机器学习笔记_数学基础_5-矩阵理论_续1_QR分解

矩阵的QR分解

实非奇异矩阵(满秩矩阵)A能够分解为正交矩阵Q和实非奇异上三角矩阵R的乘积

证明：令AA的n个列向量a1,⋯,ana_1,\cdots,a_n, 因为A非奇异，=>列向量线性无关

则列向量的施密特正交化可得，n个标准的正交列向量 q1,⋯,qNq_1,\cdots,q_N

证明：

令 A的列向量为aia_i => 对aia_i正交化得

⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪b1b2⋮bn=a1=a2−k21b1=an−kn,n−1bn−1−⋯−kn1b1 \left\{
\begin{aligned}
b_1 & = a_1 \\
b_2 & = a_2-k_{21}b_1 \\
\vdots \\
b_n &=a_n-k_{n,n-1}b_{n-1}-\cdots-k_{n1}b_1
\end{aligned}
\right.

其中kij=(ai,bj)(bj,bJ)k_{ij}=\frac{(a_i,b_j)}{(b_j,b_J)}

=>=> ⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪a1a2⋮an=b1=k21b1+b2=kn1b1+kn2b2+⋯+kn,n−1bn−1+bn \left\{
\begin{aligned}
a_1 & = b_1 \\
a_2 & = k_{21}b_1+b_2 \\
\vdots \\
a_n &=k_{n1}b_1+k_{n2}b_2+\cdots+k_{n,n-1}b_{n-1}+b_n
\end{aligned}
\right.

=> (a1,⋯,an)=(b1,⋯,bn)C(a_1,\cdots,a_n)=(b_1,\cdots,b_n)C

其中 C=⎡⎣⎢⎢⎢⎢⎢1k211⋯⋯⋱kn1kn2⋮1⎤⎦⎥⎥⎥⎥⎥C=\begin{bmatrix} 1 & k_{21} & \cdots &k_{n1} \\ & 1 &\cdots & k_{n2} \\ & & \ddots & \vdots \\ & & & 1\end{bmatrix}\quad

对(b1,b2,⋯,bn)单位化=>(b_1,b_2,\cdots,b_n)单位化 => Qi=1|bi|biQ_i=\frac{1}{|b_i|}b_i

=>=>\quad (a1,a2,⋯,an)=(b1,⋯,cn)C=(q1,⋯,qn)⎡⎣⎢⎢⎢⎢⎢|b1||b2|⋱|bn|⎤⎦⎥⎥⎥⎥⎥C(a_1,a_2,\cdots,a_n)=(b_1,\cdots,c_n)C=(q_1,\cdots,q_n)\begin{bmatrix} |b_1| & & & \\ & |b_2| & & \\ & & \ddots & \\ & & & |b_n|\end{bmatrix}C

=>=>\quad Q=(q1,⋯,qn)Q=(q_1,\cdots,q_n)

=>=>\quad R=diag(|b1|,⋯,|bn|)CR=diag(|b_1|,\cdots,|b_n|)C

C=⎡⎣⎢⎢⎢⎢⎢1k211⋯⋯⋱kn1kn2⋮1⎤⎦⎥⎥⎥⎥⎥C=\begin{bmatrix} 1 & k_{21} & \cdots &k_{n1} \\ & 1 &\cdots & k_{n2} \\ & & \ddots & \vdots \\ & & & 1\end{bmatrix}\quad

例题： 求矩阵 A=⎡⎣⎢121212121⎤⎦⎥A=\begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 &2 \\1 & 2 & 1 \end{bmatrix}\quad的QR分解

解：

令 a1=(1,2,1)Ta_1=(1,2,1)^T,a2=(2,1,2)Ta_2=(2,1,2)^T,a3=(2,2,1)Ta_3=(2,2,1)^T => 正交化可得

b1=a1=(1,2,1)Tb_1=a_1=(1,2,1)^T;b2=a2−b1=(1,−1,1)Tb_2=a_2-b_1=(1,-1,1)^T;b3=a3−13−76=(12,0,−12)Tb_3=a_3-\frac{1}{3}-\frac{7}{6}=(\frac{1}{2},0,-\frac{1}{2})^T

=> 构造矩阵Q=> Q=⎡⎣⎢⎢⎢⎢16√13√12√26√−13√016√13√−12√⎤⎦⎥⎥⎥⎥Q=\begin{bmatrix} \frac{1}{\sqrt{6}} & \frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} &\frac{1}{\sqrt{3}} \\\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \end{bmatrix}

=> R=Q=⎡⎣⎢⎢6√3√12√⎤⎦⎥⎥Q=\begin{bmatrix} \sqrt{6} & & \\ & \sqrt{3} & \\ & & \frac{1}{\sqrt{2}} \end{bmatrix}⎡⎣⎢⎢11176131⎤⎦⎥⎥\begin{bmatrix} 1&1& \frac{7}{6} \\ &1 &\frac{1}{3} \\ & & 1\end{bmatrix}