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# 高微vector-valued functions

Let A : R^n －＞ R^m be a linear map.

a.Show that |A|＜＝（m開根號）．max(j=1~m)(sum k=1~n |Ajk|).

(Hint: use max( | x1|,....., | xn|)＜＝ | x| ＜＝n開根號 max ( | x1|,....., | xn|) ).

b.Show that this inequality is an equality when the matix oh A is given by Aj1＝1 and Ajk＝0 for k＞1 (1＜＝ j ＜＝ m)

### 1 Answer

- Scharze spaceLv 71 decade agoFavorite Answer
你的|A|應該是取A的norm吧

A :R^n->R^m, linear, A=[Ajk] 表矩陣A 裡第j行第k列的元素

對於x=(x1,.....,xn)屬於R^n

||Ax||^2=||Σ(k=1~n)Ajkxk||^2=Σ(j=1~m)|Σ(k=1~n)(Ajk)(xk)|^2

<=Σ(j=1~m)[Σ(k=1~n)|Ajk||xk|]^2

<=Σ(j=1`m)[Σ(k=1~n)|Ajk|^2][Σ(k=1~n)|xk|^2] (Cauchy-scharze inequenity)

=||x||^2{Σj=1~m)Σ(k=1~n)|Ajk|^2

=>|A|<={Σ(j=1`m)[Σ(k=1~n)|Ajk|^2]}^(1/2) A is bounded operator

<=(m)^(1/2)max(j=1~m)[Σ(k=1~n)|Ajk|^2]^(1/2)

<=(m)^(1/2)max[Σ(k=1~n)|Ajk|]

b 等號成立時 Ajk 與xk 同號且存在c≠0 使得 |Ajk|=c|xk| 對於所有k x in R^n

取x=(1/c,.......0,0) 即第一項=1/c 其餘為零

則Ajk=1 for k=1 and Ajk=0 for k>1