Fudan University 2014--2015 first semester (level 14) Advanced Algebra I final exam seventh big question answer

Vii. (10 points) set \ (v\) the \ (n\) dimension on the number domain \ (\mathbb{k}\), \ (s=\{v_1,v_2,\cdots,v_m\}\) is the vector group in \ (v\), define the set \ (r_s=\{(a _1,a_2,\cdots,a_m) \in\mathbb{k}^m\,|\,a_1v_1+a_2v_2+\cdots+a_mv_m=0\}\). Then take the vector Group \ (t=\{u_1,u_2,\cdots,u_m\}\) in \ (v\). Prove:

(1) \ (r_s\) is a linear subspace of \ (\mathbb{k}^m\);

(2) The existence of linear transformation \ (\varphi\) makes \ (\varphi (v_i) =u_i\, (1\leq i\leq m) \) The sufficient and necessary conditions are \ (R_s\subseteq r_t\);

(3) The existence of linear automorphism \ (\varphi\) makes \ (\varphi (v_i) =u_i\, (1\leq i\leq m) \) The sufficient and necessary condition is \ (r_s=r_t\).

Proof method One (geometrical method)

(1) apparently \ (0\in r_s\). either fetch \ (\alpha,\beta\in r_s\), \ (k\in\mathbb{k}\), easy authentication \ (\alpha+\beta\in r_s\), \ (K\alpha\in r_s\), so \ (r_s\) is \ (\mathbb{k }^m\) linear subspace.

(2) Necessity: Linear validation by \ (\varphi\). Adequacy: It is advisable to set \ (\{v_1,\cdots,v_r\}\) is the maximal unrelated group of the vector Group \ (s\) and expand it to a set of base  \ (v\) (\{v_1,\cdot s,v_r,e_{r+1},\cdots,e_n\}\). By the linear expansion theorem, define \ (\varphi (v_i) =u_i\, (1\leq i\leq R) \), \ (\varphi (E_j) =0\, (R+1\leq j\leq n) \), and expand \ (\varphi\) to a linear transformation on \ (v\). to either \ (V_j\, (R+1\leq j\leq m) \), set \ (v_j=\lambda_1v_1+\cdots+\lambda_rv_r\), then \ ((\lambda_1,\cdots,\lambda_r,0,\cdots, 0,-1,0,\cdots,0) \in r_s\). Known by \ (R_s\subseteq r_t\) \ (u_j=\lambda_1u_1+\cdots+\lambda_ru_r\). So \[\varphi (V_j) =\varphi (\sum_{i=1}^r\lambda_iv_i) =\sum_{i=1}^r\lambda_i\varphi (v_i) =\sum_{i=1}^r\lambda_iu_i= U_j\,\, (R+1\leq j\leq m). \]

(3) Necessity: the linear verification by \ (\varphi,\varphi^{-1}\) is get. Adequacy: It is advisable to set \ (\{v_1,\cdots,v_r\}\) is a vector group \ (s\) of the Maximal independent group, and expand it to \ (v\) a set of base \ (\{v_1,\cdots,v_r,e_{r+1},\cdots,e_n\}\). Using the same proof as (2): if \ (v_j=\lambda_1v_1+\cdots+\lambda_rv_r\), then \ (u_j=\lambda_1u_1+\cdots+\lambda_ru_r\, (R+1\leq J\leq m) \). Set \ (\mu_1u_1+\cdots+\mu_ru_r=0\), then \ ((\mu_1,\cdots,\mu_r,0,\cdots,0) \in r_t\). by \ (r_s=r_t\) (\mu_1v_1+\cdots+\mu_rv_r=0\). also \ (\{v_1,\cdots,v_r\}\) linearly independent, therefore \ (\mu_1=\cdots=\mu_r=0\).  therefore \ (\{u_1,\cdots,u_r\}\) is a large independent group of vector groups \ (t\) and expands it to a group of \ (v\) (\{u_1,\cdots,u_r,f_{r+1},\cdots,f_n\}\). By the linear expansion theorem, define \ (\varphi (v_i) =u_i\, (1\leq i\leq R) \), \ (\varphi (E_j) =f_j\, (R+1\leq j\leq n) \), and expand \ (\varphi\) to \ (v\) The linear variable on Because \ (\varphi\) maps the base to the base, so \ (\varphi\) is a linear automorphism. Using the same proofs as (2) can be used (\varphi (V_j) =u_j\, (R+1\leq j\leq m) \).

Forensic Method II (algebraic method)

A set of base \ (\{e_1,e_2,\cdots,e_n\}\) of \ (v\) is taken, there is a linear isomorphism of \ (v\) to \ (n\) Willi vector space \ (\mathbb{k}_n\), which will \ ( V\in v\) maps to \ (v\) The coordinate vector of base \ (\{e_1,e_2,\cdots,e_n\}\). Set \ (\alpha_i=\eta (v_i) \), \ (\beta_i=\eta (u_i) \, (1\leq i\leq m) \), then \ (\alpha_i,\beta_i\) is the \ (n\) Willi Vector. Block construction by columns \ (N\times m\) Order matrix \[a= (\alpha_1,\alpha_2\cdots,\alpha_m), \,\,\,\,b= (\beta_1,\beta_2\cdots,\beta_m). \] In the sense of linear isomorphism, \ (r_s\) is equivalent to the solution space \ (v_a\) of the linear Equation Group \ (ax=0\), and \ (r_t\) equals to the solution space of the linear equation Group \ (bx=0\), where \ (v_b\ (x=) '). Therefore, in the sense of linear isomorphism, the subject is equivalent to proving the following conclusions:

(1) The Solution Space \ (v_a\) of the linear Equation Group \ (ax=0\) is the subspace of \ (\mathbb{k}^m\);

(2) existence \ (n\) Order matrix \ (p\) make \ (pa=b\) The sufficient and necessary condition is \ (V_a\subseteq v_b\);

(3) existence \ (n\) Order non-heterogeneous array \ (p\) makes \ (pa=b\) The sufficient and necessary condition is \ (v_a=v_b\).

prove (1) is obvious. (2) In the final review I told a geometrical proof; Its algebraic proof is also very simple, as long as the \ (ax=0\) and \ (\begin{pmatrix} a \ B \end{pmatrix}x=0\) can be obtained by the line vector \ (b\) is a linear combination of \ (a\) line vectors. (3) for the No. 208 page of the Fudan Gaodai review Question 38, the answer can refer to the Fudan high-generation white Paper 121th page Example 4.17, in the fourth Chapter review I also carefully said this proof; Its algebraic proof is also very simple, as long as the \ (ax=0\) and \ (\begin{pmatrix} A \ B \end{pmatrix}x=0\) and \ (bx=0\) The same solution can get \ (b\) line vector of the maximal independent group and \ (a\) of the line vector of the Great Independent  Groups are equivalent so that non-heterogeneous matrices can be constructed (p\). \ (\box\)

Fudan University 2014--2015 first semester (level 14) Advanced Algebra I final exam seventh big question answer