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.31

2021-10-21 13:22 作者:CharlesMa0606 0人讀過 | 我要投稿

本文內(nèi)容主要有關(guān)于線性映射及其運(yùn)算，在高代白皮書上對應(yīng)第4.2.1節(jié)

題目來自于復(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，懇請斧正！

練習(xí)題1(17級高代期末考試第六大題)? 設(shè) $M_n%5Cleft(K%5Cright)$ 為數(shù)域 $K$ 上的 $n$ 階方陣全體構(gòu)成的線性空間， $A%2CB%5Cin%20M_n%5Cleft(K%5Cright)$ , $M_n%5Cleft(K%5Cright)$ 上的線性變換 $%5Cvarphi$ 定義為 $%5Cvarphi%5Cleft(X%5Cright)%3DAXB$ .證明： $%5Cvarphi$ 是冪零線性變換的充要條件是 $A%2CB$ 中至少有一個(gè)是冪零陣.

分析與解 ?注意到 $%5Cvarphi$ 是冪零線性變換即 $%5Cexists%20m%5Cin%20N%5E%5Cast%2Cst.A%5EmXB%5Em%3DO%2C%5Cforall%20X%5Cin%20M_n%5Cleft(K%5Cright)$

于是充分性顯然，我們只需要證明必要性.

考慮基礎(chǔ)矩陣，有：

$%5Cleft(A%5EmE_%7Bij%7DB%5Em%5Cright)_%7Bkl%7D%3Da_%7Bki%7Db_%7Bjl%7D%3D0$

從而 $a_%7Bij%7Db_%7Bkl%7D%3D0%2C%5Cforall%20i%2Cj%2Ck%2Cl$ ， $%E5%85%B6%E4%B8%ADa_%7Bij%7D%2Cb_%7Bkl%7D%E6%98%AFA%5Em%2CB%5Em%E7%9A%84%E7%AC%AC%5Cleft(i%2Cj%5Cright)%2C%5Cleft(k%2Cl%5Cright)%E5%85%83%E7%B4%A0.$

若 $%5Cexists%20a_%7Bij%7D%5Cneq0%0A$ ，則 $b_%7Bkl%7D%3D0%2C%5Cforall%20k%2Cl%0A%0A$ ，于是 $B%5Em%3DO$ .

所以 $A%2CB$ 中至少有一個(gè)是冪零陣.

練習(xí)題2(17級高代I期末考試第七大題)? 設(shè) $U%2CV%2CW$ 均為數(shù)域上的非零線性空間, $%5Cvarphi%3AV%5Crightarrow%20U$ 和 $%5Cpsi%3AU%5Crightarrow%20W$ 是線性映射，滿足 $r%5Cleft(%5Cpsi%5Cvarphi%5Cright)%3Dr%5Cleft(%5Cvarphi%5Cright)$ .證明：存在線性映射 $%5Cxi%3AW%5Crightarrow%20U$ ，使得 $%5Cxi%5Cpsi%5Cvarphi%3D%5Cvarphi$ .

分析與解 ?不妨設(shè) $r%5Cleft(%5Cpsi%5Cvarphi%5Cright)%3Dr%5Cleft(%5Cvarphi%5Cright)%3Dr$ ，取 $Im%5Cvarphi$ 的一組基 $%5C%7Be_1%2C%5Ccdots%2Ce_r%5C%7D$ ， $Im%5Cpsi%5Cvarphi$ 的一組基 $%5C%7Bf_1%2C%5Ccdots%2Cf_r%5C%7D$

任取 $%5Calpha%5Cin%20Im%5Cvarphi$ ，不妨設(shè)為 $%5Calpha%3Dc_1e_1%2B%5Ccdots%2Bc_re_r$

從而：

$%5Cpsi%5Cleft(%5Calpha%5Cright)%3Dc_1%5Cpsi%5Cleft(e_1%5Cright)%2Bc_2%5Cpsi%5Cleft(e_2%5Cright)%2B%5Ccdots%2Bc_r%5Cpsi%5Cleft(e_r%5Cright)%3D%5Clambda_1f_1%2B%5Clambda_2f_2%2B%5Ccdots%2B%5Clambda_rf_r%5Cin%20Im%5Cpsi%5Cvarphi$

于是 $%5Cpsi%5Cleft(e_1%5Cright)%2C%5Cpsi%5Cleft(e_2%5Cright)%2C%5Ccdots%2C%5Cpsi%5Cleft(e_r%5Cright)$ 線性無關(guān)，即它是 $Im%5Cpsi%5Cvarphi$ 的一組基.

注意到 $e_1%2C%5Ccdots%2Ce_r%5Cin%20Im%5Cvarphi$ ，

從而存在 $r$ 個(gè)線性無關(guān)的向量 $%5Cnu_%7Bi%7D%2Cst.%20e_%7Bi%7D%3D%5Cvarphi(%5Cnu_%7Bi%7D)(%5Cforall1%5Cleq%20i%5Cleq%20r)$

我們構(gòu)造線性映射 $%5Cxi%3AW%5Crightarrow%20U$ 如下，只需要考慮 $V$ 上基向量的取值，有：

$%5Cxi%5Cpsi%5Cvarphi%5Cleft(%5Cnu_i%5Cright)%3D%5Cxi%5Cpsi%5Cleft(e_i%5Cright)%3D%5Cvarphi%5Cleft(e_i%5Cright)%2C%5Cxi%5Cpsi%5Cvarphi%5Cleft(u_i%5Cright)%3D%5Cxi%5Cpsi%5Cleft(0%5Cright)%3D0.$

于是存在線性映射 $%5Cxi%3AW%5Crightarrow%20U$ ，使得 $%5Cxi%5Cpsi%5Cvarphi%3D%5Cvarphi$ .

練習(xí)題3? 設(shè) $V$ 是有理數(shù)域上的三維線性空間， $%5Cvarphi$ 是 $V$ 上的線性變換并且滿足條件

$%5Cvarphi%5Cleft(%5Calpha%5Cright)%3D%5Cbeta%2C%5Cvarphi%5Cleft(%5Cbeta%5Cright)%3D%5Cgamma%2C%5C%20%5C%20%5Cvarphi%5Cleft(%5Cgamma%5Cright)%3D%5Calpha%2B%5Cbeta$

求證：若 $%5Calpha%5Cneq0$ ，則 $%5Calpha%2C%5Cbeta%2C%5Cgamma$ 是線性無關(guān)的向量.

分析與解 ?我們先設(shè) $%5Calpha%2C%5Cbeta$ 線性相關(guān)，即 $%5Calpha%3Dk%5Cbeta%2Ck%5Cin%20Q$ ，代入條件，有：

$%5Cgamma%3Dk%5E2%5Calpha%2Ck%5E3%5Calpha%3D%5Calpha%2Bk%5Calpha$

注意到于是得到方程 $k%5E3-k-1%3D0$ ，由高代白皮書例5.45，因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=f(0)%2Cf(1)" alt="f(0)%2Cf(1)">是奇數(shù)，所以原方程沒有有理根.從而必線性無關(guān).下設(shè) $%5Cgamma%3Dk%5Calpha%2Bl%5Cbeta%2Ck%2Cl%5Cin%20Q$ ，代入有：

$%5Cvarphi%5Cleft(%5Cgamma%5Cright)%3Dk%5Cvarphi%5Cleft(%5Calpha%5Cright)%2Bl%5Cvarphi%5Cleft(%5Cbeta%5Cright)%3Dk%5Cbeta%2Bl%5Cleft(k%5Calpha%2Bl%5Cbeta%5Cright)%3Dkl%5Calpha%2B%5Cleft(k%2Bl%5E2%5Cright)%5Cbeta%3D%5Calpha%2B%5Cbeta$

移項(xiàng)，有: $%5Cleft(kl-1%5Cright)%5Calpha%2B%5Cleft(k%2Bl%5E2-1%5Cright)%5Cbeta%3D0$ ，注意到線性無關(guān)，于是得到方程：

$l%5E3-l%2B1%3D0$

這個(gè)方程也沒有有理根，從而它們是線性無關(guān)的向量.

練習(xí)題4(14級高代I每周一題第7題) ?設(shè) $A$ 是有理數(shù)域 $Q$ 上的 $4$ 階方陣， $%5Calpha_1%2C%5Calpha_2%2C%5Calpha_3%2C%5Calpha_4$ 是 $Q$ 上的 $4$ 維列向量，滿足：

$A%5Calpha_1%3D%5Calpha_2%2CA%5Calpha_2%3D%5Calpha_3%2CA%5Calpha_3%3D%5Calpha_4%2CA%5Calpha_4%3D-%5Calpha_1-%5Calpha_2-%5Calpha_3-%5Calpha_4.$

證明：若 $%5Calpha_1%5Cneq0$ ,則 $%5C%7B%5Calpha_1%2C%5Calpha_2%2C%5Calpha_3%2C%5Calpha_4%5C%7D$ 是 $4$ 維列向量空間 $Q%5E%7B4%7D$ 的一組基.

分析與解 ?若 $%5Calpha_1%2C%5Calpha_2$ 線性相關(guān)，不妨設(shè) $%5Calpha_2%3Dk%5Calpha_1%2Ck%5Cin%20Q$ ，代入得：

$%5Calpha_3%3Dk%5E2%5Calpha_1%2C%5Calpha_4%3Dk%5E3%5Calpha_1%2CA%5Calpha_4%3Dk%5E4%5Calpha_1$

因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=%5Calpha_1%5Cneq0" alt="%5Calpha_1%5Cneq0">從而得到方程組： $k%5E4%2Bk%5E3%2Bk%5E2%2Bk%2B1%3D0$ ，這個(gè)方程沒有有理根，于是 $%5Calpha_1%2C%5Calpha_2$ 線性無關(guān).

若 $%5Calpha_1%2C%5Calpha_2%2C%5Calpha_3$ 線性相關(guān),則不妨設(shè) $%5Calpha_3%3Dk%5Calpha_1%2Bl%5Calpha_2%2Ck%2Cl%5Cin%20Q$ ，同樣代入得

$%5Calpha_4%3DA%5Calpha_3%3Dkl%5Calpha_1%2B%5Cleft(k%2Bl%5Cright)%5Calpha_2%2CA%5Calpha_4%3Dk%5Cleft(k%2Bl%5Cright)%5Calpha_1%2B%5Cleft(2kl%2Bl%5E2%5Cright)%5Calpha_2$