q}}_1^T{\mathbf{a}}_3\\ {\mathbf{q}}_2^T{\mathbf{a}}_1& {\mathbf{q}}_2^T{\mathbf{a}}_2& {\mathbf{q}}_2^T{\mathbf{a}}_3\\ {\mathbf{q}}_3^T{\mathbf{a}}_1& {\mathbf{q}}_3^T{\mathbf{a}}_2& {\mathbf{q}}_3^T{\mathbf{a}}_3\end{array}\right]=\left[\begin{array}{ccc}{\mathbf{q}}_1& {\mathbf{q}}_2& {\mathbf{q}}_3\end{array}\right]\left[\begin{array}{ccc}{\mathbf{q}}_1^T{\mathbf{a}}_1& {\mathbf{q}}_1^T{\mathbf{a}}_2& {\mathbf{q}}_1^T{\mathbf{a}}_3\\ 0& {\mathbf{q}}_2^T{\mathbf{a}}_2& {\mathbf{q}}_2^T{\mathbf{a}}_3\\ 0& 0& {\mathbf{q}}_3^T{\mathbf{a}}_3\end{array}\right]$$

QR分解要求矩阵必须列满秩，Q是正交化后的原矩阵，R是上三角矩阵，满足$R_{ij}={\mathbf{q}}_i^T{\mathbf{a}}_j$