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 = SASASA=SA。

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}SAx=0⇒ASAx=0⇒Ax=0#(∗)​

其中A x = y A\mathbf{x = y}Ax=y。易知

A x = 0 ⇒ S A x = 0 A\mathbf{x = 0 \Rightarrow}SA\mathbf{x = 0}Ax=0⇒SAx=0

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 = AASA=A

所以

S A S A = S A SASA = SASASA=SA

b. 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)pr+1​…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}x=x1​+x2​

x 2 = P x \mathbf{x}_{2} = P\mathbf{x}x2​=Px

x 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}x1​=k1​p1​+k2​p1​+…+kr​pr​

x 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}x2​=kr+1​pr+1​+…+ks​ps​

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​、pr+1​…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}x=x1​+x2​的表示是唯一的，所以当x = 0 \mathbf{x = 0}x=0时，有

x 1 = 0 x 2 = 0 \mathbf{x}_{1} = \mathbf{0\ \ \ \ \ \ }\mathbf{x}_{2}\mathbf{= 0}x1​=0x2​=0

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} = 0k1​=k2​=…=kr​=kr+1​=…ks​=0

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​、pr+1​…ps​线性无关，所以s = n s = ns=n，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