【第九章 习题25】设A ∈ L ( R n , R m ) A \in L\left( R^{n},R^{m} \right)A∈L(Rn,Rm)令A AA的秩是r rr。a. 像在定理9.32的证明中那样定义S SS。证明S A SASA是R n R^{n}Rn中的射影它的零空间是N ( A ) \mathcal{N}(A)N(A)而它的值域是R ( S ) \mathcal{R}(S)R(S)。提示根据( 68 ) (68)(68)S A S A S A SASA SASASASA。b. 用( a ) (a)(a)证明dim N ( A ) dim R ( A ) n \dim{\mathcal{N}(A)} \dim{\mathcal{R}(A)} ndimN(A)dimR(A)n【证明】a. 首先根据( 68 ) (68)(68)S A x 0 ⇒ A S A x 0 ⇒ A x 0 # ( ∗ ) \begin{array}{r} SA\mathbf{x}\mathbf{ 0 \Rightarrow}ASA\mathbf{x}\mathbf{ 0 \Rightarrow}A\mathbf{x}\mathbf{ 0\ }\#(*) \end{array}SAx0⇒ASAx0⇒Ax0#(∗)其中A x y A\mathbf{x y}Axy。易知A x 0 ⇒ S A x 0 A\mathbf{x 0 \Rightarrow}SA\mathbf{x 0}Ax0⇒SAx0所以S A SASA的零空间与A AA一致都是N ( A ) \mathcal{N}(A)N(A)。根据S SS的定义A x A\mathbf{x}Ax的值域R ( A ) \mathcal{R}(A)R(A)恰好是S SS的定义域Y 1 Y_{1}Y1所以S A SASA的值域是R ( S ) \mathcal{R}(S)R(S)。根据*因为A S A A ASA AASAA所以S A S A S A SASA SASASASAb. 根据S SS的定义dim R ( S ) dim R ( A ) \dim{\mathcal{R}(S)} \dim{\mathcal{R}(A)}dimR(S)dimR(A)于是dim N ( A ) dim R ( A ) dim N ( S A ) dim R ( S ) dim N ( S A ) dim R ( S A ) \dim{\mathcal{N}(A)} \dim{\mathcal{R}(A)} \dim{\mathcal{N(}SA)} \dim{\mathcal{R}(S)} \dim{\mathcal{N(}SA)} \dim{\mathcal{R}(SA)}dimN(A)dimR(A)dimN(SA)dimR(S)dimN(SA)dimR(SA)下面证明对于任意的射影P ∈ L ( R n ) P \in L\left( R^{n} \right)P∈L(Rn)都有dim N ( P ) dim R ( P ) n \dim{\mathcal{N}(P)} \dim{\mathcal{R}(P)} ndimN(P)dimR(P)n设向量p 1 、 p 2 … p r ∈ N ( P ) \mathbf{p}_{1}、\mathbf{p}_{2}\ldots\mathbf{p}_{r} \in \mathcal{N}(P)p1、p2…pr∈N(P)为N ( P ) \mathcal{N}(P)N(P)的一组基向量p r 1 … p s ∈ R ( P ) ( P x ) \mathbf{p}_{r 1}\ldots\mathbf{p}_{s} \in \mathcal{R}(P) \left( P\mathbf{x} \right)pr1…ps∈R(P)(Px)为R ( P ) \mathcal{R}(P)R(P)的一组基。对于任意的向量x ∈ R n \mathbf{x \in}R^{n}x∈Rn因为P PP为射影都有x x 1 x 2 \mathbf{x} \mathbf{x}_{1} \mathbf{x}_{2}xx1x2x 2 P x \mathbf{x}_{2} P\mathbf{x}x2Pxx 1 k 1 p 1 k 2 p 1 … k r p r \mathbf{x}_{1} k_{1}\mathbf{p}_{1} k_{2}\mathbf{p}_{1} \ldots k_{r}\mathbf{p}_{r}x1k1p1k2p1…krprx 2 k r 1 p r 1 … k s p s \mathbf{x}_{2} k_{r 1}\mathbf{p}_{r 1} \ldots k_{s}\mathbf{p}_{s}x2kr1pr1…ksps所以p 1 、 p 2 … p r 、 p r 1 … p s \mathbf{p}_{1}、\mathbf{p}_{2}\ldots\mathbf{p}_{r}、\mathbf{p}_{r 1}\ldots\mathbf{p}_{s}p1、p2…pr、pr1…ps生成了空间R n R^{n}Rn由于x x 1 x 2 \mathbf{x} \mathbf{x}_{1} \mathbf{x}_{2}xx1x2的表示是唯一的所以当x 0 \mathbf{x 0}x0时有x 1 0 x 2 0 \mathbf{x}_{1} \mathbf{0\ \ \ \ \ \ }\mathbf{x}_{2}\mathbf{ 0}x10x20从而k 1 k 2 … k r k r 1 … k s 0 k_{1} k_{2} \ldots k_{r} k_{r 1} \ldots k_{s} 0k1k2…krkr1…ks0也就是说p 1 、 p 2 … p r 、 p r 1 … p s \mathbf{p}_{1}、\mathbf{p}_{2}\ldots\mathbf{p}_{r}、\mathbf{p}_{r 1}\ldots\mathbf{p}_{s}p1、p2…pr、pr1…ps线性无关所以s n s nsn从而dim N ( P ) dim R ( P ) n \dim{\mathcal{N}(P)} \dim{\mathcal{R}(P)} ndimN(P)dimR(P)n由于S A SASA是R n R^{n}Rn中的射影所以dim N ( S A ) dim R ( S A ) n \dim{\mathcal{N(}SA)} \dim{\mathcal{R}(SA)} ndimN(SA)dimR(SA)n从而dim N ( A ) dim R ( A ) n \dim{\mathcal{N}(A)} \dim{\mathcal{R}(A)} ndimN(A)dimR(A)n
数学分析原理答案——第九章 习题25
【第九章 习题25】设A ∈ L ( R n , R m ) A \in L\left( R^{n},R^{m} \right)A∈L(Rn,Rm)令A AA的秩是r rr。a. 像在定理9.32的证明中那样定义S SS。证明S A SASA是R n R^{n}Rn中的射影它的零空间是N ( A ) \mathcal{N}(A)N(A)而它的值域是R ( S ) \mathcal{R}(S)R(S)。提示根据( 68 ) (68)(68)S A S A S A SASA SASASASA。b. 用( a ) (a)(a)证明dim N ( A ) dim R ( A ) n \dim{\mathcal{N}(A)} \dim{\mathcal{R}(A)} ndimN(A)dimR(A)n【证明】a. 首先根据( 68 ) (68)(68)S A x 0 ⇒ A S A x 0 ⇒ A x 0 # ( ∗ ) \begin{array}{r} SA\mathbf{x}\mathbf{ 0 \Rightarrow}ASA\mathbf{x}\mathbf{ 0 \Rightarrow}A\mathbf{x}\mathbf{ 0\ }\#(*) \end{array}SAx0⇒ASAx0⇒Ax0#(∗)其中A x y A\mathbf{x y}Axy。易知A x 0 ⇒ S A x 0 A\mathbf{x 0 \Rightarrow}SA\mathbf{x 0}Ax0⇒SAx0所以S A SASA的零空间与A AA一致都是N ( A ) \mathcal{N}(A)N(A)。根据S SS的定义A x A\mathbf{x}Ax的值域R ( A ) \mathcal{R}(A)R(A)恰好是S SS的定义域Y 1 Y_{1}Y1所以S A SASA的值域是R ( S ) \mathcal{R}(S)R(S)。根据*因为A S A A ASA AASAA所以S A S A S A SASA SASASASAb. 根据S SS的定义dim R ( S ) dim R ( A ) \dim{\mathcal{R}(S)} \dim{\mathcal{R}(A)}dimR(S)dimR(A)于是dim N ( A ) dim R ( A ) dim N ( S A ) dim R ( S ) dim N ( S A ) dim R ( S A ) \dim{\mathcal{N}(A)} \dim{\mathcal{R}(A)} \dim{\mathcal{N(}SA)} \dim{\mathcal{R}(S)} \dim{\mathcal{N(}SA)} \dim{\mathcal{R}(SA)}dimN(A)dimR(A)dimN(SA)dimR(S)dimN(SA)dimR(SA)下面证明对于任意的射影P ∈ L ( R n ) P \in L\left( R^{n} \right)P∈L(Rn)都有dim N ( P ) dim R ( P ) n \dim{\mathcal{N}(P)} \dim{\mathcal{R}(P)} ndimN(P)dimR(P)n设向量p 1 、 p 2 … p r ∈ N ( P ) \mathbf{p}_{1}、\mathbf{p}_{2}\ldots\mathbf{p}_{r} \in \mathcal{N}(P)p1、p2…pr∈N(P)为N ( P ) \mathcal{N}(P)N(P)的一组基向量p r 1 … p s ∈ R ( P ) ( P x ) \mathbf{p}_{r 1}\ldots\mathbf{p}_{s} \in \mathcal{R}(P) \left( P\mathbf{x} \right)pr1…ps∈R(P)(Px)为R ( P ) \mathcal{R}(P)R(P)的一组基。对于任意的向量x ∈ R n \mathbf{x \in}R^{n}x∈Rn因为P PP为射影都有x x 1 x 2 \mathbf{x} \mathbf{x}_{1} \mathbf{x}_{2}xx1x2x 2 P x \mathbf{x}_{2} P\mathbf{x}x2Pxx 1 k 1 p 1 k 2 p 1 … k r p r \mathbf{x}_{1} k_{1}\mathbf{p}_{1} k_{2}\mathbf{p}_{1} \ldots k_{r}\mathbf{p}_{r}x1k1p1k2p1…krprx 2 k r 1 p r 1 … k s p s \mathbf{x}_{2} k_{r 1}\mathbf{p}_{r 1} \ldots k_{s}\mathbf{p}_{s}x2kr1pr1…ksps所以p 1 、 p 2 … p r 、 p r 1 … p s \mathbf{p}_{1}、\mathbf{p}_{2}\ldots\mathbf{p}_{r}、\mathbf{p}_{r 1}\ldots\mathbf{p}_{s}p1、p2…pr、pr1…ps生成了空间R n R^{n}Rn由于x x 1 x 2 \mathbf{x} \mathbf{x}_{1} \mathbf{x}_{2}xx1x2的表示是唯一的所以当x 0 \mathbf{x 0}x0时有x 1 0 x 2 0 \mathbf{x}_{1} \mathbf{0\ \ \ \ \ \ }\mathbf{x}_{2}\mathbf{ 0}x10x20从而k 1 k 2 … k r k r 1 … k s 0 k_{1} k_{2} \ldots k_{r} k_{r 1} \ldots k_{s} 0k1k2…krkr1…ks0也就是说p 1 、 p 2 … p r 、 p r 1 … p s \mathbf{p}_{1}、\mathbf{p}_{2}\ldots\mathbf{p}_{r}、\mathbf{p}_{r 1}\ldots\mathbf{p}_{s}p1、p2…pr、pr1…ps线性无关所以s n s nsn从而dim N ( P ) dim R ( P ) n \dim{\mathcal{N}(P)} \dim{\mathcal{R}(P)} ndimN(P)dimR(P)n由于S A SASA是R n R^{n}Rn中的射影所以dim N ( S A ) dim R ( S A ) n \dim{\mathcal{N(}SA)} \dim{\mathcal{R}(SA)} ndimN(SA)dimR(SA)n从而dim N ( A ) dim R ( A ) n \dim{\mathcal{N}(A)} \dim{\mathcal{R}(A)} ndimN(A)dimR(A)n
