線代證明是線性獨立集合(22點)

Question

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請問這題該怎麼證？

他條件才給兩個感覺證不太出來？

Update:

Case1

cn<0

==>0=c1*v1+c2*v2+......+cn*vn-vn(矛盾)

為什麼會矛盾？

Case2

cn>0

==>0=c1*v1+c2*v2+......+cn*vn-vn(矛盾)

為什麼會矛盾？

Update 2:

看不懂

你直接設

vn=c1*v1+c2*v2+......+cn*vn

然後不管cn的正負都直接說

0=c1*v1+c2*v2+......+cn*vn-vn(矛盾)

是哪裡矛盾？

Answer 1

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Assume that {v1,v2,..,vn} is a dependent set, [we want to arrive at a contradiction] , then some v_* can be written as a linear combination of the rest v_j's. Without loss of generality, let's say vn (nonzero)= summation[j=1, to n-1]{cj*vj}. ----(1)

Since <vi,vj><=0 for i not equals to j , there is at least one j such that cj is negative in the above (1). Proof: Just apply inner product to (1). <vn,vn> = summation [j=1, to n-1]{cj<vn,vj>}. The left hand side is positive due to inner product; while <vn,vj> are all negative as we were told. If all cj in (1) were nonnegative a contradiction arrives.

Therefore we claim in the assumption (1), we can express vn = summation {cj*vj} + summation {dk*vk}, where the first set contains negative coefficients cj; and the second set contains positive coefficients dk. Besides, as proved already, at least one cj (negative) exist. ---(2).

Next, applying vn inner product to

vn - summation {cj*vj} = summation {dk*vk}, which is equivalent to (2),

we find that the left hand side is positive while the right hand side is going to be negative if any ck positive coefficient exists. Therefore we conclude that vn = summation {cj*vj}, where all coefficients are negative (those zero coefficients can be ommitted). ----(3)

Finally, we apply the inner product of u and vn in (3):

<u,vn> = summation {cj<u,vj>}

The left hand side is told positive but the right hand side[there is at least one term on the right] is strictly negative . This is the contradiction we expected to have. QED

Answer 2

Assume x* is vector

x*屬於u,v1,v2,......,vn,

If {v1,v2,.....,vn} is a linear dependent set

(When we know u,v1,v2,......,vn,屬於V ,(vi,u)>0 for all i ,and (vi,vj)<=0

whenever i 不等於 j )

vn=c1*v1+c2*v2+......+cn*vn

for any cn=(x*,vj);

case 1:

x*=vi

==>

cn=(vi,vj)<=0

but cn<0

==>0=c1*v1+c2*v2+......+cn*vn-vn(矛盾)

so==>

{v1,v2,.....,vn} is a linear independent set

case 2:

x*=u

==>

cn=(u,vj)>0

but cn>0

==>0=c1*v1+c2*v2+......+cn*vn-vn(矛盾)

so==>

{v1,v2,.....,vn} is a linear independent set

2010-09-05 15:28:52 補充：

若有錯請多多指教~

2010-09-05 15:33:47 補充：

0=c1*v1+c2*v2+......+cn*vn-vn(矛盾)

==>(cn-1)vn=c1*v1+c2*v2+......+c(n-1)v(n-1)

==>承Case1

==>得知 {v1,v2,.....,vn} is a linear independent set

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