復(fù)旦大學(xué)謝啟鴻老師高等代數(shù)在線習(xí)題課思考題分析與解 ep.32

2021-11-01 21:46 作者:CharlesMa0606 0人讀過(guò) | 我要投稿

本文內(nèi)容主要有關(guān)于線性同構(gòu)，在高代白皮書(shū)上對(duì)應(yīng)第4.2.2節(jié)

題目來(lái)自于復(fù)旦大學(xué)謝啟鴻教授在本站高等代數(shù)習(xí)題課的課后思考題，本文僅供學(xué)習(xí)交流

習(xí)題課視頻鏈接：復(fù)旦大學(xué)謝啟鴻高等代數(shù)習(xí)題課_嗶哩嗶哩_bilibili

本人解題水平有限，可能會(huì)有錯(cuò)誤，懇請(qǐng)斧正！

練習(xí)題1? 設(shè) $V%E5%92%8CU$ 為數(shù)域 $F$ 上的向量空間， $e_1%2Ce_2%2C%5Ccdots%2Ce_n$ 和 $f_1%2Cf_2%2C%E2%8B%AF%2Cf_m$ 分別是 $V%E5%92%8CU$ 的基.定義 $V%E5%92%8CU$ 的線性映射 $%5Cvarphi_%7Bij%7D%5Cleft(i%3D1%2C2%2C%5Ccdots%2Cn%3Bj%3D1%2C2%2C%5Ccdots%2Cm%5Cright)$ :

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

求證： $%5Cvarphi_%7Bij%7D$ 組成向量空間 $L%5Cleft(V%2CU%5Cright)$ 的一組基.若令 $%5Csigma%5Cleft(%5Cvarphi_%7Bij%7D%5Cright)%3DE_%7Bij%7D$ ，這里 $E_%7Bij%7D$ 是基礎(chǔ)矩陣，證明 $%5Csigma%3AL%5Cleft(V%2CU%5Cright)%5Crightarrow%20M_%7Bm%5Ctimes%20n%7D%5Cleft(F%5Cright)$ 是線性同構(gòu).

分析與解? 首先驗(yàn)證 $%5Cvarphi_%7Bij%7D$ 組成向量空間 $L%5Cleft(V%2CU%5Cright)$ 的一組基.

設(shè) $c_%7B11%7D%5Cvarphi_%7B11%7D%2Bc_%7B12%7D%5Cvarphi_%7B12%7D%2B%5Ccdots%2Bc_%7Bnm%7D%5Cvarphi_%7Bnm%7D%3D0%2Cc_%7Bij%7D%5Cin%20F$ ,注意到 $V$ 中任意一個(gè)向量都可以表示為 $e_1%2Ce_2%2C%5Ccdots%2Ce_n$ 的線性組合，從而不妨設(shè) $%5Calpha%3Dk_1e_1%2Bk_2e_2%2B%5Ccdots%2Bk_ne_n$ ，于是得到

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

從而

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

于是 $c_%7Bij%7D%3D0$ ，從而 $%5Cvarphi_%7Bij%7D$ 線性無(wú)關(guān).顯然對(duì)任意的線性映射 $%5Cvarphi%3AV%5Crightarrow%20U%5Cin%20L%5Cleft(V%2CU%5Cright)$ ,我們只需要考慮其在基向量上的取值，不妨設(shè)

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

顯然

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

從而對(duì)任意的 $%5Calpha%5Cin%20V%2C%5Cvarphi%5Cleft(%5Calpha%5Cright)$ 可以表示為 $%5Cvarphi_%7Bij%7D%5Cleft(e_j%5Cright)$ 的線性組合.

所以 $%5Cvarphi_%7Bij%7D$ 組成向量空間 $L%5Cleft(V%2CU%5Cright)$ 的一組基.

接下來(lái)驗(yàn)證 $%5Csigma%3AL%5Cleft(V%2CU%5Cright)%5Crightarrow%20M_%7Bm%5Ctimes%20n%7D%5Cleft(F%5Cright)$ 是線性同構(gòu).

首先這兩個(gè)空間維數(shù)相同，然后我們可以找到 $%5Csigma$ 的逆映射，只需要考慮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 $%5Csigma%3AL%5Cleft(V%2CU%5Cright)%5Crightarrow%20M_%7Bm%5Ctimes%20n%7D%5Cleft(F%5Cright)$ 是線性同構(gòu).

[Q.E.D]

注? 這個(gè)結(jié)論是十分顯然的，如果不是在這題要證明它們線性同構(gòu)的背景下，我們可以直接將第一個(gè)問(wèn)題幾何問(wèn)題代數(shù)化，研究mn維向量，但由基礎(chǔ)矩陣和mn維向量間獨(dú)特關(guān)系，我們直接研究基礎(chǔ)矩陣線性無(wú)關(guān)的性質(zhì)而立即得到 $%5Cvarphi_%7Bij%7D$ 組成向量空間 $L%5Cleft(V%2CU%5Cright)$ 的一組基.

練習(xí)題2? 設(shè) $V$ 是由幾乎處處為零的無(wú)窮實(shí)數(shù)數(shù)列（即 $%5Cleft(a_0%2Ca_1%2Ca_2%2C%5Ccdots%2Ca_n%2C%5Ccdots%5Cright)$ ,其中只有有限多個(gè) $a_i$ 不為零）組成的實(shí)向量空間， $R%5Cleft%5Bx%5Cright%5D$ 是由實(shí)系數(shù)多項(xiàng)式組成的實(shí)向量空間.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 $a%5En%5Cneq0%2Ca_s%3D0%5Cleft(%5Cforall%20s%3En%5Cright)$ ，證明 $%5Cvarphi%3AV%5Crightarrow%20R%5Cleft%5Bx%5Cright%5D$ 是線性同構(gòu).

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

對(duì)于任意的多項(xiàng)式 $f%5Cleft(x%5Cright)%3Da_0%2Ba_1x%2Ba_2x%5E2%2B%5Ccdots%2Ba_nx%5En%2C%5Cleft(a_n%5Cneq0%5Cright)%5Cin%20R%5Cleft%5Bx%5Cright%5D$ ，我們定義 $%5Cpsi%5Cleft(f%5Cleft(x%5Cright)%5Cright)%3D%5Cleft(a_0%2Ca_1%2C%5Ccdots%2Ca_n%2C0%2C0%2C%5Ccdots%5Cright)$ ，從而

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

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

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

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

于是 $%5Cpsi%3D%5Cvarphi%5E%7B-1%7D$ ，即 $%5Cvarphi%3AV%5Crightarrow%20R%5Cleft%5Bx%5Cright%5D$ 是線性同構(gòu).

[Q.E.D]

練習(xí)題3(19級(jí)高代I每周一題第12題(1))? 設(shè) $A%2CB$ 均為數(shù)域 $K$ 上的 $m%5Ctimes%20n$ 階矩陣，線性映射 $%5Cvarphi%3AM_%7Bn%5Ctimes%20m%7D%5Cleft(K%5Cright)%5Crightarrow%20M_%7Bm%5Ctimes%20n%7D%5Cleft(K%5Cright)$ 定義為 $%5Cvarphi%5Cleft(X%5Cright)%3DAXB$ .證明：若 $m%5Cneq%20n$ ，則 $%5Cvarphi$ 不是線性同構(gòu).

分析與解? 若 $%5Cvarphi$ 是線性同構(gòu)，則它是雙射.

若 $m%3Cn$ ,由滿性可得 $A%2CB$ 是行滿秩陣，否則 $r%5Cleft(AXB%5Cright)%5Cle%20min%5C%7Br%5Cleft(A%5Cright)%2Cr%5Cleft(B%5Cright)%5C%7D%3Cm$ ,從而 $A%2CB$ 都存在右逆，適合右消去律，考慮方程 $AXB%3DO$ ，則 $AX%3DO$ .因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=A" alt="A">不是列滿秩陣，由線性方程組解的判定定理， $Ax%3D0$ 不只有零解，從而 $%5Cexists%20X_0%5Cneq%20O%2Cs.t.AX_0B%3DAOB%3DO$ ，與單性矛盾!

若 $m%3En$ ,由滿性可得 $A%2CB$ 是列滿秩陣，否則 $r%5Cleft(AXB%5Cright)%5Cle%20min%5C%7Br%5Cleft(A%5Cright)%2Cr%5Cleft(B%5Cright)%5C%7D%3Cn$ ,從而 $A%2CB$ 都存在左逆，適合左消去律，考慮方程 $AXB%3DO$ ，則 $XB%3DO$ ,即 $B%5E%5Cprime%20X%5E%5Cprime%3DO$ .因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=B%5E%5Cprime" alt="B%5E%5Cprime">不是列滿秩陣，由線性方程組解的判定定理， $B%5E%5Cprime%20X%5E%5Cprime%3DO$ 不只有零解,即 $XB%3DO$ 不只有零解，從而 $%5Cexists%20X_0%5Cneq%20O%2Cs.t.AX_0B%3DAOB%3DO$ ，與單性矛盾!

綜上，若 $m%5Cneq%20n$ ，則 $%5Cvarphi$ 不是線性同構(gòu).

[Q.E.D]

事實(shí)上，我們還可以利用相抵標(biāo)準(zhǔn)型理論給出這道題的證明：

設(shè) $A%3DP_1%5Cleft(%5Cbegin%7Bmatrix%7DI_%7Br_1%7D%26O%5C%5CO%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)Q_1%2CB%3DP_2%5Cleft(%5Cbegin%7Bmatrix%7DI_%7Br_2%7D%26O%5C%5CO%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)Q_2$ ，其中 $P_i$ 是 $m$ 階非異陣， $Q_i$ 是 $n$ 階非異陣

從而 $AXB%3DP_1%5Cleft(%5Cbegin%7Bmatrix%7DI_%7Br_1%7D%26O%5C%5CO%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)Q_1XP_2%5Cleft(%5Cbegin%7Bmatrix%7DI_%7Br_2%7D%26O%5C%5CO%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)Q_2$ ,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 $K_%7B12%7D%2CK_%7B21%7D%2CK_%7B22%7D$ 不全為零,則 $X_0%5Cneq%20O$ ，但 $P_1%5Cleft(%5Cbegin%7Bmatrix%7DI_%7Br_1%7D%26O%5C%5CO%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)Q_1X_0P_2%5Cleft(%5Cbegin%7Bmatrix%7DI_%7Br_2%7D%26O%5C%5CO%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)Q_2%3DO$ .