Anne5632
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- Homework Statement
- Check that the polynomials:
- Relevant Equations
- r
I put it in echelon form but don't know where to go from there.
Do you know how to determine whether a set of vectors is a basis for a vector space? If you have a vector space of dimension 4, how many vectors do you need for a basis?Anne5632 said:Homework Statement:: Check that the polynomials:
form a basis of R3[x]. (You may use the fact that dim R3[x] = 4).
Relevant Equations:: p1(x) = -x+2x^2 +x^3
p2(x)= 2+2x^2
p3(x)= -7+x
p4(x)=3-2x+x^3
I put it in echelon form but don't know where to go from there.
Do i need to determine linear independance?Mark44 said:Do you know how to determine whether a set of vectors is a basis for a vector space? If you have a vector space of dimension 4, how many vectors do you need for a basis?
There are many similarities between a vector space of dimension n and a function space of the same dimension.
For a vector space or a function space of dimension n, you need n linearly independent vectors/functions.Anne5632 said:Do i need to determine linear independance?
Yes.Anne5632 said:if a vector space has a dim 4, I'm guessing you need 4 for a basis?
When you put the function coordinates into a matrix, you are essentially solving the equation ##c_1p_1(x) + c_2p_2(x) +c_3p_3(x) +c_4p_4(x) = 0## for the constants ##c_i##. If you end up with the identity matrix, that indicates that all the constants are 0, and there are no other solutions for these constants. This means that the four functions are linearly independent (note spelling), and since there are four of them, they constitute a basis for the space of polynomials of degree 3 or less.Anne5632 said:so does leaving it in echelon form help?
as I got a 4x4 identity matrix so all the corresponding vectors are linearly independent
how would i use that basis to find the coordinates of an equation like q(x) =x^2 − 1?Mark44 said:When you put the function coordinates into a matrix, you are essentially solving the equation ##c_1p_1(x) + c_2p_2(x) +c_3p_3(x) +c_4p_4(x) = 0## for the constants ##c_i##. If you end up with the identity matrix, that indicates that all the constants are 0, and there are no other solutions for these constants. This means that the four functions are linearly independent (note spelling), and since there are four of them, they constitute a basis for the space of polynomials of degree 3 or less.
Solve the equation ##c_1p_1(x) + c_2p_2(x) +c_3p_3(x) +c_4p_4(x) = -1 + x^2## for the constants ##c_i##. Keep in mind that you're solving for the constants, not x. You'll need to substitute in the individual basis functions. You can do this algebraically, or you can set up an augmented matrix and row reduce as I think you did before.Anne5632 said:how would i use that basis to find the coordinates of an equation like q(x) =x^2 − 1?
would I put that equation in matrices form and then sub into those constants?
I'm guessing that how you have the coordinates is probably OK, but I would need to see the exact wording of the problem to know for sure.Anne5632 said:can i leave my answer like that or do i need to substitute back into the formula which = -1+x^2
Not sure what you're asking. I usually write vector coordinates in angle brackets like this: <1, 0, 2, 3>.Anne5632 said:(is there a format to show the coordinates?)