多元线性回归方程正规方程解（Normal Equation）公式推导详细过程

待我称王封你为后i 2022-10-20 03:48 170阅读 0赞

# 多元线性方程公式 #

定义多元线性方程的损失函数如下：  
 J ( θ ) = 1 2 m ∑ i = 1 m ( y ^ ( i ) − y ( i ) ) 2              ( 1 ) J(\\theta) = \\frac\{1\}\{2m\}\\sum\_\{i=1\}^m (\\hat\{y\}^\{(i)\} - y^\{(i)\})^2~~~~~~~~~~~~(1) J(θ)=2m1i=1∑m(y^(i)−y(i))2            (1)

其中， y ^ ( i ) \\hat\{y\}^\{(i)\} y^(i) 为：  
 y ^ ( i ) = θ 0 + θ 1 X 1 ( i ) + θ 2 X 2 ( i ) + ⋯ + θ n X n ( i )            ( 2 ) \\hat\{y\}^\{(i)\} = \\theta\_0 + \\theta\_1 X\_1^\{(i)\} + \\theta\_2 X\_2^\{(i)\} + \\cdots + \\theta\_n X\_n^\{(i)\}~~~~~~~~~~(2) y^(i)=θ0\+θ1X1(i)\+θ2X2(i)\+⋯\+θnXn(i)          (2)

其中， m m m 为样本个数， n n n 为特征数量

定义向量：  
 θ = ( θ 0 , θ 1 , ⋯   , θ n ) T X ( i ) = ( X 0 ( i ) , X 1 ( i ) , X 2 ( i ) , ⋯   , X n ( i ) ) T     ， 其 中 i = ( 1 , 2 , ⋯   , m ) ， X 0 ( i ) ≡ 1 X j = ( X j ( 1 ) , X j ( 2 ) , ⋯   , X j ( m ) ) T     ， 其 中 j = ( 0 , 1 , 2 , ⋯   , n ) y = ( y ( 1 ) , y ( 2 ) , ⋯   , y ( m ) ) \\begin\{aligned\} \\theta &=(\\theta\_0,\\theta\_1, \\cdots , \\theta\_n)^T \\\\\\\\ X^\{(i)\} &= (X\_0^\{(i)\}, X\_1^\{(i)\},X\_2^\{(i)\}, \\cdots, X\_n^\{(i)\})^T~~~，其中i=(1,2,\\cdots, m)，X\_0^\{(i)\}\\equiv1 \\\\\\\\ X\_j &= (X\_j^\{(1)\}, X\_j^\{(2)\}, \\cdots, X\_j^\{(m)\})^T~~~，其中j = (0, 1,2,\\cdots, n) \\\\\\\\ y &= (y^\{(1)\}, y^\{(2)\}, \\cdots, y^\{(m)\}) \\end\{aligned\} θX(i)Xjy=(θ0,θ1,⋯,θn)T=(X0(i),X1(i),X2(i),⋯,Xn(i))T   ，其中i=(1,2,⋯,m)，X0(i)≡1=(Xj(1),Xj(2),⋯,Xj(m))T   ，其中j=(0,1,2,⋯,n)=(y(1),y(2),⋯,y(m))

定义矩阵：

X = ( X ( 1 ) , X ( 2 ) , ⋯   , X ( m ) ) T = ( X 0 , X 1 , X 2 , ⋯   , X n ) = ( X 0 ( 1 ) X 1 ( 1 ) X 2 ( 1 ) ⋯ X n ( 1 ) X 0 ( 2 ) X 1 ( 2 ) X 2 ( 2 ) ⋯ X n ( 2 ) ⋯ ⋯ X 0 ( n ) X 1 ( m ) X 2 ( m ) ⋯ X n ( m ) ) m × ( n + 1 ) \\begin\{aligned\} X = (X^\{(1)\}, X^\{(2)\}, \\cdots, X^\{(m)\})^T = (X\_0, X\_1, X\_2, \\cdots, X\_n) = \\begin\{pmatrix\} X\_0^\{(1)\} & X\_1^\{(1)\} & X\_2^\{(1)\} & \\cdots X\_n^\{(1)\} \\\\\\\\ X\_0^\{(2)\} & X\_1^\{(2)\} & X\_2^\{(2)\} & \\cdots X\_n^\{(2)\} \\\\\\\\ \\cdots & & & \\cdots \\\\\\\\ X\_0^\{(n)\} & X\_1^\{(m)\} & X\_2^\{(m)\} & \\cdots X\_n^\{(m)\} \\\\ \\end\{pmatrix\}\_\{m \\times (n+1)\} \\end\{aligned\} X=(X(1),X(2),⋯,X(m))T=(X0,X1,X2,⋯,Xn)=⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛X0(1)X0(2)⋯X0(n)X1(1)X1(2)X1(m)X2(1)X2(2)X2(m)⋯Xn(1)⋯Xn(2)⋯⋯Xn(m)⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞m×(n\+1)

**当  θ \\theta θ 取下值时，损失函数最小**：  
 θ = ( X T X ) − 1 X T y \\theta = (X^T X)^\{-1\} X^T y θ=(XTX)−1XTy

# 公式推导 #

> 温馨提示：公式推导过程不难，但很绕，请耐心…

将(2)式代入(1)式得：

J ( θ ) = 1 2 m ∑ i = 1 m ( θ 0 + θ 1 X 1 ( i ) + θ 2 X 2 ( i ) + ⋯ + θ n X n ( i ) − y ( i ) ) 2 = 1 2 m ∑ i = 1 m ( θ 0 X 0 ( i ) + θ 1 X 1 ( i ) + θ 2 X 2 ( i ) + ⋯ + θ n X n ( i ) − y ( i ) ) 2     （ 补 个 X 0 ( i ) ） = 1 2 m \[ ( θ 0 X 0 ( 1 ) + θ 1 X 1 ( 1 ) + θ 2 X 2 ( 1 ) + ⋯ + θ n X n ( 1 ) − y ( 1 ) ) 2          + ( θ 0 X 0 ( 2 ) + θ 1 X 1 ( 2 ) + θ 2 X 2 ( 2 ) + ⋯ + θ n X n ( 2 ) − y ( 2 ) ) 2          + ⋯          + ( θ 0 X 0 ( m ) + θ 1 X 1 ( 2 ) + θ 2 X 2 ( m ) + ⋯ + θ n X n ( m ) − y ( m ) ) 2 \] \\begin\{aligned\} J(\\theta) & = \\frac\{1\}\{2m\}\\sum\_\{i=1\}^m (\\theta\_0 + \\theta\_1 X\_1^\{(i)\} + \\theta\_2 X\_2^\{(i)\} + \\cdots + \\theta\_n X\_n^\{(i)\} - y^\{(i)\})^2 \\\\\\\\ & = \\frac\{1\}\{2m\}\\sum\_\{i=1\}^m (\\theta\_0X\_0^\{(i)\} + \\theta\_1 X\_1^\{(i)\} + \\theta\_2 X\_2^\{(i)\} + \\cdots + \\theta\_n X\_n^\{(i)\} - y^\{(i)\})^2 ~~~（补个X\_0^\{(i)\}）\\\\\\\\ &= \\frac\{1\}\{2m\}\[ (\\theta\_0X\_0^\{(1)\} + \\theta\_1X\_1^\{(1)\} + \\theta\_2X\_2^\{(1)\} + \\cdots + \\theta\_nX\_n^\{(1)\} - y^\{(1)\})^2 \\\\\\\\ &~~~~~~~~+ (\\theta\_0X\_0^\{(2)\} + \\theta\_1X\_1^\{(2)\}+ \\theta\_2X\_2^\{(2)\} + \\cdots + \\theta\_nX\_n^\{(2)\}- y^\{(2)\})^2 \\\\\\\\ &~~~~~~~~+ \\cdots \\\\\\\\ &~~~~~~~~+ (\\theta\_0X\_0^\{(m)\} + \\theta\_1X\_1^\{(2)\}+ \\theta\_2X\_2^\{(m)\} + \\cdots + \\theta\_nX\_n^\{(m)\}- y^\{(m)\})^2 \] \\end\{aligned\} J(θ)=2m1i=1∑m(θ0\+θ1X1(i)\+θ2X2(i)\+⋯\+θnXn(i)−y(i))2=2m1i=1∑m(θ0X0(i)\+θ1X1(i)\+θ2X2(i)\+⋯\+θnXn(i)−y(i))2   （补个X0(i)）=2m1\[(θ0X0(1)\+θ1X1(1)\+θ2X2(1)\+⋯\+θnXn(1)−y(1))2        \+(θ0X0(2)\+θ1X1(2)\+θ2X2(2)\+⋯\+θnXn(2)−y(2))2        \+⋯        \+(θ0X0(m)\+θ1X1(2)\+θ2X2(m)\+⋯\+θnXn(m)−y(m))2\]

现在对  θ \\theta θ 求偏导，下面只对  θ 1 \\theta\_1 θ1 求偏导，其他的依次类推：

∂ ∂ θ 1 J ( θ ) = 1 m \[ X 1 ( 1 ) ( θ 0 X 0 ( 1 ) + θ 1 X 1 ( 1 ) + θ 2 X 2 ( 1 ) + ⋯ + θ n X n ( 1 ) − y ( 1 ) )        + X 1 ( 2 ) ( θ 0 X 0 ( 2 ) + θ 1 X 1 ( 2 ) + θ 2 X 2 ( 2 ) + ⋯ + θ n X n ( 2 ) − y ( 2 ) )        + ⋯        + X 1 ( m ) ( θ 0 X 0 ( m ) + θ 1 X 1 ( 2 ) + θ 2 X 2 ( m ) + ⋯ + θ n X n ( m ) − y ( m ) ) \] = 1 m \[ ( θ 0 X 0 ( 1 ) X 1 ( 1 ) + θ 1 X 1 ( 1 ) X 1 ( 1 ) + θ 2 X 2 ( 1 ) X 1 ( 1 ) + ⋯ + θ n X n ( 1 ) X 1 ( 1 ) − y ( 1 ) X 1 ( 1 ) )        + ( θ 0 X 0 ( 2 ) X 1 ( 2 ) + θ 1 X 1 ( 2 ) X 1 ( 2 ) + θ 2 X 2 ( 2 ) X 1 ( 2 ) + ⋯ + θ n X n ( 2 ) X 1 ( 2 ) − y ( 2 ) X 1 ( 2 ) )        + ⋯        + ( θ 0 X 0 ( m ) X 1 ( m ) + θ 1 X 1 ( 2 ) X 1 ( m ) + θ 2 X 2 ( m ) X 1 ( m ) + ⋯ + θ n X n ( m ) − y ( m ) X 1 ( m ) ) \]       （ 把 X 1 ( i ) 乘 进 去 ） = 1 m \[ ( X 0 ( 1 ) X 1 ( 1 ) + X 0 ( 2 ) X 1 ( 2 ) + ⋯ + X 0 ( m ) X 1 ( m ) ) ⋅ θ 0        + ( X 1 ( 1 ) X 1 ( 1 ) + X 1 ( 2 ) X 1 ( 2 ) + ⋯ + X 1 ( m ) X 1 ( m ) ) ⋅ θ 1        + ( X 2 ( 1 ) X 1 ( 1 ) + X 2 ( 2 ) X 1 ( 2 ) + ⋯ + X 2 ( m ) X 1 ( m ) ) ⋅ θ 2        + ⋯        + ( X n ( 1 ) X 1 ( 1 ) + X n ( 2 ) X 1 ( 2 ) + ⋯ + X n ( m ) X 1 ( m ) ) ⋅ θ n        − ( y ( 1 ) X 1 ( 1 ) + y ( 2 ) X 1 ( 2 ) ) + ⋯ + y ( m ) X 1 ( m ) \]           （ 将 θ 提 出 来 ） = 1 m \[ θ 0 ⋅ ∑ i = 1 m X 0 ( i ) X 1 ( i ) + θ 1 ⋅ ∑ i = 1 m X 1 ( i ) X 1 ( i ) + θ 2 ⋅ ∑ i = 1 m X 2 ( i ) X 1 ( i ) + ⋯ + θ n ⋅ ∑ i = 1 m X n ( i ) X 1 ( i ) − ∑ i = 1 m y ( i ) X 1 ( i ) \] \\begin\{aligned\} \\frac\{\\partial\}\{\\partial\\theta\_1\} J(\\theta) & = \\frac\{1\}\{m\}\[ X\_1^\{(1)\} (\\theta\_0X\_0^\{(1)\} + \\theta\_1X\_1^\{(1)\} + \\theta\_2X\_2^\{(1)\} + \\cdots + \\theta\_nX\_n^\{(1)\} - y^\{(1)\}) \\\\\\\\ &~~~~~~ + X\_1^\{(2)\}(\\theta\_0X\_0^\{(2)\} + \\theta\_1X\_1^\{(2)\}+ \\theta\_2X\_2^\{(2)\} + \\cdots + \\theta\_nX\_n^\{(2)\}- y^\{(2)\}) \\\\\\\\ &~~~~~~+ \\cdots \\\\\\\\ &~~~~~~+ X\_1^\{(m)\}(\\theta\_0X\_0^\{(m)\} + \\theta\_1X\_1^\{(2)\}+ \\theta\_2X\_2^\{(m)\} + \\cdots + \\theta\_nX\_n^\{(m)\}- y^\{(m)\}) \] \\\\\\\\\\\\ & = \\frac\{1\}\{m\}\[(\\theta\_0X\_0^\{(1)\}X\_1^\{(1)\} + \\theta\_1X\_1^\{(1)\}X\_1^\{(1)\} + \\theta\_2X\_2^\{(1)\}X\_1^\{(1)\} + \\cdots + \\theta\_nX\_n^\{(1)\}X\_1^\{(1)\} - y^\{(1)\}X\_1^\{(1)\}) \\\\\\\\ &~~~~~~ + (\\theta\_0X\_0^\{(2)\}X\_1^\{(2)\} + \\theta\_1X\_1^\{(2)\}X\_1^\{(2)\}+ \\theta\_2X\_2^\{(2)\}X\_1^\{(2)\} + \\cdots + \\theta\_nX\_n^\{(2)\}X\_1^\{(2)\}- y^\{(2)\}X\_1^\{(2)\}) \\\\\\\\ &~~~~~~+ \\cdots \\\\\\\\ &~~~~~~+ (\\theta\_0X\_0^\{(m)\}X\_1^\{(m)\} + \\theta\_1X\_1^\{(2)\}X\_1^\{(m)\}+ \\theta\_2X\_2^\{(m)\}X\_1^\{(m)\} + \\cdots + \\theta\_nX\_n^\{(m)\}- y^\{(m)\}X\_1^\{(m)\}) \] ~~~~~（把 X\_1^\{(i)\} 乘进去） \\\\\\\\\\\\ & = \\frac\{1\}\{m\} \[ (X\_0^\{(1)\}X\_1^\{(1)\} + X\_0^\{(2)\}X\_1^\{(2)\} + \\cdots + X\_0^\{(m)\}X\_1^\{(m)\}) \\cdot \\theta\_0 \\\\\\\\ & ~~~~~~+(X\_1^\{(1)\}X\_1^\{(1)\} + X\_1^\{(2)\}X\_1^\{(2)\} + \\cdots + X\_1^\{(m)\}X\_1^\{(m)\}) \\cdot \\theta\_1 \\\\\\\\ & ~~~~~~+(X\_2^\{(1)\}X\_1^\{(1)\} + X\_2^\{(2)\}X\_1^\{(2)\} + \\cdots + X\_2^\{(m)\}X\_1^\{(m)\}) \\cdot \\theta\_2 \\\\\\\\ & ~~~~~~+ \\cdots \\\\\\\\ & ~~~~~~+ (X\_n^\{(1)\}X\_1^\{(1)\} + X\_n^\{(2)\}X\_1^\{(2)\} + \\cdots + X\_n^\{(m)\}X\_1^\{(m)\}) \\cdot \\theta\_n\\\\\\\\ & ~~~~~~- (y^\{(1)\} X\_1^\{(1)\} + y^\{(2)\} X\_1^\{(2)\}) + \\cdots + y^\{(m)\} X\_1^\{(m)\}\] ~~~~~~~~~（将\\theta 提出来）\\\\\\\\\\\\ & = \\frac\{1\}\{m\} \[\\theta\_0 \\cdot \\sum\_\{i=1\}^m X\_0^\{(i)\}X\_1^\{(i)\} + \\theta\_1 \\cdot \\sum\_\{i=1\}^m X\_1^\{(i)\}X\_1^\{(i)\} + \\theta\_2 \\cdot \\sum\_\{i=1\}^m X\_2^\{(i)\}X\_1^\{(i)\} + \\cdots + \\theta\_n \\cdot \\sum\_\{i=1\}^m X\_n^\{(i)\}X\_1^\{(i)\} - \\sum\_\{i=1\}^m y^\{(i)\}X\_1^\{(i)\}\] \\end\{aligned\} ∂θ1∂J(θ)=m1\[X1(1)(θ0X0(1)\+θ1X1(1)\+θ2X2(1)\+⋯\+θnXn(1)−y(1))      \+X1(2)(θ0X0(2)\+θ1X1(2)\+θ2X2(2)\+⋯\+θnXn(2)−y(2))      \+⋯      \+X1(m)(θ0X0(m)\+θ1X1(2)\+θ2X2(m)\+⋯\+θnXn(m)−y(m))\]=m1\[(θ0X0(1)X1(1)\+θ1X1(1)X1(1)\+θ2X2(1)X1(1)\+⋯\+θnXn(1)X1(1)−y(1)X1(1))      \+(θ0X0(2)X1(2)\+θ1X1(2)X1(2)\+θ2X2(2)X1(2)\+⋯\+θnXn(2)X1(2)−y(2)X1(2))      \+⋯      \+(θ0X0(m)X1(m)\+θ1X1(2)X1(m)\+θ2X2(m)X1(m)\+⋯\+θnXn(m)−y(m)X1(m))\]     （把X1(i)乘进去）=m1\[(X0(1)X1(1)\+X0(2)X1(2)\+⋯\+X0(m)X1(m))⋅θ0      \+(X1(1)X1(1)\+X1(2)X1(2)\+⋯\+X1(m)X1(m))⋅θ1      \+(X2(1)X1(1)\+X2(2)X1(2)\+⋯\+X2(m)X1(m))⋅θ2      \+⋯      \+(Xn(1)X1(1)\+Xn(2)X1(2)\+⋯\+Xn(m)X1(m))⋅θn      −(y(1)X1(1)\+y(2)X1(2))\+⋯\+y(m)X1(m)\]         （将θ提出来）=m1\[θ0⋅i=1∑mX0(i)X1(i)\+θ1⋅i=1∑mX1(i)X1(i)\+θ2⋅i=1∑mX2(i)X1(i)\+⋯\+θn⋅i=1∑mXn(i)X1(i)−i=1∑my(i)X1(i)\]

我们先对  ∑ i = 1 m X a ( i ) X b ( i ) \\sum\_\{i=1\}^m X\_a^\{(i)\}X\_b^\{(i)\} ∑i=1mXa(i)Xb(i) 做下研究：  
 ∑ i = 1 m X a ( i ) X b ( i ) = ( X a ( 1 ) , X a ( 2 ) , ⋯   , X a ( n ) ) ⋅ ( X b ( 1 ) X b ( 2 ) ⋯ X b ( n ) ) = X a T ⋅ X b = X b T ⋅ X a             ( 3 ) \\sum\_\{i=1\}^m X\_a^\{(i)\}X\_b^\{(i)\} = (X\_a^\{(1)\}, X\_a^\{(2)\}, \\cdots , X\_a^\{(n)\}) \\cdot \\begin\{pmatrix\} X\_b^\{(1)\} \\\\\\\\ X\_b^\{(2)\}\\\\\\\\ \\cdots\\\\\\\\ X\_b^\{(n)\}\\\\ \\end\{pmatrix\} = X\_a^T \\cdot X\_b = X\_b^T\\cdot X\_a ~~~~~~~~~~~(3) i=1∑mXa(i)Xb(i)=(Xa(1),Xa(2),⋯,Xa(n))⋅⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛Xb(1)Xb(2)⋯Xb(n)⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞=XaT⋅Xb=XbT⋅Xa           (3)

将(3)式代入  ∂ ∂ θ 1 J ( θ ) \\frac\{\\partial\}\{\\partial\\theta\_1\} J(\\theta) ∂θ1∂J(θ) 得

∂ ∂ θ 1 J ( θ ) = 1 m \[ θ 0 ⋅ ∑ i = 1 m X 0 ( i ) X 1 ( i ) + θ 1 ⋅ ∑ i = 1 m X 1 ( i ) X 1 ( i ) + θ 2 ⋅ ∑ i = 1 m X 2 ( i ) X 1 ( i ) + ⋯ + θ n ⋅ ∑ i = 1 m X n ( i ) X 1 ( i ) − ∑ i = 1 m y ( i ) X 1 ( i ) \] = 1 m ( X 0 T ⋅ X 1 ⋅ θ 0 + X 1 T ⋅ X 1 ⋅ θ 1 + X 2 T ⋅ X 1 ⋅ θ 2 + ⋯ + X n T ⋅ X 1 ⋅ θ n − X 1 T ⋅ y ) \\begin\{aligned\} \\frac\{\\partial\}\{\\partial\\theta\_1\} J(\\theta) & = \\frac\{1\}\{m\} \[\\theta\_0 \\cdot \\sum\_\{i=1\}^m X\_0^\{(i)\}X\_1^\{(i)\} + \\theta\_1 \\cdot \\sum\_\{i=1\}^m X\_1^\{(i)\}X\_1^\{(i)\} + \\theta\_2 \\cdot \\sum\_\{i=1\}^m X\_2^\{(i)\}X\_1^\{(i)\} + \\cdots + \\theta\_n \\cdot \\sum\_\{i=1\}^m X\_n^\{(i)\}X\_1^\{(i)\} - \\sum\_\{i=1\}^m y^\{(i)\}X\_1^\{(i)\}\] \\\\\\\\ & = \\frac\{1\}\{m\} (X\_0^T \\cdot X\_1 \\cdot \\theta\_0 + X\_1^T \\cdot X\_1 \\cdot \\theta\_1 + X\_2^T \\cdot X\_1 \\cdot \\theta\_2 +\\cdots + X\_n^T \\cdot X\_1 \\cdot \\theta\_n - X\_1^T\\cdot y) \\\\\\\\ \\end\{aligned\} ∂θ1∂J(θ)=m1\[θ0⋅i=1∑mX0(i)X1(i)\+θ1⋅i=1∑mX1(i)X1(i)\+θ2⋅i=1∑mX2(i)X1(i)\+⋯\+θn⋅i=1∑mXn(i)X1(i)−i=1∑my(i)X1(i)\]=m1(X0T⋅X1⋅θ0\+X1T⋅X1⋅θ1\+X2T⋅X1⋅θ2\+⋯\+XnT⋅X1⋅θn−X1T⋅y)

令  ∂ ∂ θ 1 J ( θ ) = 0 \\frac\{\\partial\}\{\\partial\\theta\_1\} J(\\theta)=0 ∂θ1∂J(θ)=0 ，得到如下等式：

( X 0 T X 1 , X 1 T X 1 , X 2 T X 1 ⋯   , X n T X 1 ) ⋅ ( θ 0 θ 1 ⋯ θ n ) = X 1 T ⋅ y (X\_0^T X\_1, X\_1^T X\_1, X\_2^T X\_1 \\cdots, X\_n^T X\_1) \\cdot \\begin\{pmatrix\} \\theta\_0 \\\\\\\\ \\theta\_1 \\\\\\\\ \\cdots\\\\\\\\ \\theta\_n\\\\ \\end\{pmatrix\} = X\_1^T\\cdot y (X0TX1,X1TX1,X2TX1⋯,XnTX1)⋅⎝⎜⎜⎜⎜⎜⎜⎜⎜⎛θ0θ1⋯θn⎠⎟⎟⎟⎟⎟⎟⎟⎟⎞=X1T⋅y

与  ∂ ∂ θ 1 J ( θ ) = 0 \\frac\{\\partial\}\{\\partial\\theta\_1\} J(\\theta)=0 ∂θ1∂J(θ)=0 同理，对  θ 0 , θ 2 , θ 3 , ⋯   , θ n \\theta\_0, \\theta\_2, \\theta\_3, \\cdots, \\theta\_n θ0,θ2,θ3,⋯,θn 做偏导等零，最终得：

( X 0 T X 0 , X 1 T X 0 , X 2 T X 0 ⋯   , X n T X 0 X 0 T X 1 , X 1 T X 1 , X 2 T X 1 ⋯   , X n T X 1 X 0 T X 2 , X 1 T X 2 , X 2 T X 2 ⋯   , X n T X 2 ⋯ X 0 T X n , X 1 T X n , X 2 T X n ⋯   , X n T X n ) ⋅ ( θ 0 θ 1 θ 2 ⋯ θ n ) = ( X 0 T ⋅ y X 1 T ⋅ y X 2 T ⋅ y ⋯ X n T ⋅ y ) \\begin\{pmatrix\} X\_0^T X\_0, X\_1^T X\_0, X\_2^T X\_0 \\cdots, X\_n^T X\_0 \\\\\\\\ X\_0^T X\_1, X\_1^T X\_1, X\_2^T X\_1 \\cdots, X\_n^T X\_1 \\\\\\\\ X\_0^T X\_2, X\_1^T X\_2, X\_2^T X\_2 \\cdots, X\_n^T X\_2 \\\\\\\\ \\cdots\\\\\\\\ X\_0^T X\_n, X\_1^T X\_n, X\_2^T X\_n \\cdots, X\_n^T X\_n\\\\ \\end\{pmatrix\} \\cdot \\begin\{pmatrix\} \\theta\_0 \\\\\\\\ \\theta\_1 \\\\\\\\ \\theta\_2 \\\\\\\\ \\cdots\\\\\\\\ \\theta\_n\\\\ \\end\{pmatrix\} = \\begin\{pmatrix\} X\_0^T\\cdot y \\\\\\\\ X\_1^T\\cdot y \\\\\\\\ X\_2^T\\cdot y \\\\\\\\ \\cdots\\\\\\\\ X\_n^T\\cdot y\\\\ \\end\{pmatrix\} ⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛X0TX0,X1TX0,X2TX0⋯,XnTX0X0TX1,X1TX1,X2TX1⋯,XnTX1X0TX2,X1TX2,X2TX2⋯,XnTX2⋯X0TXn,X1TXn,X2TXn⋯,XnTXn⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞⋅⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛θ0θ1θ2⋯θn⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞=⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛X0T⋅yX1T⋅yX2T⋅y⋯XnT⋅y⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞

左右两边进行处理，得（由(3)式子知  X a T ⋅ X b = X b T ⋅ X a X\_a^T \\cdot X\_b = X\_b^T\\cdot X\_a XaT⋅Xb=XbT⋅Xa，这里交换了一下）：  
 ( X 0 T X 1 T X 2 T ⋯ X n T ) ⋅ ( X 0 , X 1 , X 2 , ⋯   , X n ) ⋅ ( θ 0 θ 1 θ 2 ⋯ θ n ) = ( X 0 T ⋅ y X 1 T ⋅ y X 2 T ⋅ y ⋯ X n T ⋅ y ) \\begin\{pmatrix\} X\_0^T \\\\\\\\ X\_1^T \\\\\\\\ X\_2^T \\\\\\\\ \\cdots\\\\\\\\ X\_n^T\\\\ \\end\{pmatrix\} \\cdot (X\_0, X\_1, X\_2, \\cdots, X\_n) \\cdot \\begin\{pmatrix\} \\theta\_0 \\\\\\\\ \\theta\_1 \\\\\\\\ \\theta\_2 \\\\\\\\ \\cdots\\\\\\\\ \\theta\_n\\\\ \\end\{pmatrix\} = \\begin\{pmatrix\} X\_0^T\\cdot y \\\\\\\\ X\_1^T\\cdot y \\\\\\\\ X\_2^T\\cdot y \\\\\\\\ \\cdots\\\\\\\\ X\_n^T\\cdot y\\\\ \\end\{pmatrix\} ⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛X0TX1TX2T⋯XnT⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞⋅(X0,X1,X2,⋯,Xn)⋅⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛θ0θ1θ2⋯θn⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞=⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛X0T⋅yX1T⋅yX2T⋅y⋯XnT⋅y⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞

将上式继续变换，得：  
 X T ⋅ X ⋅ θ = X T ⋅ y X^T \\cdot X \\cdot \\theta = X^T \\cdot y XT⋅X⋅θ=XT⋅y

两边同时乘  ( X T ⋅ X ) − 1 (X^T \\cdot X)^\{-1\} (XT⋅X)−1 得最终结果：  
 θ = ( X T ⋅ X ) − 1 ⋅ X T ⋅ y \\theta = (X^T \\cdot X)^\{-1\} \\cdot X^T\\cdot y θ=(XT⋅X)−1⋅XT⋅y

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# 参考资料 #

[考研必备数学公式大全][Link 1]: https://blog.csdn.net/zhaohongfei\_358/article/details/106039576

[机器学习纸上谈兵之线性回归][Link 2]: https://blog.csdn.net/zhaohongfei\_358/article/details/117967229

[Link 1]: https://blog.csdn.net/zhaohongfei_358/article/details/106039576
[Link 2]: https://blog.csdn.net/zhaohongfei_358/article/details/117967229