满秩分解
如果矩阵
的行（列）向量组线性无关，则称为行（列）满秩矩阵。
定理：设是
矩阵，A的秩为
，则存在
列满秩Ｆ和
行满秩
，使
，称为矩阵的满秩分解。
设%2C%20G%3D(q_1%2Cq_2%2C%5Ccdots%2Cq_r)%5ET)
，则
![A=\left( \matrix{ p_1,p_2,\cdots,p_r }\right) \left[ \matrix{ q_1^T\\q_2^T\\\cdots\\q_r^T }\right]=\sum_{i=1}^np_iq_i^T](https://math.jianshu.com/math?formula=A%3D%5Cleft(%20%5Cmatrix%7B%20p_1%2Cp_2%2C%5Ccdots%2Cp_r%20%7D%5Cright)%20%5Cleft%5B%20%5Cmatrix%7B%20q_1%5ET%5C%5Cq_2%5ET%5C%5C%5Ccdots%5C%5Cq_r%5ET%20%7D%5Cright%5D%3D%5Csum_%7Bi%3D1%7D%5Enp_iq_i%5ET)
例：设![A=\left(\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5\right)=\left[\matrix{2&1&6&1&0\\3&2&10&1&0\\2&3&10&-1&3\\4&4&16&0&1}\right]](https://math.jianshu.com/math?formula=A%3D%5Cleft(%5Calpha_1%2C%5Calpha_2%2C%5Calpha_3%2C%5Calpha_4%2C%5Calpha_5%5Cright)%3D%5Cleft%5B%5Cmatrix%7B2%261%266%261%260%5C%5C3%262%2610%261%260%5C%5C2%263%2610%26-1%263%5C%5C4%264%2616%260%261%7D%5Cright%5D)
，求的满秩分解。
解：先用初等行变换将化为简化阶梯型矩阵。
![\left[\matrix{2&1&6&1&0\\3&2&10&1&0\\2&3&10&-1&3\\4&4&16&0&1}\right]\rightarrow\left[\matrix{2&1&6&1&0\\1&1&4&0&0&\\0&2&4&-2&3\\0&2&4&-2&1}\right]\rightarrow\left[\matrix{1&0&2&1&0\\0&1&2&-1&0\\0&0&0&0&1\\0&0&0&0&0}\right]\rightarrow(\beta_1,\beta_2,\beta_3,\beta_4,\beta_5)=\left(\matrix{G\\0}\right)](https://math.jianshu.com/math?formula=%5Cleft%5B%5Cmatrix%7B2%261%266%261%260%5C%5C3%262%2610%261%260%5C%5C2%263%2610%26-1%263%5C%5C4%264%2616%260%261%7D%5Cright%5D%5Crightarrow%5Cleft%5B%5Cmatrix%7B2%261%266%261%260%5C%5C1%261%264%260%260%26%5C%5C0%262%264%26-2%263%5C%5C0%262%264%26-2%261%7D%5Cright%5D%5Crightarrow%5Cleft%5B%5Cmatrix%7B1%260%262%261%260%5C%5C0%261%262%26-1%260%5C%5C0%260%260%260%261%5C%5C0%260%260%260%260%7D%5Cright%5D%5Crightarrow(%5Cbeta_1%2C%5Cbeta_2%2C%5Cbeta_3%2C%5Cbeta_4%2C%5Cbeta_5)%3D%5Cleft(%5Cmatrix%7BG%5C%5C0%7D%5Cright))
其中![G=\left[\matrix{1&0&2&1&0\\0&1&2&-1&0\\0&0&0&0&1}\right]](https://math.jianshu.com/math?formula=G%3D%5Cleft%5B%5Cmatrix%7B1%260%262%261%260%5C%5C0%261%262%26-1%260%5C%5C0%260%260%260%261%7D%5Cright%5D)
是
行满秩矩阵；
线性无关，且
。由于初等行变换保持列向量线性组合关系，故
线性无关，且
。
取![F=(\alpha_1,\alpha_2,\alpha_5)=\left[\matrix{2&1&3\\3&2&0\\2&3&3\\4&4&1}\right]](https://math.jianshu.com/math?formula=F%3D(%5Calpha_1%2C%5Calpha_2%2C%5Calpha_5)%3D%5Cleft%5B%5Cmatrix%7B2%261%263%5C%5C3%262%260%5C%5C2%263%263%5C%5C4%264%261%7D%5Cright%5D)
是
列满秩矩阵，
则![FG=(\alpha_1,\alpha_2,\alpha_5)\left[\matrix{1&0&2&1&0\\0&1&2&-1&0\\0&0&0&0&1}\right]=(\alpha_1,\alpha_2,2\alpha_1+2\alpha_2,\alpha_1-\alpha_2,\alpha_5)=(\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5)=A](https://math.jianshu.com/math?formula=FG%3D(%5Calpha_1%2C%5Calpha_2%2C%5Calpha_5)%5Cleft%5B%5Cmatrix%7B1%260%262%261%260%5C%5C0%261%262%26-1%260%5C%5C0%260%260%260%261%7D%5Cright%5D%3D(%5Calpha_1%2C%5Calpha_2%2C2%5Calpha_1%2B2%5Calpha_2%2C%5Calpha_1-%5Calpha_2%2C%5Calpha_5)%3D(%5Calpha_1%2C%5Calpha_2%2C%5Calpha_3%2C%5Calpha_4%2C%5Calpha_5)%3DA)
定理二：（正交满秩分解定理）设是
阶实矩阵，的秩为，则存在列正交矩阵
和行满秩矩阵的阶矩阵
，使得
。其中，列正交含义为
。
