cynosure 發問時間： 科學數學 · 1 0 年前

線性代數，第6,7,8題

http://tw.myblog.yahoo.com/jw!uQyf3r2cGRndUOIZDwnr...

1 個解答

• 1 0 年前
最佳解答

請問一下:你的第八題:detined by T(v)=vN,是這樣的嗎?

2006-04-28 00:31:42 補充：

6.a)T is linear,let f(x),g(x)屬於D,a屬於純量T(af+g)=(af+g)'(x)=(af)'(x)+g'(x)=af'(x)+g'(x)=aT(f)+T(g),∴T is linearb)T is not one to one,Let f(x)=3,g(x)=4T(f)=f'(x)=0,T(g)=g'(x)=0=>T(f)=T(g)=0,but f≠g,so T is not one to onec)T is onto,For all f'(x)屬於D,since f'(x) is differentiable,so f'(x) is continuous,根據微積分基本定理,∫f'(x)dx exist,and f(x)+C=∫f'(x)dx即存在f(x)屬於D,使得T(f)=f'(x),so T is ontod)Let T(f)=f'(x)=0=>f is constant function=>ker(T)=span{1}7.這題很簡單,只要把定義向量空間的那十個條件拿出來檢驗就行了,然後再比較對或錯,因為這樣寫下去就太多了8.T(v)=vB,對於所有v屬於V,要證明T是isomorphism,只要證明T是one to one and onto 的線性轉換T is linear:for all v1,v2屬於V,a屬於純量,Let v1=Σk=1nαkbk,v2=Σk=1nβkbk,αk,βk都屬於純量,fork=1,..,n則T(av1+v2)=(av1+v2)B=(Σk=1n(aαk+βk)bk)B=[aα1+β1]                                         [aα2+β2]                                         [......]                                         [aαn+βn]=a[α1]+[β1]＝a(v1)B+(v2)B=aT(v1)+t(v2),∴T is linear  [α2] [β2]  [..] [...]  [αn] [βn]T is one to one,if T(v1)=T(v2)=>(v1)B=(v2)B=>[α1]=[β1]  [α2]=[β2]  [..]=[..]  [αn]=[βn]＝＞αk=βk,k=1,2,...,n∴v1=v2=>T is one to oneT is onto:For all [a1]                  [a2]                  [..]                  [an]屬於Rn,取v=Σk=1nak*bk,則T(v)=(v)B=[a1]                                                        [a2]                                                        [..]                                                        [an]∴T is onto,Hence T is isomorphism

參考資料： me