《线性代数入门》，梁鑫等编著，清华大学出版社，2022 年.
考虑其中任意一个向量 a \boldsymbol{a} a a \boldsymbol{a} a b \boldsymbol{b} b a = [ x y ] \boldsymbol{a}=\begin{bmatrix}x\\y\\\end{bmatrix} a = [ x y ] b = [ x cos π 2 − y sin π 2 x sin π 2 + y cos π 2 ] = [ − x − y ] \boldsymbol{b}=\begin{bmatrix}x\cos{\cfrac{\pi}{2}}-y\sin{\cfrac{\pi}{2}}\\x\sin{\cfrac{\pi}{2}}+y\cos{\cfrac{\pi}{2}}\\\end{bmatrix}=\begin{bmatrix}-x\\-y\\\end{bmatrix} b = ⎣ ⎢ ⎡ x cos 2 π − y sin 2 π x sin 2 π + y cos 2 π ⎦ ⎥ ⎤ = [ − x − y ] a + b = [ x y ] + [ − x − y ] = 0 \boldsymbol{a}+\boldsymbol{b}=\begin{bmatrix}x\\y\\\end{bmatrix}+\begin{bmatrix}-x\\-y\\\end{bmatrix}=\boldsymbol{0} a + b = [ x y ] + [ − x − y ] = 0 0 \boldsymbol{0} 0
按照 1 中所述可知，余下 11 个向量之和为与 2 点方向相反方向的向量，即 8 点方向向量.由 12 点方向向量，可表示 8 点方向向量为 [ − sin π 3 cos π 3 ] = [ − 3 2 − 1 2 ] \begin{bmatrix}-\sin{\cfrac{\pi}{3}}\\\cos{\cfrac{\pi}{3}}\\\end{bmatrix}=\begin{bmatrix}-\cfrac{\sqrt{3}}{2}\\-\cfrac{1}{2}\\\end{bmatrix} ⎣ ⎢ ⎡ − sin 3 π cos 3 π ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ − 2 3 − 2 1 ⎦ ⎥ ⎤ [ − 3 2 − 1 2 ] \begin{bmatrix}-\cfrac{\sqrt{3}}{2}\\-\cfrac{1}{2}\\\end{bmatrix} ⎣ ⎢ ⎡ − 2 3 − 2 1 ⎦ ⎥ ⎤
设原有向量为 [ x 1 y 1 ] , [ x 2 y 2 ] , ⋯ , [ x 12 y 12 ] \begin{bmatrix}x_1\\y_1\\\end{bmatrix},\begin{bmatrix}x_2\\y_2\\\end{bmatrix},\cdots ,\begin{bmatrix}x_{12}\\y_{12}\\\end{bmatrix} [ x 1 y 1 ] , [ x 2 y 2 ] , ⋯ , [ x 1 2 y 1 2 ] ∑ i = 1 12 [ x i y i ] = [ x 1 + x 2 + ⋯ + x 12 y 1 + y 2 + ⋯ + y 12 ] = 0 \sum\limits_{i=1}^{12}\begin{bmatrix}x_i\\y_i\\\end{bmatrix}=\begin{bmatrix}x_1+x_2+\cdots +x_{12}\\y_1+y_2+\cdots +y_{12}\\\end{bmatrix}=\boldsymbol{0} i = 1 ∑ 1 2 [ x i y i ] = [ x 1 + x 2 + ⋯ + x 1 2 y 1 + y 2 + ⋯ + y 1 2 ] = 0 y y y ∑ i = 1 12 [ x i y i + 1 ] = [ x 1 + x 2 + ⋯ + x 12 y 1 + y 2 + ⋯ + y 12 + 12 ] = [ 0 12 ] \sum\limits_{i=1}^{12}\begin{bmatrix}x_i\\y_i+1\\\end{bmatrix}=\begin{bmatrix}x_1+x_2+\cdots +x_{12}\\y_1+y_2+\cdots +y_{12}+12\\\end{bmatrix}=\begin{bmatrix}0\\12\\\end{bmatrix} i = 1 ∑ 1 2 [ x i y i + 1 ] = [ x 1 + x 2 + ⋯ + x 1 2 y 1 + y 2 + ⋯ + y 1 2 + 1 2 ] = [ 0 1 2 ]
共线.
证明 既然 [ a b ] \begin{bmatrix}a\\b\\\end{bmatrix} [ a b ] [ c d ] \begin{bmatrix}c\\d\\\end{bmatrix} [ c d ] [ a b ] = k [ c d ] \begin{bmatrix}a\\b\\\end{bmatrix}=k\begin{bmatrix}c\\d\\\end{bmatrix} [ a b ] = k [ c d ] k ∈ R k\in\R k ∈ R a = k c a=kc a = k c b = k d b=kd b = k d m = c d ∈ R m=\cfrac{c}{d}\in\R m = d c ∈ R [ a c ] = [ k c c ] = m [ k d d ] = m [ b d ] \begin{bmatrix}a\\c\\\end{bmatrix}=\begin{bmatrix}kc\\c\\\end{bmatrix}=m\begin{bmatrix}kd\\d\\\end{bmatrix}=m\begin{bmatrix}b\\d\\\end{bmatrix} [ a c ] = [ k c c ] = m [ k d d ] = m [ b d ] [ a c ] \begin{bmatrix}a\\c\\\end{bmatrix} [ a c ] [ b d ] \begin{bmatrix}b\\d\\\end{bmatrix} [ b d ]
验证即可，此处略.
相等表达式有 ( b + a ) + c (\boldsymbol{b}+\boldsymbol{a})+\boldsymbol{c} ( b + a ) + c c + ( a + b ) \boldsymbol{c}+(\boldsymbol{a}+\boldsymbol{b}) c + ( a + b ) c + ( b + a ) \boldsymbol{c}+(\boldsymbol{b}+\boldsymbol{a}) c + ( b + a )
相等表达式有 a + b + c + d \boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}+\boldsymbol{d} a + b + c + d ( a + b ) + c + d (\boldsymbol{a}+\boldsymbol{b})+\boldsymbol{c}+\boldsymbol{d} ( a + b ) + c + d a + ( b + c ) + d \boldsymbol{a}+(\boldsymbol{b}+\boldsymbol{c})+\boldsymbol{d} a + ( b + c ) + d a + b + ( c + d ) \boldsymbol{a}+\boldsymbol{b}+(\boldsymbol{c}+\boldsymbol{d}) a + b + ( c + d ) ( a + b + c ) + d (\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c})+\boldsymbol{d} ( a + b + c ) + d a + ( b + c + d ) \boldsymbol{a}+(\boldsymbol{b}+\boldsymbol{c}+\boldsymbol{d}) a + ( b + c + d ) ( a + ( b + c ) ) + d (\boldsymbol{a}+(\boldsymbol{b}+\boldsymbol{c}))+\boldsymbol{d} ( a + ( b + c ) ) + d a + ( b + ( c + d ) ) \boldsymbol{a}+(\boldsymbol{b}+(\boldsymbol{c}+\boldsymbol{d})) a + ( b + ( c + d ) ) a + ( ( b + c ) + d ) \boldsymbol{a}+((\boldsymbol{b}+\boldsymbol{c})+\boldsymbol{d}) a + ( ( b + c ) + d )
不是. 因为 f ( x 1 ) + f ( x 2 ) = x 1 + 1 + x 2 + 1 = x 1 + x 2 + 2 f(x_1)+f(x_2)=x_1+1+x_2+1=x_1+x_2+2 f ( x 1 ) + f ( x 2 ) = x 1 + 1 + x 2 + 1 = x 1 + x 2 + 2 f ( x 1 + x 2 ) = x 1 + x 2 + 1 f(x_1+x_2)=x_1+x_2+1 f ( x 1 + x 2 ) = x 1 + x 2 + 1
是.
是.
不是. 因为 f ( x 1 ) + f ( x 2 ) = 1 + 1 = 2 f(x_1)+f(x_2)=1+1=2 f ( x 1 ) + f ( x 2 ) = 1 + 1 = 2 f ( x 1 + x 2 ) = 1 f(x_1+x_2)=1 f ( x 1 + x 2 ) = 1
不是. 因为 f ( x 1 ) + f ( x 2 ) = x 1 2 + x 2 2 f(x_1)+f(x_2)=x_1^2+x_2^2 f ( x 1 ) + f ( x 2 ) = x 1 2 + x 2 2 f ( x 1 + x 2 ) = ( x 1 + x 2 ) 2 = x 1 2 + 2 x 1 x 2 + x 2 2 f(x_1+x_2)=(x_1+x_2)^2=x_1^2+2x_1x_2+x_2^2 f ( x 1 + x 2 ) = ( x 1 + x 2 ) 2 = x 1 2 + 2 x 1 x 2 + x 2 2
不是. 因为 f ( x 1 ) + f ( x 2 ) = 2 x 1 + 2 x 2 f(x_1)+f(x_2)=2^{x_1}+2^{x_2} f ( x 1 ) + f ( x 2 ) = 2 x 1 + 2 x 2 f ( x 1 + x 2 ) = 2 x 1 + x 2 f(x_1+x_2)=2^{x_1+x_2} f ( x 1 + x 2 ) = 2 x 1 + x 2
是.
不是. 因为 f ( [ x 1 y 1 ] ) + f ( [ x 2 y 2 ] ) = [ x 1 + 1 y 1 − x 1 2 x 1 ] + [ x 2 + 1 y 2 − x 2 2 x 2 ] = [ x 1 + x 2 + 2 y 1 + y 2 − x 1 − x 2 2 x 1 + 2 x 2 ] f(\begin{bmatrix}x_1\\y_1\end{bmatrix})+f(\begin{bmatrix}x_2\\y_2\end{bmatrix})=\begin{bmatrix}x_1+1\\y_1-x_1\\2x_1\end{bmatrix}+\begin{bmatrix}x_2+1\\y_2-x_2\\2x_2\end{bmatrix}=\begin{bmatrix}x_1+x_2+2\\y_1+y_2-x_1-x_2\\2x_1+2x_2\end{bmatrix} f ( [ x 1 y 1 ] ) + f ( [ x 2 y 2 ] ) = ⎣ ⎡ x 1 + 1 y 1 − x 1 2 x 1 ⎦ ⎤ + ⎣ ⎡ x 2 + 1 y 2 − x 2 2 x 2 ⎦ ⎤ = ⎣ ⎡ x 1 + x 2 + 2 y 1 + y 2 − x 1 − x 2 2 x 1 + 2 x 2 ⎦ ⎤ f ( [ x 1 + x 2 y 1 + y 2 ] ) = [ x 1 + x 2 + 1 y 1 + y 2 − x 1 − x 2 2 x 1 + 2 x 2 ] f(\begin{bmatrix}x_1+x_2\\y_1+y_2\end{bmatrix})=\begin{bmatrix}x_1+x_2+1\\y_1+y_2-x_1-x_2\\2x_1+2x_2\end{bmatrix} f ( [ x 1 + x 2 y 1 + y 2 ] ) = ⎣ ⎡ x 1 + x 2 + 1 y 1 + y 2 − x 1 − x 2 2 x 1 + 2 x 2 ⎦ ⎤
证明 因为 f f f f ( x ) = f ( 1 ⋅ x ) = f ( 1 ) ⋅ x f(x)=f(1\cdot x)=f(1)\cdot x f ( x ) = f ( 1 ⋅ x ) = f ( 1 ) ⋅ x f ( 1 ) ∈ R f(1)\in\R f ( 1 ) ∈ R k = f ( 1 ) k=f(1) k = f ( 1 ) k k k f ( x ) = k x f(x)=kx f ( x ) = k x
证明 先证充分性. 既然 f f f f ( [ x 1 y 1 z 1 ] + [ x 2 y 2 z 2 ] ) = [ g ( x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) h ( x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) ] = f ( [ x 1 y 1 z 1 ] ) + f ( [ x 2 y 2 z 2 ] ) = [ g ( x 1 , y 1 , z 1 ) h ( x 1 , y 1 , z 1 ) ] + [ g ( x 2 , y 2 , z 2 ) h ( x 2 , y 2 , z 2 ) ] f(\begin{bmatrix}x_1\\y_1\\z_1\end{bmatrix}+\begin{bmatrix}x_2\\y_2\\z_2\end{bmatrix})=\begin{bmatrix}g(x_1+x_2,y_1+y_2,z_1+z_2)\\h(x_1+x_2,y_1+y_2,z_1+z_2)\end{bmatrix}=f(\begin{bmatrix}x_1\\y_1\\z_1\end{bmatrix})+f(\begin{bmatrix}x_2\\y_2\\z_2\end{bmatrix})=\begin{bmatrix}g(x_1,y_1,z_1)\\h(x_1,y_1,z_1)\end{bmatrix}+\begin{bmatrix}g(x_2,y_2,z_2)\\h(x_2,y_2,z_2)\end{bmatrix} f ( ⎣ ⎡ x 1 y 1 z 1 ⎦ ⎤ + ⎣ ⎡ x 2 y 2 z 2 ⎦ ⎤ ) = [ g ( x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) h ( x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) ] = f ( ⎣ ⎡ x 1 y 1 z 1 ⎦ ⎤ ) + f ( ⎣ ⎡ x 2 y 2 z 2 ⎦ ⎤ ) = [ g ( x 1 , y 1 , z 1 ) h ( x 1 , y 1 , z 1 ) ] + [ g ( x 2 , y 2 , z 2 ) h ( x 2 , y 2 , z 2 ) ] f ( k [ x y z ] ) = [ g ( k x , k y , k z ) h ( k x , k y , k z ) ] = k f ( [ x y z ] ) = k [ g ( x , y , z ) h ( x , y , z ) ] f(k\begin{bmatrix}x\\y\\z\end{bmatrix})=\begin{bmatrix}g(kx,ky,kz)\\h(kx,ky,kz)\end{bmatrix}=kf(\begin{bmatrix}x\\y\\z\end{bmatrix})=k\begin{bmatrix}g(x,y,z)\\h(x,y,z)\end{bmatrix} f ( k ⎣ ⎡ x y z ⎦ ⎤ ) = [ g ( k x , k y , k z ) h ( k x , k y , k z ) ] = k f ( ⎣ ⎡ x y z ⎦ ⎤ ) = k [ g ( x , y , z ) h ( x , y , z ) ] g , h g,h g , h
再证必要性. 根据上述证明中的等式易得 f f f
不存在. 考虑 f ( x 2 ) = b 2 f(x_2)=b_2 f ( x 2 ) = b 2 f ( x 2 ) = f ( x 1 + x 3 ) = f ( x 1 ) + f ( x 3 ) = b 1 + b 3 f(x_2)=f(x_1+x_3)=f(x_1)+f(x_3)=b_1+b_3 f ( x 2 ) = f ( x 1 + x 3 ) = f ( x 1 ) + f ( x 3 ) = b 1 + b 3 b 2 ≠ b 1 + b 3 b_2\not ={b_1+b_3} b 2 = b 1 + b 3
是.
是.
不是.
f f f f ( [ b 1 b 2 b 3 ] ) = [ a 1 a 2 a 3 ] ⋅ [ b 1 b 2 b 3 ] = a 1 b 1 + a 2 b 2 + a 3 b 3 f(\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix})=\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}\cdot\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}=a_1b_1+a_2b_2+a_3b_3 f ( ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ ) = ⎣ ⎡ a 1 a 2 a 3 ⎦ ⎤ ⋅ ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ = a 1 b 1 + a 2 b 2 + a 3 b 3 f ( [ b 1 b 2 b 3 ] + [ c 1 c 2 c 3 ] ) = [ a 1 a 2 a 3 ] ⋅ [ b 1 + c 1 b 2 + c 2 b 3 + c 3 ] = a 1 b 1 + a 2 b 2 + a 3 b 3 + a 1 c 1 + a 2 c 2 + a 3 c 3 = [ a 1 a 2 a 3 ] ⋅ [ b 1 b 2 b 3 ] + [ a 1 a 2 a 3 ] ⋅ [ c 1 c 2 c 3 ] = f ( [ b 1 b 2 b 3 ] ) + f ( [ c 1 c 2 c 3 ] ) f(\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}+\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix})=\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}\cdot\begin{bmatrix}b_1+c_1\\b_2+c_2\\b_3+c_3\end{bmatrix}=a_1b_1+a_2b_2+a_3b_3+a_1c_1+a_2c_2+a_3c_3=\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}\cdot\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}+\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}\cdot\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=f(\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix})+f(\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}) f ( ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ + ⎣ ⎡ c 1 c 2 c 3 ⎦ ⎤ ) = ⎣ ⎡ a 1 a 2 a 3 ⎦ ⎤ ⋅ ⎣ ⎡ b 1 + c 1 b 2 + c 2 b 3 + c 3 ⎦ ⎤ = a 1 b 1 + a 2 b 2 + a 3 b 3 + a 1 c 1 + a 2 c 2 + a 3 c 3 = ⎣ ⎡ a 1 a 2 a 3 ⎦ ⎤ ⋅ ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ + ⎣ ⎡ a 1 a 2 a 3 ⎦ ⎤ ⋅ ⎣ ⎡ c 1 c 2 c 3 ⎦ ⎤ = f ( ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ ) + f ( ⎣ ⎡ c 1 c 2 c 3 ⎦ ⎤ ) f ( k [ b 1 b 2 b 3 ] ) = [ a 1 a 2 a 3 ] ⋅ [ k b 1 k b 2 k b 3 ] = k a 1 b 1 + k a 2 b 2 + k a 3 b 3 = k ( a 1 b 1 + a 2 b 2 + a 3 b 3 ) = k f ( [ b 1 b 2 b 3 ] ) f(k\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix})=\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}\cdot\begin{bmatrix}kb_1\\kb_2\\kb_3\end{bmatrix}=ka_1b_1+ka_2b_2+ka_3b_3=k(a_1b_1+a_2b_2+a_3b_3)=kf(\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}) f ( k ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ ) = ⎣ ⎡ a 1 a 2 a 3 ⎦ ⎤ ⋅ ⎣ ⎡ k b 1 k b 2 k b 3 ⎦ ⎤ = k a 1 b 1 + k a 2 b 2 + k a 3 b 3 = k ( a 1 b 1 + a 2 b 2 + a 3 b 3 ) = k f ( ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ ) f f f
g g g g ( [ b 1 b 2 b 3 ] ) = [ a 1 a 2 a 3 ] × [ b 1 b 2 b 3 ] = [ a 2 b 3 − a 3 b 2 a 3 b 1 − a 1 b 3 a 1 b 2 − a 2 b 1 ] g(\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix})=\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}\times\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}=\begin{bmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{bmatrix} g ( ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ ) = ⎣ ⎡ a 1 a 2 a 3 ⎦ ⎤ × ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ = ⎣ ⎡ a 2 b 3 − a 3 b 2 a 3 b 1 − a 1 b 3 a 1 b 2 − a 2 b 1 ⎦ ⎤ g ( [ b 1 b 2 b 3 ] + [ c 1 c 2 c 3 ] ) = [ a 1 a 2 a 3 ] × [ b 1 + c 1 b 2 + c 2 b 3 + c 3 ] = [ a 2 ( b 3 + c 3 ) − a 3 ( b 2 + c 2 ) a 3 ( b 1 + c 1 ) − a 1 ( b 3 + c 3 ) a 1 ( b 2 + c 2 ) − a 2 ( b 1 + c 1 ) ] = [ a 2 b 3 − a 3 b 2 a 3 b 1 − a 1 b 3 a 1 b 2 − a 2 b 1 ] + [ a 2 c 3 − a 3 c 2 a 3 c 1 − a 1 c 3 a 1 c 2 − a 2 c 1 ] = g ( [ b 1 b 2 b 3 ] ) + g ( [ c 1 c 2 c 3 ] ) g(\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}+\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix})=\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}\times\begin{bmatrix}b_1+c_1\\b_2+c_2\\b_3+c_3\end{bmatrix}=\begin{bmatrix}a_2(b_3+c_3)-a_3(b_2+c_2)\\a_3(b_1+c_1)-a_1(b_3+c_3)\\a_1(b_2+c_2)-a_2(b_1+c_1)\end{bmatrix}=\begin{bmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{bmatrix}+\begin{bmatrix}a_2c_3-a_3c_2\\a_3c_1-a_1c_3\\a_1c_2-a_2c_1\end{bmatrix}=g(\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix})+g(\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}) g ( ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ + ⎣ ⎡ c 1 c 2 c 3 ⎦ ⎤ ) = ⎣ ⎡ a 1 a 2 a 3 ⎦ ⎤ × ⎣ ⎡ b 1 + c 1 b 2 + c 2 b 3 + c 3 ⎦ ⎤ = ⎣ ⎡ a 2 ( b 3 + c 3 ) − a 3 ( b 2 + c 2 ) a 3 ( b 1 + c 1 ) − a 1 ( b 3 + c 3 ) a 1 ( b 2 + c 2 ) − a 2 ( b 1 + c 1 ) ⎦ ⎤ = ⎣ ⎡ a 2 b 3 − a 3 b 2 a 3 b 1 − a 1 b 3 a 1 b 2 − a 2 b 1 ⎦ ⎤ + ⎣ ⎡ a 2 c 3 − a 3 c 2 a 3 c 1 − a 1 c 3 a 1 c 2 − a 2 c 1 ⎦ ⎤ = g ( ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ ) + g ( ⎣ ⎡ c 1 c 2 c 3 ⎦ ⎤ ) g ( k [ b 1 b 2 b 3 ] ) = [ a 1 a 2 a 3 ] × [ k b 1 k b 2 k b 3 ] = [ k a 2 b 3 − k a 3 b 2 k a 3 b 1 − k a 1 b 3 k a 1 b 2 − k a 2 b 1 ] = k [ a 2 b 3 − a 3 b 2 a 3 b 1 − a 1 b 3 a 1 b 2 − a 2 b 1 ] = k g ( [ b 1 b 2 b 3 ] ) g(k\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix})=\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}\times\begin{bmatrix}kb_1\\kb_2\\kb_3\end{bmatrix}=\begin{bmatrix}ka_2b_3-ka_3b_2\\ka_3b_1-ka_1b_3\\ka_1b_2-ka_2b_1\end{bmatrix}=k\begin{bmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{bmatrix}=kg(\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}) g ( k ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ ) = ⎣ ⎡ a 1 a 2 a 3 ⎦ ⎤ × ⎣ ⎡ k b 1 k b 2 k b 3 ⎦ ⎤ = ⎣ ⎡ k a 2 b 3 − k a 3 b 2 k a 3 b 1 − k a 1 b 3 k a 1 b 2 − k a 2 b 1 ⎦ ⎤ = k ⎣ ⎡ a 2 b 3 − a 3 b 2 a 3 b 1 − a 1 b 3 a 1 b 2 − a 2 b 1 ⎦ ⎤ = k g ( ⎣ ⎡ b 1 b 2 b 3 ⎦ ⎤ ) g g g
确定. S = 1 S=1 S = 1
确定. S = 1 S=1 S = 1
确定. S = 0 S=0 S = 0
确定. S = 3 S=3 S = 3
确定. S = 1 S=1 S = 1 错切变换不会改变图形的面积 ）
证明 f ( [ x 1 y 1 ] ) + f ( [ x 2 y 2 ] ) = [ − y 1 + 1 x 1 + 2 ] + [ − y 2 + 1 x 2 + 2 ] = [ − y 1 − y 2 + 2 x 1 + x 2 + 4 ] f(\begin{bmatrix}x_1\\y_1\end{bmatrix})+f(\begin{bmatrix}x_2\\y_2\end{bmatrix})=\begin{bmatrix}-y_1+1\\x_1+2\end{bmatrix}+\begin{bmatrix}-y_2+1\\x_2+2\end{bmatrix}=\begin{bmatrix}-y_1-y_2+2\\x_1+x_2+4\end{bmatrix} f ( [ x 1 y 1 ] ) + f ( [ x 2 y 2 ] ) = [ − y 1 + 1 x 1 + 2 ] + [ − y 2 + 1 x 2 + 2 ] = [ − y 1 − y 2 + 2 x 1 + x 2 + 4 ] f ( [ x 1 y 1 ] + [ x 2 y 2 ] ) = [ − y 1 − y 2 + 1 x 1 + x 2 + 2 ] f(\begin{bmatrix}x_1\\y_1\end{bmatrix}+\begin{bmatrix}x_2\\y_2\end{bmatrix})=\begin{bmatrix}-y_1-y_2+1\\x_1+x_2+2\end{bmatrix} f ( [ x 1 y 1 ] + [ x 2 y 2 ] ) = [ − y 1 − y 2 + 1 x 1 + x 2 + 2 ] f ( [ x 1 y 1 ] ) + f ( [ x 2 y 2 ] ) ≠ f ( [ x 1 y 1 ] + [ x 2 y 2 ] ) f(\begin{bmatrix}x_1\\y_1\end{bmatrix})+f(\begin{bmatrix}x_2\\y_2\end{bmatrix})\not ={f(\begin{bmatrix}x_1\\y_1\end{bmatrix}+\begin{bmatrix}x_2\\y_2\end{bmatrix})} f ( [ x 1 y 1 ] ) + f ( [ x 2 y 2 ] ) = f ( [ x 1 y 1 ] + [ x 2 y 2 ] ) f f f 不妨令 g ( [ x y ] ) = [ x + 1 y + 1 ] , h ( [ x y ] ) = [ k x y ] g(\begin{bmatrix}x\\y\end{bmatrix})=\begin{bmatrix}x+1\\y+1\end{bmatrix},h(\begin{bmatrix}x\\y\end{bmatrix})=\begin{bmatrix}kx\\y\end{bmatrix} g ( [ x y ] ) = [ x + 1 y + 1 ] , h ( [ x y ] ) = [ k x y ] f ( [ x y ] ) = g ∘ h ( [ x y ] ) = g ( [ k x y ] ) = [ k x + 1 y + 1 ] f(\begin{bmatrix}x\\y\end{bmatrix})=g\circ h(\begin{bmatrix}x\\y\end{bmatrix})=g(\begin{bmatrix}kx\\y\end{bmatrix})=\begin{bmatrix}kx+1\\y+1\end{bmatrix} f ( [ x y ] ) = g ∘ h ( [ x y ] ) = g ( [ k x y ] ) = [ k x + 1 y + 1 ] 仿射变换 ）
是线性映射.
证明 形如 f ( x + y ) = f ( x ) + f ( y ) f(x+y)=f(x)+f(y) f ( x + y ) = f ( x ) + f ( y ) 加性 Cauchy 方程 ，下面给出求解过程，显然其求解结果可以证明本题：
若 x , y ∈ Z + x,y \in \Z^+ x , y ∈ Z + x = 1 , y = 1 x=1,y=1 x = 1 , y = 1 f ( 2 ) = 2 f ( 1 ) f(2)=2f(1) f ( 2 ) = 2 f ( 1 ) x = 2 , y = 1 x=2,y=1 x = 2 , y = 1 f ( 3 ) = f ( 2 ) + f ( 1 ) = 3 f ( 1 ) f(3)=f(2)+f(1)=3f(1) f ( 3 ) = f ( 2 ) + f ( 1 ) = 3 f ( 1 ) f ( n ) = n f ( 1 ) f(n)=nf(1) f ( n ) = n f ( 1 )
若 x , y ∈ Z x,y \in \Z x , y ∈ Z x = 1 , y = 0 x=1,y=0 x = 1 , y = 0 f ( 1 ) = f ( 0 ) + f ( 1 ) f(1)=f(0)+f(1) f ( 1 ) = f ( 0 ) + f ( 1 ) f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 0 = f ( 0 ) = f ( x + ( − x ) ) = f ( x ) + f ( − x ) 0=f(0)=f(x+(-x))=f(x)+f(-x) 0 = f ( 0 ) = f ( x + ( − x ) ) = f ( x ) + f ( − x ) f ( x ) f(x) f ( x ) ∀ x ∈ Z \forall x\in\Z ∀ x ∈ Z f ( x ) = x f ( 1 ) f(x)=xf(1) f ( x ) = x f ( 1 )
若 x , y ∈ Q x,y \in \mathbb{Q} x , y ∈ Q x = y = p q x=y=\cfrac{p}{q} x = y = q p p , q ∈ Z p,q\in\Z p , q ∈ Z f ( 2 p q ) = 2 f ( p q ) f(\cfrac{2p}{q})=2f(\cfrac{p}{q}) f ( q 2 p ) = 2 f ( q p ) q f ( p q ) = f ( p q q ) = f ( p ) = p f ( 1 ) qf(\cfrac{p}{q})=f(\cfrac{pq}{q})=f(p)=pf(1) q f ( q p ) = f ( q p q ) = f ( p ) = p f ( 1 ) f ( p q ) = p q f ( 1 ) f(\cfrac{p}{q})=\cfrac{p}{q}f(1) f ( q p ) = q p f ( 1 )
接下来推广至 R \R R f f f lim x → x 0 f ( x ) = f ( x 0 ) \lim\limits_{x\rightarrow x_0}f(x)=f(x_0) x → x 0 lim f ( x ) = f ( x 0 ) x 0 ∈ R x_0\in\R x 0 ∈ R x 0 x_0 x 0 q 1 , q 2 , ⋯ , q n q_1, q_2,\cdots,q_n q 1 , q 2 , ⋯ , q n f ( x 0 ) = lim n → + ∞ f ( q n ) = lim n → + ∞ q n f ( 1 ) = x 0 f ( 1 ) f(x_0)=\lim\limits_{n\rightarrow+\infin}f(q_n)=\lim\limits_{n\rightarrow+\infin}q_nf(1)=x_0f(1) f ( x 0 ) = n → + ∞ lim f ( q n ) = n → + ∞ lim q n f ( 1 ) = x 0 f ( 1 )
