秩一矩阵的分解

$$a=\left[\begin{array}{c}{a}_1\\ ⋮\\ a_n\end{array}\right],b=\left[\begin{array}{c}{b}_1\\ ⋮\\ b_n\end{array}\right]$$
$$a^{T}b=\sum _{i=1}^n{a}_i{b}_i,a{b}^T=\left[\begin{array}{ccc}{a}_1{b}_1& \cdots & a_1{b}_n\\ \cdots & \cdots & \cdots \\ a_n{b}_1& \cdots & a_n{b}_n\end{array}\right],$$
$a{b}^T$的秩为1，其所有行或列都线性相关，只有一个非零特征值：$\lambda (a{b}^T)=a^{T}b$.

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列变换

$$\left[\begin{array}{ccc}{a}_1{b}_1& \cdots & a_1{b}_n\\ \cdots & \cdots & \cdots \\ a_n{b}_1& \cdots & a_n{b}_n\end{array}\right]\left[\begin{array}{ccccc}1& -\frac{b_2}{b_1}& -\frac{b_3}{b_1}& \cdots & -\frac{b_n}{b_1}\\ & 1& 0& \cdots & 0\\ & & \ddots & \ddots & ⋮\\ & & & 1& 0\\ & & & & 1\end{array}\right]=\left[\begin{array}{ccc}{a}_1{b}_1& \cdots & 0\\ \cdots & \cdots & \cdots \\ a_n{b}_1& \cdots & 0\end{array}\right]$$

行变换

$$\left[\begin{array}{ccccc}1& & & & \\ -\frac{a_2}{a_1}& 1& & & \\ -\frac{a_3}{a_1}& 0& 1& & \\ ⋮& ⋮& \ddots & \ddots & \\ -\frac{a_n}{a_1}& 0& \cdots & 0& 1\end{array}\right]\left[\begin{array}{ccc}{a}_1{b}_1& \cdots & 0\\ \cdots & \cdots & \cdots \\ a_n{b}_1& \cdots & 0\end{array}\right]=\left[\begin{array}{cccc}{a}_1{b}_1& 0& \cdots & 0\\ 0& 0& \cdots & 0\\ \cdots & \cdots & \cdots & \cdots \\ 0& 0& \cdots & 0\end{array}\right]$$
记
$$P_{a1}=\left[\begin{array}{ccccc}1& & & & \\ -\frac{a_2}{a_1}& 1& & & \\ -\frac{a_3}{a_1}& 0& 1& & \\ ⋮& ⋮& \ddots & \ddots & \\ -\frac{a_n}{a_1}& 0& \cdots & 0& 1\end{array}\right],P_{b1}=\left[\begin{array}{ccccc}1& -\frac{b_2}{b_1}& -\frac{b_3}{b_1}& \cdots & -\frac{b_n}{b_1}\\ & 1& 0& \cdots & 0\\ & & \ddots & \ddots & ⋮\\ & & & 1& 0\\ & & & & 1\end{array}\right]$$

特征分解与对角化

$$(a{b}^T)a=a(b^{T}a)=(a^{T}b)a$$
说明 $a$ 是特征值 $a^{T}b$ 对应的特征向量，剩下 $n-1$个特征向量只要满足和 $b$正交即可！