# 成大線代考古一題

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

(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

Update:

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W_1可拆成W_2⊕W_2,..

Update 2:

PTT我問過了，就是沒有人回答這題

Update 3:

Rating

(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.

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