Hi joannehuang!!!
1. If (p0,p1,p2,p3) is a basis of p3(F), then every polynomial in
p3(F) can be expressed in this way in a unique form:
p = a0*p0 + a1*p1 + a2*p2 + a3*p3 ;
If all the polynomials p0,p1,p2,p3 give the value 0 when evaluated at
2, then
pi = (x-2)*qi for i=1 to 3.
Then
p = (x-2)*a0*q0 + (X-2)*a1*q1 + (x-2)*a2*q2 + (x-2)*a3*q3
= (x-2)*(a0*q0 + a1*q1 + a2*q2 + a3*q3);
Then every p of p3(F) give the value 0 when evaluated at 2, that is
not true, so the initial hypothesis (p0,p1,p2,p3 give the value 0 when
evaluated at 2) is not true.
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2. We need to remember that dim(v) = dim(Im(T)) + dim(Ker(T))
To start we need to find a basis of Ker(T), the second step is to
expand the basis of Ker(T) to a basis of V, for example if (k1,k2) is
a basis of Ker(T) we need to find a basis of V (v1,v2,v3,v4,v5) where
v4=k1 and v52=k2.
The vectors found to complete the basis of V from the basis of Ker(T)
(in the example v1,v2,v3) generate the subespace U of V such that if
T’ is the restriction of T to U, then T’: U->Image(T) is a bijection.
We can proof that easily:
Supose that dim(Im(T))=n , then we have n vectors Vi such that are
linearly independent. We need to prove that if Ti = T(Vi) for all i=1
to n , then
(T1,T2,....,Tn) is a basis of Im(T).
Note that if v is a vector of U different to 0 then v is not a vector
of
Ker(T), in other words T(v) is different to 0 for all v of U different
to 0.
0 = a1*T1 + a2*T2 + ... + an*Tn =
= a1*T(V1) + a2*T(V2) + ... + an*T(Vn) =
= T(a1*V1) + T(a2*V2) + ... + T(an*Vn) =
= T(a1*V1 + a2*V2 + ... + an*Vn) [eq.1]
X = (a1*V1 + a2*V2 + ... + an*Vn) is a vector of U, then if T(X) = 0
we have that X = 0 ;then:
(a1*V1 + a2*V2 + ... + an*Vn) = 0 , because (V1,V2,...,Vn) are
linearly independent a1 = a2 = ... = an = 0 , then (T1,T2, ...,Tn) are
linearly independent.
We only need to prove that (T1,T2, ...,Tn) generate Im(T).
If Y is of Im(T), then exists a vector X of U such that T(X) = Y .
As X is of U then:
X = a1*V1 + a2*V2 + ... + an*Vn ; then
Y = T(X)
= T((a1*V1 + a2*V2 + ... + an*Vn)) =
= T(a1*V1) + T(a2*V2) + ... + T(an*Vn)=
= a1*T(V1) + a2*T(V2) + ... + an*T(Vn)=
= a1*T1 + a2*T2 + ... + an*Tn ;
Then (T1,T2, ...,Tn) generates Im(T).
Tī is surjective on Im(T) proof is trivial.
If X , Y of U satisfied T(X) = T(Y) then T(X) - T(Y) = 0,
then T(X-Y) = 0 , because (X-Y) is a vector of U then (X-Y) = 0, then
X = Y , we conclude that Tī is injective.
So if T’ is the restriction of T to U, then T’: U->Image(T) is a
bijection.
--------------------------------------
I hope this helps. I did this using my own knowledge.
If you need a clarification, please post a request of it.
Best Regards.
livioflores-ga |
Request for Answer Clarification by
joannehuang-ga
on
27 Feb 2003 17:21 PST
Hi livioflores-ga:
I really appreicate your answer, can I ask another question regarding
to question 1 . How can I prove or disprove all the polynomials p0,p1
p2,p3 have degree 3. The following answer is what I think, but I'm not
quite is correct or not? ( I increased price to $12) Thanks!!
The given space has dimension four, so we must find four basis
vectors. Each degree ( 0 through 3 ) must be represented in the
basis, but the basis vectors needn't have precisely degrees 0,1,2,3
(degree means the highest power that appears in the polynomial). For
example, the set {1,x,X^3+X^2,X^3} is linearly independent.
Furthermore, any polynomial P belong to P4 (R)can be expressed in
terms of this basis:
P(X)=a0+a1x+a2x^2+a2x^2+a3x^3 = a0.1 +a1x+a2(x^3+x^2)+(a3-a2)x^3
thus the set {1,x,X^3+X^2,X^3}is a basis
Thus, the set is a basis.
|
Clarification of Answer by
livioflores-ga
on
27 Feb 2003 20:58 PST
Hello joannehuang, I am here again!!
Except for a pair of typos your proposed answer to the question "How
can I prove or disprove that all the polynomials p0,p1,p2,p3 have
degree 3?" is correct.
In order to state this more formaly we need to clarify the question:
a) If you are asking for true or false the following statement:
"If (p0,p1,p2,p3) is a basis of p3(R), then all this polynomial have
degree 3"
The answer is false, and your proposed answer can fit for disprove the
statement, but only as a presentation of an example showing the
falsity of the statement. With your answer you are showing that the
condition "all this polynomial have degree 3" is not necessary.
b) If you are asking for the possibility of that "all this polynomial
have degree 3"
The answer is yes there are basis of p3(R), (p0,p1,p2,p3) such that
"all this polynomial have degree 3".
For example:
(p0 = x^3, p1 = x^3+x^2, p2 = x^3+x, p3 = x^3+1) is a basis of p3(R);
In effect, if p(X) is a polynomial of p3(R) then:
p(X) = a3*x^3 + a2*x^2 + a1*x + a0 =
= a3*x^3 - a2*x^3 + a2*x^3 + a2*x^2 - a1*x^3 + a1*x^3 + a1*x -
a0*x^3 +
+ a0*x^3 + a0 =
= (a3-a2-a1-a0)*x^3 + a2*(x^3+x^2) + a1*(x^3+x) + a0*(x^3+1) =
= (a3-a2-a1-a0)*p0 + a2*p1 + a1*p2 + a0*p3;
Then (p0,p1,p2,p3) is a linearly independent set and generates p3(R),
then it is a basis of p3(R) and all of this polynomial have degree 3.
I hope this help, if you need more clarifications please post a
request of a clarification.
PS: I told you that you have the choice to add a tip to the researcher
if you think that the answer deserves the prize. I want to add to this
that you can ask a question and select the researcher who will answer
it by adding "for myselectedresearcher-ga" in the Subject or the
description of the question.
Best Regards.
livioflores-ga
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