复旦大学谢启鸿老师高等代数在线习题课思考题分析与解 ep.32

2021-11-01 21:46 作者:CharlesMa0606 0人读过 | 我要投稿

本文内容主要有关于线性同构，在高代白皮书上对应第4.2.2节

题目来自于复旦大学谢启鸿教授在本站高等代数习题课的课后思考题，本文仅供学习交流

本人解题水平有限，可能会有错误，恳请斧正！

练习题1 设 $V%E5%92%8CU$ 为数域 $F$ 上的向量空间， $e_1%2Ce_2%2C%5Ccdots%2Ce_n$ 和 $f_1%2Cf_2%2C%E2%8B%AF%2Cf_m$ 分别是 $V%E5%92%8CU$ 的基.定义 $V%E5%92%8CU$ 的线性映射 $%5Cvarphi_%7Bij%7D%5Cleft(i%3D1%2C2%2C%5Ccdots%2Cn%3Bj%3D1%2C2%2C%5Ccdots%2Cm%5Cright)$ :

$%5Cvarphi_%7Bij%7D%5Cleft(e_j%5Cright)%3Df_i%2C%5Cvarphi_%7Bij%7D%5Cleft(e_k%5Cright)%3D0%5Cleft(%5Cforall%20k%5Cneq%20j%5Cright)$

求证： $%5Cvarphi_%7Bij%7D$ 组成向量空间 $L%5Cleft(V%2CU%5Cright)$ 的一组基.若令 $%5Csigma%5Cleft(%5Cvarphi_%7Bij%7D%5Cright)%3DE_%7Bij%7D$ ，这里 $E_%7Bij%7D$ 是基础矩阵，证明 $%5Csigma%3AL%5Cleft(V%2CU%5Cright)%5Crightarrow%20M_%7Bm%5Ctimes%20n%7D%5Cleft(F%5Cright)$ 是线性同构.

分析与解 首先验证 $%5Cvarphi_%7Bij%7D$ 组成向量空间 $L%5Cleft(V%2CU%5Cright)$ 的一组基.

设 $c_%7B11%7D%5Cvarphi_%7B11%7D%2Bc_%7B12%7D%5Cvarphi_%7B12%7D%2B%5Ccdots%2Bc_%7Bnm%7D%5Cvarphi_%7Bnm%7D%3D0%2Cc_%7Bij%7D%5Cin%20F$ ,注意到 $V$ 中任意一个向量都可以表示为 $e_1%2Ce_2%2C%5Ccdots%2Ce_n$ 的线性组合，从而不妨设 $%5Calpha%3Dk_1e_1%2Bk_2e_2%2B%5Ccdots%2Bk_ne_n$ ，于是得到

$%5Cleft(c_%7B11%7Dk_1%2Bc_%7B21%7Dk_2%2B%5Ccdots%2Bc_%7Bn1%7Dk_n%5Cright)f_1%2B%5Ccdots%2B%5Cleft(c_%7B1m%7Dk_1%2B%5Ccdots%2Bc_%7Bnm%7Dk_n%5Cright)f_m%3D0$

从而

$c_%7B11%7Dk_1%2Bc_%7B21%7Dk_2%2B%5Ccdots%2Bc_%7Bn1%7Dk_n%3D%5Ccdots%3Dc_%7B1m%7Dk_1%2B%5Ccdots%2Bc_%7Bnm%7Dk_n%3D0%2C%5Cforall%20k_i%5Cin%20F$

于是 $c_%7Bij%7D%3D0$ ，从而 $%5Cvarphi_%7Bij%7D$ 线性无关.显然对任意的线性映射 $%5Cvarphi%3AV%5Crightarrow%20U%5Cin%20L%5Cleft(V%2CU%5Cright)$ ,我们只需要考虑其在基向量上的取值，不妨设

$%5Cvarphi%5Cleft(e_i%5Cright)%3Dl_%7Bi1%7Df_1%2Bl_%7Bi2%7Df_2%2B%5Ccdots%2Bl_%7Bim%7Df_m%2C%5Cforall1%5Cle%20i%5Cle%20n%2Cl_%7Bij%7D%5Cin%20F$

显然

$%5Cvarphi%5Cleft(e_i%5Cright)%3Dl_%7Bi1%7D%5Cvarphi_%7B1i%7D%5Cleft(e_i%5Cright)%2Bl_%7Bi2%7D%5Cvarphi_%7B2i%7D%5Cleft(e_i%5Cright)%2B%5Ccdots%2Bl_%7Bim%7D%5Cvarphi_%7Bmi%7D%5Cleft(e_i%5Cright)$

从而对任意的 $%5Calpha%5Cin%20V%2C%5Cvarphi%5Cleft(%5Calpha%5Cright)$ 可以表示为 $%5Cvarphi_%7Bij%7D%5Cleft(e_j%5Cright)$ 的线性组合.

所以 $%5Cvarphi_%7Bij%7D$ 组成向量空间 $L%5Cleft(V%2CU%5Cright)$ 的一组基.

接下来验证 $%5Csigma%3AL%5Cleft(V%2CU%5Cright)%5Crightarrow%20M_%7Bm%5Ctimes%20n%7D%5Cleft(F%5Cright)$ 是线性同构.

首先这两个空间维数相同，然后我们可以找到 $%5Csigma$ 的逆映射，只需要考虑其在基向量上的取值： $%5Csigma%5E%7B-1%7D%5Cleft(E_%7Bij%7D%5Cright)%3D%5Cvarphi_%7Bij%7D%2C%5Cforall1%5Cle%20i%5Cle%20n%2C1%5Cle%20j%5Cle%20m$ ，于是 $%5Csigma%3AL%5Cleft(V%2CU%5Cright)%5Crightarrow%20M_%7Bm%5Ctimes%20n%7D%5Cleft(F%5Cright)$ 是线性同构.

[Q.E.D]

注这个结论是十分显然的，如果不是在这题要证明它们线性同构的背景下，我们可以直接将第一个问题几何问题代数化，研究mn维向量，但由基础矩阵和mn维向量间独特关系，我们直接研究基础矩阵线性无关的性质而立即得到 $%5Cvarphi_%7Bij%7D$ 组成向量空间 $L%5Cleft(V%2CU%5Cright)$ 的一组基.

练习题2 设 $V$ 是由几乎处处为零的无穷实数数列（即 $%5Cleft(a_0%2Ca_1%2Ca_2%2C%5Ccdots%2Ca_n%2C%5Ccdots%5Cright)$ ,其中只有有限多个 $a_i$ 不为零）组成的实向量空间， $R%5Cleft%5Bx%5Cright%5D$ 是由实系数多项式组成的实向量空间.若令 $%5Cvarphi%5Cleft(a_0%2Ca_1%2Ca_2%2C%5Ccdots%2Ca_n%2C%5Ccdots%5Cright)%3Da_0%2Ba_1x%2Ba_2x%5E2%2B%5Ccdots%2Ba_nx%5En$ ，其中 $a%5En%5Cneq0%2Ca_s%3D0%5Cleft(%5Cforall%20s%3En%5Cright)$ ，证明 $%5Cvarphi%3AV%5Crightarrow%20R%5Cleft%5Bx%5Cright%5D$ 是线性同构.

分析与解 容易找到 $%5Cvarphi%3AV%5Crightarrow%20R%5Cleft%5Bx%5Cright%5D$ 的逆映射 $%5Cpsi%3AR%5Cleft%5Bx%5Cright%5D%5Crightarrow%20V$

对于任意的多项式 $f%5Cleft(x%5Cright)%3Da_0%2Ba_1x%2Ba_2x%5E2%2B%5Ccdots%2Ba_nx%5En%2C%5Cleft(a_n%5Cneq0%5Cright)%5Cin%20R%5Cleft%5Bx%5Cright%5D$ ，我们定义 $%5Cpsi%5Cleft(f%5Cleft(x%5Cright)%5Cright)%3D%5Cleft(a_0%2Ca_1%2C%5Ccdots%2Ca_n%2C0%2C0%2C%5Ccdots%5Cright)$ ，从而

$%5Cpsi%5Cvarphi%5Cleft(a_0%2Ca_1%2C%5Ccdots%2Ca_n%2C%5Ccdots%5Cright)%3D%5Cpsi%5Cleft(a_0%2Ba_1x%2B%5Ccdots%2Ba_nx%5En%5Cright)%3D%5Cleft(a_0%2Ca_1%2C%5Ccdots%2Ca_n%2C%5Ccdots%5Cright)$

其中 $a%5En%5Cneq0%2Ca_s%3D0%5Cleft(%5Cforall%20s%3En%5Cright)$ ，从而 $%5Cpsi%5Cvarphi%3DId_V$

$%5Cvarphi%5Cpsi%5Cleft(a_0%2Ba_1x%2B%5Ccdots%2Ba_nx%5En%5Cright)%3D%5Cvarphi%5Cleft(a_0%2Ca_1%2C%5Ccdots%2Ca_n%2C0%2C0%2C%5Ccdots%5Cright)%3Da_0%2Ba_1x%2B%5Ccdots%2Ba_nx%5En$

从而 $%5Cvarphi%5Cpsi%3DId_%7BR%5Cleft%5Bx%5Cright%5D%7D$

于是 $%5Cpsi%3D%5Cvarphi%5E%7B-1%7D$ ，即 $%5Cvarphi%3AV%5Crightarrow%20R%5Cleft%5Bx%5Cright%5D$ 是线性同构.

[Q.E.D]

练习题3(19级高代I每周一题第12题(1)) 设 $A%2CB$ 均为数域 $K$ 上的 $m%5Ctimes%20n$ 阶矩阵，线性映射 $%5Cvarphi%3AM_%7Bn%5Ctimes%20m%7D%5Cleft(K%5Cright)%5Crightarrow%20M_%7Bm%5Ctimes%20n%7D%5Cleft(K%5Cright)$ 定义为 $%5Cvarphi%5Cleft(X%5Cright)%3DAXB$ .证明：若 $m%5Cneq%20n$ ，则 $%5Cvarphi$ 不是线性同构.

分析与解 若 $%5Cvarphi$ 是线性同构，则它是双射.

若 $m%3Cn$ ,由满性可得 $A%2CB$ 是行满秩阵，否则 $r%5Cleft(AXB%5Cright)%5Cle%20min%5C%7Br%5Cleft(A%5Cright)%2Cr%5Cleft(B%5Cright)%5C%7D%3Cm$ ,从而 $A%2CB$ 都存在右逆，适合右消去律，考虑方程 $AXB%3DO$ ，则 $AX%3DO$ .因为 $A$ 不是列满秩阵，由线性方程组解的判定定理， $Ax%3D0$ 不只有零解，从而 $%5Cexists%20X_0%5Cneq%20O%2Cs.t.AX_0B%3DAOB%3DO$ ，与单性矛盾!

若 $m%3En$ ,由满性可得 $A%2CB$ 是列满秩阵，否则 $r%5Cleft(AXB%5Cright)%5Cle%20min%5C%7Br%5Cleft(A%5Cright)%2Cr%5Cleft(B%5Cright)%5C%7D%3Cn$ ,从而 $A%2CB$ 都存在左逆，适合左消去律，考虑方程 $AXB%3DO$ ，则 $XB%3DO$ ,即 $B%5E%5Cprime%20X%5E%5Cprime%3DO$ .因为 $B%5E%5Cprime$ 不是列满秩阵，由线性方程组解的判定定理， $B%5E%5Cprime%20X%5E%5Cprime%3DO$ 不只有零解,即 $XB%3DO$ 不只有零解，从而 $%5Cexists%20X_0%5Cneq%20O%2Cs.t.AX_0B%3DAOB%3DO$ ，与单性矛盾!

综上，若 $m%5Cneq%20n$ ，则 $%5Cvarphi$ 不是线性同构.

[Q.E.D]

事实上，我们还可以利用相抵标准型理论给出这道题的证明：

设 $A%3DP_1%5Cleft(%5Cbegin%7Bmatrix%7DI_%7Br_1%7D%26O%5C%5CO%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)Q_1%2CB%3DP_2%5Cleft(%5Cbegin%7Bmatrix%7DI_%7Br_2%7D%26O%5C%5CO%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)Q_2$ ，其中 $P_i$ 是 $m$ 阶非异阵， $Q_i$ 是 $n$ 阶非异阵

从而 $AXB%3DP_1%5Cleft(%5Cbegin%7Bmatrix%7DI_%7Br_1%7D%26O%5C%5CO%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)Q_1XP_2%5Cleft(%5Cbegin%7Bmatrix%7DI_%7Br_2%7D%26O%5C%5CO%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)Q_2$ ,令 $X_0%3DQ_1%5E%7B-1%7D%5Cleft(%5Cbegin%7Bmatrix%7DO%26K_%7B12%7D%5C%5CK_%7B21%7D%26K_%7B22%7D%5C%5C%5Cend%7Bmatrix%7D%5Cright)P_2%5E%7B-1%7D$ ，其中 $K_%7B12%7D%2CK_%7B21%7D%2CK_%7B22%7D$ 不全为零,则 $X_0%5Cneq%20O$ ，但 $P_1%5Cleft(%5Cbegin%7Bmatrix%7DI_%7Br_1%7D%26O%5C%5CO%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)Q_1X_0P_2%5Cleft(%5Cbegin%7Bmatrix%7DI_%7Br_2%7D%26O%5C%5CO%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)Q_2%3DO$ .