TheWrathOfMath
Junior Member
- Joined
- Mar 31, 2022
- Messages
- 162
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 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?
For instance:
a+bx+c(x^2-2x)+dx^3+e(-2x^3) = 0
a(-x^4)+b(x^4)+c(-x^4-x^2)+d(x^4)+e(-x^4) =0
I am pretty sure it is wrong, though, because then the matrix A will simply contain two rows of only the entry "1"...
Perhaps I need to group and take out like variables, so that:
1(a)+x(b-2c)+x^2(c)+x^3(d-2e) = 0
x^2(-c)+x^4(-a+b-c+d-e) = 0
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 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?
For instance:
a+bx+c(x^2-2x)+dx^3+e(-2x^3) = 0
a(-x^4)+b(x^4)+c(-x^4-x^2)+d(x^4)+e(-x^4) =0
I am pretty sure it is wrong, though, because then the matrix A will simply contain two rows of only the entry "1"...
Perhaps I need to group and take out like variables, so that:
1(a)+x(b-2c)+x^2(c)+x^3(d-2e) = 0
x^2(-c)+x^4(-a+b-c+d-e) = 0
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