此练习推荐给所有读者。
问题 1
将向量投影到 M {\displaystyle M} N {\displaystyle N}
( 3 − 2 ) , M = { ( x y ) | x + y = 0 } , N = { ( x y ) | − x − 2 y = 0 } {\displaystyle {\begin{pmatrix}3\\-2\end{pmatrix}},\quad M=\{{\begin{pmatrix}x\\y\end{pmatrix}}\,{\big |}\,x+y=0\},\quad N=\{{\begin{pmatrix}x\\y\end{pmatrix}}\,{\big |}\,-x-2y=0\}} ( 1 2 ) , M = { ( x y ) | x − y = 0 } , N = { ( x y ) | 2 x + y = 0 } {\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}},\quad M=\{{\begin{pmatrix}x\\y\end{pmatrix}}\,{\big |}\,x-y=0\},\quad N=\{{\begin{pmatrix}x\\y\end{pmatrix}}\,{\big |}\,2x+y=0\}} ( 3 0 1 ) , M = { ( x y z ) | x + y = 0 } , N = { c ⋅ ( 1 0 1 ) | c ∈ R } {\displaystyle {\begin{pmatrix}3\\0\\1\end{pmatrix}},\quad M=\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\,{\big |}\,x+y=0\},\quad N=\{c\cdot {\begin{pmatrix}1\\0\\1\end{pmatrix}}\,{\big |}\,c\in \mathbb {R} \}}
答案
当子空间的基 B M = ⟨ ( 1 − 1 ) ⟩ B N = ⟨ ( 2 − 1 ) ⟩ {\displaystyle B_{M}=\langle {\begin{pmatrix}1\\-1\end{pmatrix}}\rangle \qquad B_{N}=\langle {\begin{pmatrix}2\\-1\end{pmatrix}}\rangle } B = B M ⌢ B N = ⟨ ( 1 − 1 ) , ( 2 − 1 ) ⟩ {\displaystyle B=B_{M}\!{\mathbin {{}^{\frown }}}\!B_{N}=\langle {\begin{pmatrix}1\\-1\end{pmatrix}},{\begin{pmatrix}2\\-1\end{pmatrix}}\rangle } ( 3 − 2 ) = 1 ⋅ ( 1 − 1 ) + 1 ⋅ ( 2 − 1 ) {\displaystyle {\begin{pmatrix}3\\-2\end{pmatrix}}=1\cdot {\begin{pmatrix}1\\-1\end{pmatrix}}+1\cdot {\begin{pmatrix}2\\-1\end{pmatrix}}} M {\displaystyle M} N {\displaystyle N} proj M , N ( ( 3 − 2 ) ) = ( 1 − 1 ) {\displaystyle {\mbox{proj}}_{M,N}({\begin{pmatrix}3\\-2\end{pmatrix}})={\begin{pmatrix}1\\-1\end{pmatrix}}} 当基底 B M = ⟨ ( 1 1 ) ⟩ B N ⟨ ( 1 − 2 ) ⟩ {\displaystyle B_{M}=\langle {\begin{pmatrix}1\\1\end{pmatrix}}\rangle \qquad B_{N}\langle {\begin{pmatrix}1\\-2\end{pmatrix}}\rangle } ( 1 2 ) = ( 4 / 3 ) ⋅ ( 1 1 ) − ( 1 / 3 ) ⋅ ( 1 − 2 ) {\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}}=(4/3)\cdot {\begin{pmatrix}1\\1\end{pmatrix}}-(1/3)\cdot {\begin{pmatrix}1\\-2\end{pmatrix}}} M {\displaystyle M} proj M , N ( ( 1 2 ) ) = ( 4 / 3 4 / 3 ) {\displaystyle {\mbox{proj}}_{M,N}({\begin{pmatrix}1\\2\end{pmatrix}})={\begin{pmatrix}4/3\\4/3\end{pmatrix}}} 对于这些基底 B M = ⟨ ( 1 − 1 0 ) , ( 0 0 1 ) ⟩ B N = ⟨ ( 1 0 1 ) ⟩ {\displaystyle B_{M}=\langle {\begin{pmatrix}1\\-1\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\1\end{pmatrix}}\rangle \qquad B_{N}=\langle {\begin{pmatrix}1\\0\\1\end{pmatrix}}\rangle } ( 3 0 1 ) = 0 ⋅ ( 1 − 1 0 ) − 2 ⋅ ( 0 0 1 ) + 3 ⋅ ( 1 0 1 ) {\displaystyle {\begin{pmatrix}3\\0\\1\end{pmatrix}}=0\cdot {\begin{pmatrix}1\\-1\\0\end{pmatrix}}-2\cdot {\begin{pmatrix}0\\0\\1\end{pmatrix}}+3\cdot {\begin{pmatrix}1\\0\\1\end{pmatrix}}} proj M , N ( ( 3 0 1 ) ) = ( 0 0 − 2 ) {\displaystyle {\mbox{proj}}_{M,N}({\begin{pmatrix}3\\0\\1\end{pmatrix}})={\begin{pmatrix}0\\0\\-2\end{pmatrix}}}
此练习推荐给所有读者。
问题 2
求 M ⊥ {\displaystyle M^{\perp }}
M = { ( x y ) | x + y = 0 } {\displaystyle M=\{{\begin{pmatrix}x\\y\end{pmatrix}}\,{\big |}\,x+y=0\}} M = { ( x y ) | − 2 x + 3 y = 0 } {\displaystyle M=\{{\begin{pmatrix}x\\y\end{pmatrix}}\,{\big |}\,-2x+3y=0\}} M = { ( x y ) | x − y = 0 } {\displaystyle M=\{{\begin{pmatrix}x\\y\end{pmatrix}}\,{\big |}\,x-y=0\}} M = { 0 → } {\displaystyle M=\{{\vec {0}}\,\}} M = { ( x y ) | x = 0 } {\displaystyle M=\{{\begin{pmatrix}x\\y\end{pmatrix}}\,{\big |}\,x=0\}} M = { ( x y z ) | − x + 3 y + z = 0 } {\displaystyle M=\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\,{\big |}\,-x+3y+z=0\}} M = { ( x y z ) | x = 0 and y + z = 0 } {\displaystyle M=\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\,{\big |}\,x=0{\text{ and }}y+z=0\}}
答案
如同在示例 3.5 中，我们可以通过仅仅找出垂直于 M {\displaystyle M}
参数化得到 M = { c ⋅ ( − 1 1 ) | c ∈ R } {\displaystyle M=\{c\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}\,{\big |}\,c\in \mathbb {R} \}} M ⊥ { ( u v ) | 0 = ( u v ) ⋅ ( − 1 1 ) } = { ( u v ) | 0 = − u + v } {\displaystyle M^{\perp }\{{\begin{pmatrix}u\\v\end{pmatrix}}\,{\big |}\,0={\begin{pmatrix}u\\v\end{pmatrix}}\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}\}=\{{\begin{pmatrix}u\\v\end{pmatrix}}\,{\big |}\,0=-u+v\}} M ⊥ = { k ⋅ ( 1 1 ) | k ∈ R } {\displaystyle M^{\perp }=\{k\cdot {\begin{pmatrix}1\\1\end{pmatrix}}\,{\big |}\,k\in \mathbb {R} \}} 如先前部分的答案所示， M {\displaystyle M} M = { c ⋅ ( 3 / 2 1 ) | c ∈ R } B M = ⟨ ( 3 / 2 1 ) ⟩ {\displaystyle M=\{c\cdot {\begin{pmatrix}3/2\\1\end{pmatrix}}\,{\big |}\,c\in \mathbb {R} \}\qquad B_{M}=\langle {\begin{pmatrix}3/2\\1\end{pmatrix}}\rangle } M ⊥ {\displaystyle M^{\perp }} M ⊥ = { ( u v ) | ( 3 / 2 ) ⋅ u + 1 ⋅ v = 0 } = { k ⋅ ( − 2 / 3 1 ) | k ∈ R } {\displaystyle M^{\perp }=\{{\begin{pmatrix}u\\v\end{pmatrix}}\,{\big |}\,(3/2)\cdot u+1\cdot v=0\}=\{k\cdot {\begin{pmatrix}-2/3\\1\end{pmatrix}}\,{\big |}\,k\in \mathbb {R} \}} 对 M {\displaystyle M} M = { c ⋅ ( 1 1 ) | c ∈ R } B M = ⟨ ( 1 1 ) ⟩ {\displaystyle M=\{c\cdot {\begin{pmatrix}1\\1\end{pmatrix}}\,{\big |}\,c\in \mathbb {R} \}\qquad B_{M}=\langle {\begin{pmatrix}1\\1\end{pmatrix}}\rangle } M ⊥ {\displaystyle M^{\perp }} B M {\displaystyle B_{M}} M ⊥ = { ( u v ) | u + v = 0 } = { k ⋅ ( − 1 1 ) | k ∈ R } {\displaystyle M^{\perp }=\{{\begin{pmatrix}u\\v\end{pmatrix}}\,{\big |}\,u+v=0\}=\{k\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}\,{\big |}\,k\in \mathbb {R} \}} 空间中的每个向量都垂直于零向量，所以 M ⊥ = R n {\displaystyle M^{\perp }=\mathbb {R} ^{n}} 对 M {\displaystyle M} M = { y ⋅ ( 0 1 ) | y ∈ R } B M = ⟨ ( 0 1 ) ⟩ {\displaystyle M=\{y\cdot {\begin{pmatrix}0\\1\end{pmatrix}}\,{\big |}\,y\in \mathbb {R} \}\qquad B_{M}=\langle {\begin{pmatrix}0\\1\end{pmatrix}}\rangle } M ⊥ = { ( u v ) | 0 ⋅ u + 1 ⋅ v = 0 } = { k ⋅ ( 1 0 ) | k ∈ R } {\displaystyle M^{\perp }=\{{\begin{pmatrix}u\\v\end{pmatrix}}\,{\big |}\,0\cdot u+1\cdot v=0\}=\{k\cdot {\begin{pmatrix}1\\0\end{pmatrix}}\,{\big |}\,k\in \mathbb {R} \}} ( y -axis ) ⊥ = x -axis {\displaystyle (y{\text{-axis}})^{\perp }=x{\text{-axis}}} 使用参数方程很容易找到 M {\displaystyle M} M = { c ⋅ ( 3 1 0 ) + d ⋅ ( 1 0 1 ) | c , d ∈ R } B M = ⟨ ( 3 1 0 ) , ( 1 0 1 ) ⟩ {\displaystyle M=\{c\cdot {\begin{pmatrix}3\\1\\0\end{pmatrix}}+d\cdot {\begin{pmatrix}1\\0\\1\end{pmatrix}}\,{\big |}\,c,d\in \mathbb {R} \}\qquad B_{M}=\langle {\begin{pmatrix}3\\1\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\1\end{pmatrix}}\rangle } M ⊥ {\displaystyle M^{\perp }} 3 u + v = 0 u + w = 0 → − ( 1 / 3 ) ρ 1 + ρ 2 3 u + v = 0 − ( 1 / 3 ) v + w = 0 {\displaystyle {\begin{array}{*{3}{rc}r}3u&+&v&&&=&0\\u&&&+&w&=&0\end{array}}\;{\xrightarrow[{}]{-(1/3)\rho _{1}+\rho _{2}}}\;{\begin{array}{*{3}{rc}r}3u&+&v&&&=&0\\&&-(1/3)v&+&w&=&0\end{array}}} M ⊥ = { k ⋅ ( − 1 3 1 ) | k ∈ R } {\displaystyle M^{\perp }=\{k\cdot {\begin{pmatrix}-1\\3\\1\end{pmatrix}}\,{\big |}\,k\in \mathbb {R} \}} 这里， M {\displaystyle M} M = { c ⋅ ( 0 − 1 1 ) | c ∈ R } B M = ⟨ ( 0 − 1 1 ) ⟩ {\displaystyle M=\{c\cdot {\begin{pmatrix}0\\-1\\1\end{pmatrix}}\,{\big |}\,c\in \mathbb {R} \}\qquad B_{M}=\langle {\begin{pmatrix}0\\-1\\1\end{pmatrix}}\rangle } M ⊥ {\displaystyle M^{\perp }} M ⊥ = { ( u v w ) | 0 ⋅ u − 1 ⋅ v + 1 ⋅ w = 0 } = { j ⋅ ( 1 0 0 ) + k ⋅ ( 0 1 1 ) | j , k ∈ R } {\displaystyle M^{\perp }=\{{\begin{pmatrix}u\\v\\w\end{pmatrix}}\,{\big |}\,0\cdot u-1\cdot v+1\cdot w=0\}=\{j\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}+k\cdot {\begin{pmatrix}0\\1\\1\end{pmatrix}}\,{\big |}\,j,k\in \mathbb {R} \}}
此练习推荐给所有读者。
问题 4
我们有三种方法可以找到向量在直线上的正交投影，第一种方法是本节第一小节中给出的定义 1.1 方法，第二种方法是通过示例 3.2 和3.3 使用空间基底表示向量，然后保留 M {\displaystyle M} 定理 3.8 。对于这些情况，请使用所有三种方法。
v → = ( 1 − 3 ) , M = { ( x y ) | x + y = 0 } {\displaystyle {\vec {v}}={\begin{pmatrix}1\\-3\end{pmatrix}},\quad M=\{{\begin{pmatrix}x\\y\end{pmatrix}}\,{\big |}\,x+y=0\}} v → = ( 0 1 2 ) , M = { ( x y z ) | x + z = 0 and y = 0 } {\displaystyle {\vec {v}}={\begin{pmatrix}0\\1\\2\end{pmatrix}},\quad M=\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\,{\big |}\,x+z=0{\text{ and }}y=0\}}
答案
参数化给出以下结果。 M = { c ⋅ ( − 1 1 ) | c ∈ R } {\displaystyle M=\{c\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}\,{\big |}\,c\in \mathbb {R} \}} M {\displaystyle M} s → = ( − 1 1 ) {\displaystyle {\vec {s}}={\begin{pmatrix}-1\\1\end{pmatrix}}} 定义 1.1 公式给出以下结果。 proj [ s → ] ( ( 1 − 3 ) ) = ( 1 − 3 ) ⋅ ( − 1 1 ) ( − 1 1 ) ⋅ ( − 1 1 ) ⋅ ( − 1 1 ) = − 4 2 ⋅ ( − 1 1 ) = ( 2 − 2 ) {\displaystyle {\mbox{proj}}_{[{\vec {s}}\,]}({\begin{pmatrix}1\\-3\end{pmatrix}})={\frac {{\begin{pmatrix}1\\-3\end{pmatrix}}\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}}{{\begin{pmatrix}-1\\1\end{pmatrix}}\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}}}\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}={\frac {-4}{2}}\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}={\begin{pmatrix}2\\-2\end{pmatrix}}} B M = ⟨ ( − 1 1 ) ⟩ {\displaystyle B_{M}=\langle {\begin{pmatrix}-1\\1\end{pmatrix}}\rangle } 示例 3.5 和 3.6 中，我们只需找到垂直于基底中所有向量的向量） M ⊥ = { ( u v ) | − 1 ⋅ u + 1 ⋅ v = 0 } = { k ⋅ ( 1 1 ) | k ∈ R } B M ⊥ = ⟨ ( 1 1 ) ⟩ {\displaystyle M^{\perp }=\{{\begin{pmatrix}u\\v\end{pmatrix}}\,{\big |}\,-1\cdot u+1\cdot v=0\}=\{k\cdot {\begin{pmatrix}1\\1\end{pmatrix}}\,{\big |}\,k\in \mathbb {R} \}\qquad B_{M^{\perp }}=\langle {\begin{pmatrix}1\\1\end{pmatrix}}\rangle } ( 1 − 3 ) = − 2 ⋅ ( − 1 1 ) − 1 ⋅ ( 1 1 ) {\displaystyle {\begin{pmatrix}1\\-3\end{pmatrix}}=-2\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}-1\cdot {\begin{pmatrix}1\\1\end{pmatrix}}} M {\displaystyle M} proj M , M ⊥ ( ( 1 − 3 ) ) = ( 2 − 2 ) {\displaystyle {\mbox{proj}}_{M,M^{\perp }}({\begin{pmatrix}1\\-3\end{pmatrix}})={\begin{pmatrix}2\\-2\end{pmatrix}}} 1 × 1 {\displaystyle 1\!\times \!1} 1 × 1 {\displaystyle 1\!\times \!1} A ( A t r a n s A ) − 1 A t r a n s = ( − 1 1 ) ( ( − 1 1 ) ( − 1 1 ) ) − 1 ( − 1 1 ) = ( − 1 1 ) ( 2 ) − 1 ( − 1 1 ) {\displaystyle A\left({{A}^{\rm {trans}}}A\right)^{-1}{{A}^{\rm {trans}}}={\begin{pmatrix}-1\\1\end{pmatrix}}\left({\begin{pmatrix}-1&1\end{pmatrix}}{\begin{pmatrix}-1\\1\end{pmatrix}}\right)^{-1}{\begin{pmatrix}-1&1\end{pmatrix}}={\begin{pmatrix}-1\\1\end{pmatrix}}{\begin{pmatrix}2\end{pmatrix}}^{-1}{\begin{pmatrix}-1&1\end{pmatrix}}} = ( − 1 1 ) ( 1 / 2 ) ( − 1 1 ) = ( − 1 1 ) ( − 1 / 2 1 / 2 ) = ( 1 / 2 − 1 / 2 − 1 / 2 1 / 2 ) {\displaystyle ={\begin{pmatrix}-1\\1\end{pmatrix}}{\begin{pmatrix}1/2\end{pmatrix}}{\begin{pmatrix}-1&1\end{pmatrix}}={\begin{pmatrix}-1\\1\end{pmatrix}}{\begin{pmatrix}-1/2&1/2\end{pmatrix}}={\begin{pmatrix}1/2&-1/2\\-1/2&1/2\end{pmatrix}}} proj M ( ( 1 − 3 ) ) = ( 1 / 2 − 1 / 2 − 1 / 2 1 / 2 ) ( 1 − 3 ) = ( 2 − 2 ) {\displaystyle {\mbox{proj}}_{M}({\begin{pmatrix}1\\-3\end{pmatrix}})={\begin{pmatrix}1/2&-1/2\\-1/2&1/2\end{pmatrix}}{\begin{pmatrix}1\\-3\end{pmatrix}}={\begin{pmatrix}2\\-2\end{pmatrix}}} 参数化给出以下结果。 M = { c ⋅ ( − 1 0 1 ) | c ∈ R } {\displaystyle M=\{c\cdot {\begin{pmatrix}-1\\0\\1\end{pmatrix}}\,{\big |}\,c\in \mathbb {R} \}} ( 0 1 2 ) ⋅ ( − 1 0 1 ) ( − 1 0 1 ) ⋅ ( − 1 0 1 ) ⋅ ( − 1 0 1 ) = 2 2 ⋅ ( − 1 0 1 ) = ( − 1 0 1 ) {\displaystyle {\frac {{\begin{pmatrix}0\\1\\2\end{pmatrix}}\cdot {\begin{pmatrix}-1\\0\\1\end{pmatrix}}}{{\begin{pmatrix}-1\\0\\1\end{pmatrix}}\cdot {\begin{pmatrix}-1\\0\\1\end{pmatrix}}}}\cdot {\begin{pmatrix}-1\\0\\1\end{pmatrix}}={\frac {2}{2}}\cdot {\begin{pmatrix}-1\\0\\1\end{pmatrix}}={\begin{pmatrix}-1\\0\\1\end{pmatrix}}} M ⊥ {\displaystyle M^{\perp }} M ⊥ = { ( u v w ) | − u + w = 0 } = { j ⋅ ( 1 0 1 ) + k ⋅ ( 0 1 0 ) | j , k ∈ R } {\displaystyle M^{\perp }=\{{\begin{pmatrix}u\\v\\w\end{pmatrix}}\,{\big |}\,-u+w=0\}=\{j\cdot {\begin{pmatrix}1\\0\\1\end{pmatrix}}+k\cdot {\begin{pmatrix}0\\1\\0\end{pmatrix}}\,{\big |}\,j,k\in \mathbb {R} \}} B M {\displaystyle B_{M}} B M ⊥ {\displaystyle B_{M^{\perp }}} ( 0 1 2 ) = 1 ⋅ ( − 1 0 1 ) + 1 ⋅ ( 1 0 1 ) + 1 ⋅ ( 0 1 0 ) {\displaystyle {\begin{pmatrix}0\\1\\2\end{pmatrix}}=1\cdot {\begin{pmatrix}-1\\0\\1\end{pmatrix}}+1\cdot {\begin{pmatrix}1\\0\\1\end{pmatrix}}+1\cdot {\begin{pmatrix}0\\1\\0\end{pmatrix}}} M {\displaystyle M} proj M ( ( 0 1 2 ) ) = 1 ⋅ ( − 1 0 1 ) = ( − 1 0 1 ) {\displaystyle {\mbox{proj}}_{M}({\begin{pmatrix}0\\1\\2\end{pmatrix}})=1\cdot {\begin{pmatrix}-1\\0\\1\end{pmatrix}}={\begin{pmatrix}-1\\0\\1\end{pmatrix}}} A ( A t r a n s A ) − 1 A t r a n s = ( − 1 0 1 ) ( ( − 1 0 1 ) ( − 1 0 1 ) ) − 1 ( − 1 0 1 ) = ( − 1 0 1 ) ( 2 ) − 1 ( − 1 0 1 ) {\displaystyle A\left({{A}^{\rm {trans}}}A\right)^{-1}{{A}^{\rm {trans}}}={\begin{pmatrix}-1\\0\\1\end{pmatrix}}{\bigl (}{\begin{pmatrix}-1&0&1\end{pmatrix}}{\begin{pmatrix}-1\\0\\1\end{pmatrix}}{\bigr )}^{-1}{\begin{pmatrix}-1&0&1\end{pmatrix}}={\begin{pmatrix}-1\\0\\1\end{pmatrix}}{\begin{pmatrix}2\end{pmatrix}}^{-1}{\begin{pmatrix}-1&0&1\end{pmatrix}}} = ( − 1 0 1 ) ( 1 / 2 ) ( − 1 0 1 ) = ( − 1 0 1 ) ( − 1 / 2 0 1 / 2 ) = ( 1 / 2 0 − 1 / 2 0 0 0 − 1 / 2 0 1 / 2 ) {\displaystyle ={\begin{pmatrix}-1\\0\\1\end{pmatrix}}{\begin{pmatrix}1/2\end{pmatrix}}{\begin{pmatrix}-1&0&1\end{pmatrix}}={\begin{pmatrix}-1\\0\\1\end{pmatrix}}{\begin{pmatrix}-1/2&0&1/2\end{pmatrix}}={\begin{pmatrix}1/2&0&-1/2\\0&0&0\\-1/2&0&1/2\end{pmatrix}}} proj M ( ( 0 1 2 ) ) ( 1 / 2 0 − 1 / 2 0 0 0 − 1 / 2 0 1 / 2 ) ( 0 1 2 ) = ( − 1 0 1 ) {\displaystyle {\mbox{proj}}_{M}({\begin{pmatrix}0\\1\\2\end{pmatrix}}){\begin{pmatrix}1/2&0&-1/2\\0&0&0\\-1/2&0&1/2\end{pmatrix}}{\begin{pmatrix}0\\1\\2\end{pmatrix}}={\begin{pmatrix}-1\\0\\1\end{pmatrix}}}
问题 5
验证 定义 3.1 中的投影操作是定义良好的。也就是说，在 示例 3.2 和 3.3 中，答案是否依赖于所选基？
答案
不，将向量分解为 v → = m → + n → {\displaystyle {\vec {v}}={\vec {m}}+{\vec {n}}} m → ∈ M {\displaystyle {\vec {m}}\in M} n → ∈ N {\displaystyle {\vec {n}}\in N}
问题 6
什么是到零子空间的正交投影？
答案
向量到子空间的正交投影是该子空间中的一个元素。由于零子空间只有一个元素，即 0 → {\displaystyle {\vec {0}}} 0 → {\displaystyle {\vec {0}}}
此练习推荐给所有读者。 此练习推荐给所有读者。
问题 10
证明如果一个向量垂直于一个集合中的每个向量，那么它也垂直于该集合的生成空间中的每个向量。
答案
设 v → {\displaystyle {\vec {v}}} w → ∈ S {\displaystyle {\vec {w}}\in S} v → ⋅ ( c 1 w → 1 + ⋯ + c n w → n ) = v → ⋅ ( c 1 w → 1 ) + ⋯ + v → ⋅ ( c n ⋅ w → n ) = c 1 ( v → ⋅ w → 1 ) + ⋯ + c n ( v → ⋅ w → n ) = c 1 ⋅ 0 + ⋯ + c n ⋅ 0 = 0 {\displaystyle {\vec {v}}\cdot (c_{1}{\vec {w}}_{1}+\dots +c_{n}{\vec {w}}_{n})={\vec {v}}\cdot (c_{1}{\vec {w}}_{1})+\dots +{\vec {v}}\cdot (c_{n}\cdot {\vec {w}}_{n})=c_{1}({\vec {v}}\cdot {\vec {w}}_{1})+\dots +c_{n}({\vec {v}}\cdot {\vec {w}}_{n})=c_{1}\cdot 0+\dots +c_{n}\cdot 0=0}
问题 11
判断对错：子空间与其正交补集的交集是平凡的。
答案
正确；唯一与自身正交的向量是零向量。
问题 12
证明正交补集的维数之和等于整个空间的维数。
答案
这直接来自 引理 3.7 中的陈述，即空间是这两个空间的直和。
此练习推荐给所有读者。 此练习推荐给所有读者。
问题 14
设 M , N {\displaystyle M,N} R n {\displaystyle \mathbb {R} ^{n}}
证明两次正交运算抵消： ( M ⊥ ) ⊥ = M {\displaystyle (M^{\perp })^{\perp }=M} 证明 M ⊆ N {\displaystyle M\subseteq N} N ⊥ ⊆ M ⊥ {\displaystyle N^{\perp }\subseteq M^{\perp }} 证明 ( M + N ) ⊥ = M ⊥ ∩ N ⊥ {\displaystyle (M+N)^{\perp }=M^{\perp }\cap N^{\perp }}
答案
我们将证明这些集合是相互包含的， M ⊆ ( M ⊥ ) ⊥ {\displaystyle M\subseteq (M^{\perp })^{\perp }} ( M ⊥ ) ⊥ ⊆ M {\displaystyle (M^{\perp })^{\perp }\subseteq M} m → ∈ M {\displaystyle {\vec {m}}\in M} m → {\displaystyle {\vec {m}}} v → ∈ M ⊥ {\displaystyle {\vec {v}}\in M^{\perp }} m → ∈ ( M ⊥ ) ⊥ {\displaystyle {\vec {m}}\in (M^{\perp })^{\perp }} v → ∈ ( M ⊥ ) ⊥ {\displaystyle {\vec {v}}\in (M^{\perp })^{\perp }} 引理 3.7 的证明表明 R n = M ⊕ M ⊥ {\displaystyle \mathbb {R} ^{n}=M\oplus M^{\perp }} ⟨ κ → 1 , … , κ → k , κ → k + 1 , … , κ → n ⟩ {\displaystyle \langle {\vec {\kappa }}_{1},\dots ,{\vec {\kappa }}_{k},{\vec {\kappa }}_{k+1},\dots ,{\vec {\kappa }}_{n}\rangle } ⟨ κ → 1 , … , κ → k ⟩ {\displaystyle \langle {\vec {\kappa }}_{1},\dots ,{\vec {\kappa }}_{k}\rangle } M {\displaystyle M} M ⊥ {\displaystyle M^{\perp }} v → = proj [ κ → 1 ] ( v → ) + ⋯ + proj [ κ → n ] ( v → ) {\displaystyle {\vec {v}}={\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({{\vec {v}}\,})+\dots +{\mbox{proj}}_{[{\vec {\kappa }}_{n}]}({{\vec {v}}\,})} v → ∈ ( M ⊥ ) ⊥ {\displaystyle {\vec {v}}\in (M^{\perp })^{\perp }} M ⊥ {\displaystyle M^{\perp }} v → = proj [ κ → 1 ] ( v → ) + ⋯ + proj [ κ → k ] ( v → ) {\displaystyle {\vec {v}}={\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({{\vec {v}}\,})+\dots +{\mbox{proj}}_{[{\vec {\kappa }}_{k}]}({{\vec {v}}\,})} M {\displaystyle M} v → ∈ M {\displaystyle {\vec {v}}\in M} 备注 。这里有一个更巧妙的方法来处理第二部分：将空间写成 M ⊕ M ⊥ {\displaystyle M\oplus M^{\perp }} M ⊥ ⊕ ( M ⊥ ) ⊥ {\displaystyle M^{\perp }\oplus (M^{\perp })^{\perp }} M ⊆ ( M ⊥ ) ⊥ {\displaystyle M\subseteq (M^{\perp })^{\perp }} M {\displaystyle M} ( M ⊥ ) ⊥ {\displaystyle (M^{\perp })^{\perp }} M {\displaystyle M} ( M ⊥ ) ⊥ {\displaystyle (M^{\perp })^{\perp }} 因为 M ⊆ N {\displaystyle M\subseteq N} N {\displaystyle N} v → {\displaystyle {\vec {v}}} M {\displaystyle M} N ⊥ ⊆ M ⊥ {\displaystyle N^{\perp }\subseteq M^{\perp }} 我们将再次通过互相包含来证明集合相等。第一个方向很简单；任何垂直于 v → {\displaystyle {\vec {v}}} M + N = { m → + n → | m → ∈ M , n → ∈ N } {\displaystyle M+N=\{{\vec {m}}+{\vec {n}}\,{\big |}\,{\vec {m}}\in M,\,{\vec {n}}\in N\}} m → + 0 → {\displaystyle {\vec {m}}+{\vec {0}}} M {\displaystyle M} 0 → + n → {\displaystyle {\vec {0}}+{\vec {n}}} N {\displaystyle N} ( M + N ) ⊥ ⊆ M ⊥ ∩ N ⊥ {\displaystyle (M+N)^{\perp }\subseteq M^{\perp }\cap N^{\perp }} v → ∈ M ⊥ ∩ N ⊥ {\displaystyle {\vec {v}}\in M^{\perp }\cap N^{\perp }} c m → + d n → {\displaystyle c{\vec {m}}+d{\vec {n}}} v → ⋅ ( c m → + d n → ) = c ⋅ ( v → ⋅ m → ) + d ⋅ ( v → ⋅ n → ) = c ⋅ 0 + d ⋅ 0 = 0 {\displaystyle {\vec {v}}\cdot (c{\vec {m}}+d{\vec {n}})=c\cdot ({\vec {v}}\cdot {\vec {m}})+d\cdot ({\vec {v}}\cdot {\vec {n}})=c\cdot 0+d\cdot 0=0}
此练习推荐给所有读者。
问题 15
本小节中的材料使我们能够表达线性映射的范围空间和零空间之间尚未见过的几何关系。
表示 f : R 3 → R {\displaystyle f:\mathbb {R} ^{3}\to \mathbb {R} } ( v 1 v 2 v 3 ) ↦ 1 v 1 + 2 v 2 + 3 v 3 {\displaystyle {\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\mapsto 1v_{1}+2v_{2}+3v_{3}} ( 1 2 3 ) {\displaystyle {\begin{pmatrix}1\\2\\3\end{pmatrix}}} N ( f ) ⊥ {\displaystyle {\mathcal {N}}(f)^{\perp }} 将其推广到适用于任何 f : R n → R {\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} } 表示 f : R 3 → R 2 {\displaystyle f:\mathbb {R} ^{3}\to \mathbb {R} ^{2}} ( v 1 v 2 v 3 ) ↦ ( 1 v 1 + 2 v 2 + 3 v 3 4 v 1 + 5 v 2 + 6 v 3 ) {\displaystyle {\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\mapsto {\begin{pmatrix}1v_{1}+2v_{2}+3v_{3}\\4v_{1}+5v_{2}+6v_{3}\end{pmatrix}}} ( 1 2 3 ) , ( 4 5 6 ) {\displaystyle {\begin{pmatrix}1\\2\\3\end{pmatrix}},\;{\begin{pmatrix}4\\5\\6\end{pmatrix}}} N ( f ) ⊥ {\displaystyle {\mathcal {N}}(f)^{\perp }} 问题 14 的第三项。） 将其推广到适用于任何 f : R n → R m {\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} ^{m}}
这个结果及其相关结果被称为线性代数的基本定理（参见Strang 1993 ）。
答案
表示 ( v 1 v 2 v 3 ) ⟼ f 1 v 1 + 2 v 2 + 3 v 3 {\displaystyle {\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}{\stackrel {f}{\longmapsto }}1v_{1}+2v_{2}+3v_{3}} R e p E 3 , E 1 ( f ) = ( 1 2 3 ) {\displaystyle {\rm {Rep}}_{{\mathcal {E}}_{3},{\mathcal {E}}_{1}}(f)={\begin{pmatrix}1&2&3\end{pmatrix}}} f {\displaystyle f} N ( f ) = { ( v 1 v 2 v 3 ) | 1 v 1 + 2 v 2 + 3 v 3 = 0 } = { ( v 1 v 2 v 3 ) | ( 1 2 3 ) ⋅ ( v 1 v 2 v 3 ) = 0 } {\displaystyle {\mathcal {N}}(f)=\{{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\,{\big |}\,1v_{1}+2v_{2}+3v_{3}=0\}=\{{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\,{\big |}\,{\begin{pmatrix}1\\2\\3\end{pmatrix}}\cdot {\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}=0\}} N ( f ) ⊥ = [ { ( 1 2 3 ) } ] {\displaystyle {\mathcal {N}}(f)^{\perp }=[\{{\begin{pmatrix}1\\2\\3\end{pmatrix}}\}]} 推广到任何 f : R n → R {\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} } h → {\displaystyle {\vec {h}}} ( v 1 ⋮ v n ) ⟼ f h 1 v 1 + ⋯ + h n v n {\displaystyle {\begin{pmatrix}v_{1}\\\vdots \\v_{n}\end{pmatrix}}{\stackrel {f}{\longmapsto }}h_{1}v_{1}+\dots +h_{n}v_{n}} h → ∈ N ( f ) ⊥ {\displaystyle {\vec {h}}\in {\mathcal {N}}(f)^{\perp }} f {\displaystyle f} h → {\displaystyle {\vec {h}}} 当然， R e p E 3 , E 2 ( f ) = ( 1 2 3 4 5 6 ) {\displaystyle {\rm {Rep}}_{{\mathcal {E}}_{3},{\mathcal {E}}_{2}}(f)={\begin{pmatrix}1&2&3\\4&5&6\end{pmatrix}}} N ( f ) { ( v 1 v 2 v 3 ) | ( 1 2 3 4 5 6 ) ( v 1 v 2 v 3 ) = ( 0 0 ) } {\displaystyle {\mathcal {N}}(f)\{{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\,{\big |}\,\;{\begin{pmatrix}1&2&3\\4&5&6\end{pmatrix}}{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}\;\}} ( 1 2 3 ) , ( 4 5 6 ) ∈ N ( f ) ⊥ {\displaystyle {\begin{pmatrix}1\\2\\3\end{pmatrix}},\,{\begin{pmatrix}4\\5\\6\end{pmatrix}}\in {\mathcal {N}}(f)^{\perp }} N ( f ) ⊥ {\displaystyle {\mathcal {N}}(f)^{\perp }} R n {\displaystyle \mathbb {R} ^{n}} M = [ { ( 1 2 3 ) } ] N = [ { ( 4 5 6 ) } ] {\displaystyle M=[\{{\begin{pmatrix}1\\2\\3\end{pmatrix}}\}]\qquad N=[\{{\begin{pmatrix}4\\5\\6\end{pmatrix}}\}]} 问题 14 的第三项中，如提示中所建议的那样。 如上所述，从具体情况推广到一般情况很容易：对于任何 f : R n → R m {\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} ^{m}} H {\displaystyle H} ( v 1 ⋮ v n ) ⟼ f ( h 1 , 1 v 1 + h 1 , 2 v 2 + ⋯ + h 1 , n v n ⋮ h m , 1 v 1 + h m , 2 v 2 + ⋯ + h m , n v n ) {\displaystyle {\begin{pmatrix}v_{1}\\\vdots \\v_{n}\end{pmatrix}}{\stackrel {f}{\longmapsto }}{\begin{pmatrix}h_{1,1}v_{1}+h_{1,2}v_{2}+\dots +h_{1,n}v_{n}\\\vdots \\h_{m,1}v_{1}+h_{m,2}v_{2}+\dots +h_{m,n}v_{n}\end{pmatrix}}} m {\displaystyle m} H {\displaystyle H} h → 1 = ( h 1 , 1 h 1 , 2 ⋮ h 1 , n ) , … h → m = ( h m , 1 h m , 2 ⋮ h m , n ) {\displaystyle {\vec {h}}_{1}={\begin{pmatrix}h_{1,1}\\h_{1,2}\\\vdots \\h_{1,n}\end{pmatrix}},\dots {\vec {h}}_{m}={\begin{pmatrix}h_{m,1}\\h_{m,2}\\\vdots \\h_{m,n}\end{pmatrix}}} N ( f ) ⊥ = [ { h → 1 , … , h → m } ] {\displaystyle {\mathcal {N}}(f)^{\perp }=[\{{\vec {h}}_{1},\dots ,{\vec {h}}_{m}\}]} Strang 1993 ) 中，此空间被描述为 H {\displaystyle H}
问题 16
定义投影 为线性变换 t : V → V {\displaystyle t:V\to V} ( t ∘ t ) ( v → ) = t ( v → ) {\displaystyle (t\circ t)\,({\vec {v}})=t({\vec {v}})} v → ∈ V {\displaystyle {\vec {v}}\in V}
证明：将向量正交投影到直线上，具有该性质。 证明：将向量投影到子空间上，具有该性质。 证明：对于任何这样的 t {\displaystyle t} B = ⟨ β → 1 , … , β → n ⟩ {\displaystyle B=\langle {\vec {\beta }}_{1},\ldots ,{\vec {\beta }}_{n}\rangle } V {\displaystyle V} t ( β → i ) = { β → i i = 1 , 2 , … , r 0 → i = r + 1 , r + 2 , … , n {\displaystyle t({\vec {\beta }}_{i})={\begin{cases}{\vec {\beta }}_{i}&i=1,2,\dots ,\,r\\{\vec {0}}&i=r+1,r+2,\dots ,\,n\end{cases}}} r {\displaystyle r} t {\displaystyle t} 由此推断，每个投影都是沿子空间的投影。 同样可以推断，每个投影都有一个表示形式 R e p B , B ( t ) = ( I Z Z Z ) {\displaystyle {\rm {Rep}}_{B,B}(t)=\left({\begin{array}{c|c}I&Z\\\hline Z&Z\end{array}}\right)}
答案
首先要注意，如果一个向量 v → {\displaystyle {\vec {v}}} v → {\displaystyle {\vec {v}}} s → {\displaystyle {\vec {s}}} ( v → ⋅ s → / s → ⋅ s → ) ⋅ s → {\displaystyle ({\vec {v}}\cdot {\vec {s}}/{\vec {s}}\cdot {\vec {s}})\cdot {\vec {s}}} { k ⋅ v → | k ∈ R } {\displaystyle \{k\cdot {\vec {v}}\,{\big |}\,k\in \mathbb {R} \}} v → = 0 → {\displaystyle {\vec {v}}={\vec {0}}} ( v → ⋅ v → / v → ⋅ v → ) ⋅ v → {\displaystyle ({\vec {v}}\cdot {\vec {v}}/{\vec {v}}\cdot {\vec {v}})\cdot {\vec {v}}} v → {\displaystyle {\vec {v}}} proj ℓ ( v → ) {\displaystyle {\mbox{proj}}_{\ell }({\vec {v}})} 这里的论点与上一项中的论点类似。对于 V = M ⊕ N {\displaystyle V=M\oplus N} v → = m → + n → {\displaystyle {\vec {v}}={\vec {m}}+{\vec {n}}} proj M , N ( v → ) = m → {\displaystyle {\mbox{proj}}_{M,N}({{\vec {v}}\,})={\vec {m}}} proj M , N ( m → ) = m → {\displaystyle {\mbox{proj}}_{M,N}({\vec {m}})={\vec {m}}} M {\displaystyle M} M {\displaystyle M} N {\displaystyle N} m → = m → + 0 → {\displaystyle {\vec {m}}={\vec {m}}+{\vec {0}}} N {\displaystyle N} M {\displaystyle M} As suggested by the prior items, the condition gives that t {\displaystyle t} β → 1 {\displaystyle {\vec {\beta }}_{1}} β → r {\displaystyle {\vec {\beta }}_{r}} t {\displaystyle t} M {\displaystyle M} dim ( M ) = r {\displaystyle \dim(M)=r} N {\displaystyle N} t {\displaystyle t} V = M ⊕ N {\displaystyle V=M\oplus N} M ∩ N = { 0 → } {\displaystyle M\cap N=\{{\vec {0}}\}} M + N = V {\displaystyle M+N=V} m → {\displaystyle {\vec {m}}} v → ∈ V {\displaystyle {\vec {v}}\in V} t ( v → ) = m → {\displaystyle t({\vec {v}})={\vec {m}}} t {\displaystyle t} t ( m → ) = ( t ∘ t ) ( v → ) = t ( v → ) = m → {\displaystyle t({\vec {m}})=(t\circ t)\,({\vec {v}})=t({\vec {v}})={\vec {m}}} t ( m → ) = 0 → {\displaystyle t({\vec {m}})={\vec {0}}} v → {\displaystyle {\vec {v}}} t ( v → ) = t ( t ( v → ) ) {\displaystyle t({\vec {v}})=t(t({\vec {v}}))} t ( v → − t ( v → ) ) = 0 → {\displaystyle t({\vec {v}}-t({\vec {v}}))={\vec {0}}} v → = t ( v → ) + ( v → − t ( v → ) ) {\displaystyle {\vec {v}}=t({\vec {v}})+({\vec {v}}-t({\vec {v}}))} B = ⟨ β → 1 , … , β → n ⟩ {\displaystyle B=\langle {\vec {\beta }}_{1},\ldots ,{\vec {\beta }}_{n}\rangle } V {\displaystyle V} ⟨ β → 1 , … , β → r ⟩ {\displaystyle \langle {\vec {\beta }}_{1},\ldots ,{\vec {\beta }}_{r}\rangle } M {\displaystyle M} ⟨ β → r + 1 , … , β → n ⟩ {\displaystyle \langle {\vec {\beta }}_{r+1},\ldots ,{\vec {\beta }}_{n}\rangle } N {\displaystyle N} 每个投影（在本练习中定义）都是投影到其值域并沿着其零空间进行投影。 这也直接来自第三项。
Strang, Gilbert (1993), "线性代数基本定理", 美国数学月刊 , 美国数学学会: 848–855 Strang, Gilbert (1980), 线性代数及其应用 (第 2 版), Hartcourt Brace Javanovich 
![{\displaystyle {\begin{array}{*{3}{rc}r}3u&+&v&&&=&0\\u&&&+&w&=&0\end{array}}\;{\xrightarrow[{}]{-(1/3)\rho _{1}+\rho _{2}}}\;{\begin{array}{*{3}{rc}r}3u&+&v&&&=&0\\&&-(1/3)v&+&w&=&0\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9aba2fb1f7833438bedd9c17071d40c4936e0dea)
![{\displaystyle {\mbox{proj}}_{[{\vec {s}}\,]}({\begin{pmatrix}1\\-3\end{pmatrix}})={\frac {{\begin{pmatrix}1\\-3\end{pmatrix}}\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}}{{\begin{pmatrix}-1\\1\end{pmatrix}}\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}}}\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}={\frac {-4}{2}}\cdot {\begin{pmatrix}-1\\1\end{pmatrix}}={\begin{pmatrix}2\\-2\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ab8c36b1ba917e7a3dfd6caa965ad861269a18f)
![{\displaystyle {\vec {v}}={\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({{\vec {v}}\,})+\dots +{\mbox{proj}}_{[{\vec {\kappa }}_{n}]}({{\vec {v}}\,})={\frac {{\vec {v}}\cdot {\vec {\kappa }}_{1}}{{\vec {\kappa }}_{1}\cdot {\vec {\kappa }}_{1}}}\cdot {\vec {\kappa }}_{1}+\dots +{\frac {{\vec {v}}\cdot {\vec {\kappa }}_{n}}{{\vec {\kappa }}_{n}\cdot {\vec {\kappa }}_{n}}}\cdot {\vec {\kappa }}_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8fe1b3e5e8515281dce05e454783abc8d50c2349)
![{\displaystyle {\vec {v}}={\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({{\vec {v}}\,})+\dots +{\mbox{proj}}_{[{\vec {\kappa }}_{n}]}({{\vec {v}}\,})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60f2b61a471c291b1d2d064cb0a57887e77bc73e)
![{\displaystyle {\vec {v}}={\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({{\vec {v}}\,})+\dots +{\mbox{proj}}_{[{\vec {\kappa }}_{k}]}({{\vec {v}}\,})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f14a7fab8047e6f3be3e5da8f2e419199ce045e)
![{\displaystyle {\mathcal {N}}(f)^{\perp }=[\{{\begin{pmatrix}1\\2\\3\end{pmatrix}}\}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d754510f8453f245f0583c06ad45e2e6981d3585)
![{\displaystyle M=[\{{\begin{pmatrix}1\\2\\3\end{pmatrix}}\}]\qquad N=[\{{\begin{pmatrix}4\\5\\6\end{pmatrix}}\}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18756942da312249ae314a7ca2e1cd43bfea08dd)
![{\displaystyle {\mathcal {N}}(f)^{\perp }=[\{{\vec {h}}_{1},\dots ,{\vec {h}}_{m}\}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8653570926540892bb808449181ef23b66fc0385)
