Lighthope asked in 科學數學 · 1 decade ago

成大線代考古一題

http://*****/FLExg

這題的問題是第1跟3小題

第1是我可以直接說選擇兩線性獨立向量a,b使得f(a,b)≠0嗎?

第3                           ⊥          ⊥

我的想法是這樣,利用第2小題的結果,猜想V可以拆成W_1⊕W_1, W_1可拆成W_2⊕W_2,..

一直這樣做下去                              ⊥

因為V是有限維度,所以會在某一步停下來,變成V=W_1⊕W_2⊕…⊕W_n-1⊕W_n-1,其中

     ⊥        ⊥

dim(W_n-1)=1或2.令W_n=W_n-1.{x_i,y_i}是W_i的基底,1≦i≦n-1.

如果真能這樣,會發現滿足題目下面所要求的結果

那個f(x_j,y_j)應該是打錯,其實是f(x_i,y_j).

現在問題有兩個

(1)要怎麼證明dim(W_n)=2?

(2)要怎麼證明W_i∩(ΣW_j)={0}對所有i都成立?

j≠i

請高手幫忙,謝謝

Update:

第5行後面

          ⊥

W_1可拆成W_2⊕W_2,..

第10行後面 1小於等於 i 小於等於 n-1

Update 2:

PTT我問過了,就是沒有人回答這題

針對第3小題是否能再詳細解釋呢?我有點不懂

Update 3:

回答者就甘心捏

2 Answers

Rating
  • 1 decade ago
    Favorite Answer

    (a)

    因 non-degenerate: f(u,v)= 0, for v in V implies u=0

    故 u≠0時, 必存在某一個 v in V, f(u, v)=k≠0,

    取 x=u, y=v/k => f(x, y)=f(u, v/k)= 1/k * f(u, v) (bilinear)

    => f(x, y)= 1

    x, y linearly indep.?

    設 ax+by=0,

    f(0,x)=f(ax+by, x)= af(x, x)+bf(y, x) (bilinear)0=

    = 0 + b*1 ( skew-symmetgric =>f(x,x)=0 )

    => b=0

    f(0, y)=0=f(ax+by, y)= af(x,y)+bf(y,y)= a*1+0 => a=0

    ie. ax+by=0 => a=b=0 , so, x, y are linearly indep.

    (b)

    For all v in V, let v=[ v- f(v,y)x - f(v,x)y] + f(v,y)x+f(v,x)y

    (1) f(v,y)x+ f(v,x)y is a form of px+qy , so f(v,y)x+f(v,x)y in W

    (2) Let v'= v- f(v,y)x- f(v, x)y. Does f(v', w)=0 for all w in W?

    (i) f(v', x)= f(v - px- qy, x) (p=f(v,y), q=f(v,x))

    = f(v,x)- p f(x,x) - q f(y, x) (bilinear)

    = q - p*0 - q*1 =0

    (ii) Similarly, f(v', y) = 0

    by (i),(ii) and bilinear property of f => f(v', ax+by)=0

    so, v'= f(v,y)x+ f(v,x)y in W' (orthogonal complement of W)

    (3) Does the intersection of W and W' empty?

    Let v in the inteersection of W and W', then v= ax+by

    f(ax+by, x)= a f(x,x)+ bf(y,x)= 0 + b*(-1) = -b

    while v in W' , then f(v, x)=0 => - b =0,

    similarly, a=0 => v=0

    ie. the intersection of W and W' = { 0 }

    By (1),(2),(3) => V is the direct sum of W and W'

    (C)

    (i) dim(V) is finite.

    (ii) From Q(b) => V= W + W' (direct sum)

    By induction on dim(V), then

    V= W + W' + W'' + ... ( finite direct sum of n subspace)

    and W=span{x, y}, W'= span{x', y'}, ...

    so {x, y, x', y', .... }( 2n vectors) is a basis of V

    and dim(V)= 2n

    2009-03-01 14:34:50 補充:

    是否也在 PTT 提問呢?

    2009-03-03 14:08:43 補充:

    By(a),(b): f is non-degenerate on V =>exist dim(W)=2 and V=W+W'

    If f is non-degenerate on W' => W'=W1+W1' =>W=W+W1+W1' , ... etc.

    So, we should f is non-degenerate on W'.

    Select any vector from W', set it be u.

    2009-03-03 14:08:52 補充:

    Since V=W+W' , so, any vector v in V must be the form w+w'.

    If f(u, w')=0 for all w' in W', then

    f(u, v)=f(u, w+w')= f(u, w)+ f(u, w') = 0 + f(u, w')= 0 + 0

    If f(u, w')=0 for all w' in W', then f(u, v)=0 for all v in V => u=0 (non-degenerate)

    2009-03-03 14:08:57 補充:

    Thus, f is non-degenerate on W'.

    By(a), (b) => W=W1+W1' (direct sum)

    We can continue the process V=W+W', W'=W1+W1', ...

    =>V=W+W1+W2+ ... (all are direct sum, and dim(Wi)=2 ), that proved this problem.

  • 1 decade ago

    網站有毒 >"< Orz

    防毒軟體不給我開

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