《线性代数应该这样学》课内验证3-线性映射

作者: Raow1 | 来源:发表于2021-09-24 00:24 被阅读0次

CHAPTER 3 Linear Maps

Make sure you verify that each of the functions defined below is indeed a linear map:
- zero: $0 \in \mathcal{L}(V,W)$ is defined by $0v=0$ .
- identity: $I \in \mathcal{L}(V,V)$ is defined by $Iv=v$ .
- multiplication by $x^2$ : $T \in \mathcal{L}(\mathcal{P}(\mathbb{R}),\mathcal{P}(\mathbb{R}))$ is defined by $(Tp)(x)=x^2p(x)$ for $x \in \mathbb{R}$ .
- backwar shift: $T \in \mathcal{L}(\mathbb{F}^{\infty}, \mathbb{F}^{\infty})$ is defined by $T(x_1,x_2,x_3,\cdots)=(x_2,x_3,\cdots)$ .
- from $\mathbb{F}^{n}$ to $\mathbb{F}^{m}$ : $T \in \mathcal{L}(\mathbb{F}^{n}, \mathbb{F}^{m})$ is defined by $T(x_1, \cdots,x_n)=(A_{1,1}x_1+\cdots+A_{1,n}x_n,\cdots,A_{m,1}x_1+\cdots+A_{m,n}x_n)$ .

Proof

zero. Additivity: for all $u,v \in V$ , $0(u+v)=0=0u+0v$ .

homogeneity: for all $\lambda \in \mathbb{F}$ and all $v \in V$ , $0(\lambda v)=0=\lambda 0= \lambda 0(v)$ .
identity: Additivity: for all $u,v \in V$ , $I(u+v)=u+v=I(u)+I(v)$ .

homogeneity: for all $\lambda \in \mathbb{F}$ and all $v \in V$ , $I(\lambda v)=\lambda v=\lambda I(v)$ .
multiplication by $x^2$ : Additivity: for all $u,v \in \mathcal{P}(\mathbb{R})$ , $T(u+v)(x)=x^2(u+v)=x^2u+x^2v=T(u)+T(v)$

homogeneity: for all $\lambda \in \mathbb{F}$ and all $v \in \mathcal{P}(\mathbb{R})$ , $T(\lambda v)(x)=\lambda x^2 v=\lambda T(v)$ .
backwar shift: Additivity: for all $u,v \in \mathbb{F}^{\infty}$ , $T(u+v)=T((u_1+v_1,u_2+v_2,u_3+v_3,\cdots))=(u_2+v_2,u_3+v_3,\cdots)=(u_2,u_3,\cdots)+(v_2,v_3,\cdots)=T(u)+T(v)$

homogeneity: for all $\lambda \in \mathbb{F}$ and all $v \in \mathbb{F}^{\infty}$ , $T(\lambda v)=T((\lambda v_1,\lambda v_2,\lambda v_3,\cdots))=(\lambda v_2,\lambda v_3,\cdots)=\lambda (v_2,v_3,\cdots)=\lambda T(v)$
from $\mathbb{F}^{n}$ to $\mathbb{F}^{m}$ : Additivity: for all $u,v \in \mathbb{F}^{n}$ ,
$\begin{align*} T(u+v) &= T(u_1+v_1,\cdots,u_n+v_n)=(A_{1,1}(u_1+v_1)+\cdots+A_{1,n}(u_n+v_n),\cdots,A_{m,1}(u_1+v_1)+\cdots+A_{m,n}(u_n+v_n)) \\ &= (A_{1,1}u_1+\cdots+A_{1,n}u_n+A_{1,1}v_1+\cdots+A_{1,n}v_n , \cdots,A_{m,1}u_1+\cdots+A_{m,n}u_n+A_{m,1}v_1+\cdots+A_{m,n}v_n) \\ &= (A_{1,1}u_1+\cdots+A_{1,n}u_n ,\cdots , A_{m,1}u_1+\cdots+A_{m,n}u_n)+(A_{1,1}v_1+\cdots+A_{1,n}v_n ,\cdots , A_{m,1}v_1+\cdots+A_{m,n}v_n) \\ &= T(u)+T(v) \end{align*}$
homogeneity: for all $\lambda \in \mathbb{F}$ and all $v \in \mathbb{F}^{n}$ ,
$\begin{align*} T(\lambda v)&=T(\lambda v_1+\cdots+\lambda v_n) \\ &= (\lambda A_{1,1}v_1+\cdots+\lambda A_{1,n}v_n ,\cdots , \lambda A_{m,1}v_1+\cdots+\lambda A_{m,n}v_n) \\ &= \lambda (A_{1,1}v_1+\cdots+A_{1,n}v_n ,\cdots , A_{m,1}v_1+\cdots+A_{m,n}v_n) \\ &= \lambda T(v) \end{align*}$

You should verify that $ST$ is indeed a linear map from $U$ to $W$ whenever $T \in \mathcal{L}(U,V)$ and $S \in \mathcal{L}(V,W)$ .

Proof Additivity: for all $u,v \in U$ , $(ST)(u+v)=S(T(u+v))=S(T(u)+T(v))=S(T(u))+S(T(v))=(ST)(u)+(ST)(v)$

homogeneity: for all $\lambda \in \mathbb{F}$ and all $u\in U$ , $(ST)(\lambda u)=S(T(\lambda u))=S(\lambda Tu)=\lambda S(Tu)=\lambda (ST)(u)$

The reader should verify that $\pi$ is indeed a linear map.

Proof Additivity: for all $v,w \in V$ , $\pi (v+w)=(v+w)+U=(v+U)+(w+U)=\pi(v)+\pi(w)$

homogeneity: for all $\lambda \in \mathbb{F}$ and all $v \in V$ , $\pi (\lambda v)=\lambda v+U=\lambda (v+U)=\lambda \pi(v)$

The routine verification that $\tilde{T}$ is linear is left to the reader.

Proof Additivity: for all $u,v \in V$ , $\tilde{T}(u+\text{null }T +v+\text{null }T)=T(u+v)=T(u)+T(v)=\tilde{T}(u+\text{null }T)+\tilde{T}(v+\text{null }T)$

homogeneity: for all $\lambda \in \mathbb{F}$ and all $v \in V$ , $\tilde{T}(\lambda v+\text{null }T)=T(\lambda v)=\lambda T(v)=\lambda \tilde{T}(v+\text{null }T)$

You should construct the proof outlined in the paragraph above, even though a slicker proof is presented here. Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$ . Then $\dim U + \dim U^0 =\dim V$ .

Proof Let $u_1,\cdots,u_m$ be a basis of $U$ ; thus $\dim U =m$ . The linearly independent list $u_1,\cdots,u_m$ can be extended to a basis $u_1,\cdots,u_m,\cdots, u_n$ of $V$ .

Thus $\dim V=n$ . To complete the proof, we need only show that $U^0$ is finite-dimensional and $\dim U^0=n-m$ .

Suppose $\varphi_1,\cdots , \varphi_m,\cdots,\varphi_n$ is a basis of $V'$ , $\varphi \in U^0$ . We can write
$\varphi =c_1\varphi_1+\cdots+c_m\varphi_m+\cdots+c_n\varphi_n$
Because $u_1 \in U$ and $\varphi \in U^0$ , we have
$0=\varphi(u_1)=(c_1\varphi_1+\cdots+c_m\varphi_m+\cdots+c_n\varphi_n)(u_1)=c_1$
Similarly, $c_2,\cdots,c_m$ all are $0$ . Hence $\varphi=c_{m+1}\varphi_{m+1}+\cdots+c_n\varphi_n$ .

Thus $\varphi \in \text{span}(\varphi_{m+1},\cdots,\varphi_n)$ , which shows that $U^0 \subset \text{span}(\varphi_{m+1},\cdots,\varphi_n)$ .

Suppose $\varphi \in \text{span}(\varphi_{m+1},\cdots,\varphi_n)$ . Then there exist $c_{m+1},\cdots,c_n$ such that $\varphi=c_{m+1}\varphi_{m+1}+\cdots+c_n\varphi_n$ . If $(a_1,\cdots,a_m,0,\cdots,0) \in U$ , then
$\varphi(a_1,\cdots,a_m,0,\cdots,0)=0$
Thus $\varphi \in U^0$ , which shows that $\text{span}(\varphi_{m+1},\cdots,\varphi_n) \subset U^0$ .

Hence $U^0=\text{span}(\varphi_{m+1},\cdots,\varphi_n)$ . We also know that $\varphi_{1},\cdots,\varphi_n$ is a basis of $V'$ . Therefore $\varphi_{m+1},\cdots,\varphi_{n}$ is linearly independent.

Hence it can be a basis of $U^0$ . Thus $\dim U^0=n-m$ .

《线性代数应该这样学》课内验证3-线性映射

CHAPTER 3 Linear Maps

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