Find the Basis

T

TheWrathOfMath

Junior Member
Joined
Mar 31, 2022
Messages
162
V=P4[x] (the subspace of all polynomials of degree 4 or less).
Find the basis of the polynomial subspace W, which includes all polynomials in V such as that when multiplied by (x-2), the product will be an odd degree polynomial.

I only managed to write:
---------------------------?
W={p(x}∈V | p(x)*(x-2) = Σ an*x^(2n+1)
--------------------------n=0
Looking at the p(x)*(x-2) part:

(a+bx+cx^2+dx^3+ex^4)*(x-2) =​
= ax+bx^2+cx^3+dx^4+ex^5-2a-2bx-2cx^2-2dx^3-2ex^4 =​
= -2a+(a-2b)x+(b-2c)x^2+(c-2d)x^3+(d-2e)x^4+ex^5​
.Now, I know that the above expression must be a polynomial with an odd degree.​
The condition is met in the following cases:
e=/=0​
e=0 and c-2d=/=0​
e=0 and c-2d=0 and a-2b=/=0​

Two questions:
1) Is it preferable to write the sigma notation in words instead, since there is no numerical upper limit for the "largest odd number"?
2) How do I proceed to find the basis? Perhaps I have done something wrong or went about this in a complicated manner for no good reason?
 
Why do you require [imath]e \neq 0[/imath], [imath]c-2d \neq 0[/imath], etc. ?

I am guessing that by "odd degree polynomial" you mean polynomials with odd degree components only, i.e., all even degree coefficients must be 0. Is my guess correct?
 
blamocur said:
Why do you require [imath]e \neq 0[/imath], [imath]c-2d \neq 0[/imath], etc. ?

I am guessing that by "odd degree polynomial" you mean polynomials with odd degree components only, i.e., all even degree coefficients must be 0. Is my guess correct?
Click to expand...
Oh, oops. You are right. Only odd degree components.
 
blamocur said:
But I still don't understand why you require all even degree coefficients to be non-zero.
Click to expand...
I misread the original question.
In this case, I require all even degree coefficients of the resulting polynomial to be equal to zero.
First I need to check how this limitation would impact p(x) in W.
 
blamocur said:
But I still don't understand why you require all even degree coefficients to be non-zero.
Click to expand...
Okay, I think that the basis for W is {2x+x^2, 2x^3+x^4}.
I started out by writing:

(a+bx+cx^2+dx^3+ex^4)(x-2) = (alpha)+(beta)x+(gamma)x^2+(delta)x^3+(epsilon)x^4+(zeta)x^5

I distributed the left side, grouped like-terms and took a common factor x...x^5 out.
I then compared the coefficients on both sides:

-2a=alpha
a-2b=beta
b-2c=gamma
c-2d=delta
d-2e=epsilon
e=zeta

We know that alpha, gamma, epsilon= 0, therefore:

-2a=alpha=0
b-2c=gamma=0
d-2e=epsilon=0

Therefore:

a=0
b=2c
d=2e

Therefore every vector in W is of the form:
p(x)= 2cx+cx^2+2ex^3+ex^4

c(2x+x^2)+e(2x^3+x^4)

Therefore the basis for W is:
{2x+x^2, 2x^3+x^4}
 
TheWrathOfMath said:
Okay, I think that the basis for W is {2x+x^2, 2x^3+x^4}.
I started out by writing:

(a+bx+cx^2+dx^3+ex^4)(x-2) = (alpha)+(beta)x+(gamma)x^2+(delta)x^3+(epsilon)x^4+(zeta)x^5

I distributed the left side, grouped like-terms and took a common factor x...x^5 out.
I then compared the coefficients on both sides:

-2a=alpha
a-2b=beta
b-2c=gamma
c-2d=delta
d-2e=epsilon
e=zeta

We know that alpha, gamma, epsilon= 0, therefore:

-2a=alpha=0
b-2c=gamma=0
d-2e=epsilon=0

Therefore:

a=0
b=2c
d=2e

Therefore every vector in W is of the form:
p(x)= 2cx+cx^2+2ex^3+ex^4

c(2x+x^2)+e(2x^3+x^4)

Therefore the basis for W is:
{2x+x^2, 2x^3+x^4}
Click to expand...
Agree!
 
blamocur said:
Agree!
Click to expand...
Now I found that the basis for U={q(x)∈V | q(-1)=0} ⊆ V=P4[x] is {1-x^4, x+x^4, x^2-x^4, x^3+x^4} .
I found it by letting q(x) = f+gx+hx^2+jx^3+kx^4, used the constraint q(-1)=0 to find that k= -f+g-h+j, and substituted it in q(x), simplified, and obtained the basis.

I was asked to find the intersection of U and W.

I let v∈ U∩W, therefore:

v∈W, hence:
v= c(2x+x^2)+e(2x^3+x^4)

v∈U, hence:
a(1-x^4)+b(x+x^4)+c(x^2-x^4)+d(x^3+x^4)
*Notice that I changed the coefficients from f,g,h,j to a,b,c,d.

v=v, therefore v-v=0 :

a(1-x^4)+b(x+x^4)+c(x^2-x^4)+d(x^3+x^4)-c(2x+x^2)-e(2x^3+x^4) = (0,0).

Now I obtained the following equations:

a+bx+cx^2+dx^3-2cx-2ex^3=0
-ax^4+bx^4-cx^4+dx^4-cx^2-ex^4=0

I thought about attempting to solve the system of equations and expressing some variables using other (free) variables.

How do I do that using a matrix?
I am slightly confused about how will the matrix look like.
Ax=b
Will A be a matrix of the coefficients, x a 1x2 matrix of the variables, and b a 1x2 zero matrix?
If so, don't I need to rearrange my equations, since I have the coefficient c, for instance, appearing in two different terms in the first and second equations?
 
I am confused: where do U and V come from? Is this from a different problem? If so, can you post a complete problem description. Also, I believe the guidelines for the forum require separate threads for different problems.
 
blamocur said:
I am confused: where do U and V come from? Is this from a different problem? If so, can you post a complete problem description. Also, I believe the guidelines for the forum require separate threads for different problems.
Click to expand...
It is not a different problem. It is the second part of the same problem.
Nevertheless, I can start a new thread for the second part, if it's preferable.
 
blamocur said:
I am confused: where do U and V come from? Is this from a different problem? If so, can you post a complete problem description. Also, I believe the guidelines for the forum require separate threads for different problems.
Click to expand...

U∩W

V= P4[x] The the basis for W, which is the linear subspace of all polynomials in V such as that when multiplied by (x-2), the resulting polynomial will include only odd degree exponents. is {2x+x^2, 2x^3+x^4} The basis for U={q(x)∈V | q(-1)=0} ⊆ V=P4[x] is {1-x^4, x+x^4, x^2-x^4, x^3+x^4} . I...
www.freemathhelp.com

I started a new thread.
 
You must log in or register to reply here.
Top