Check that the polynomials form a basis of R3[x]

[Anne5632]

Homework Statement

Check that the polynomials:

Relevant Equations

r

I put it in echelon form but don't know where to go from there.

 

[Mark44]

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 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.

 

[Anne5632]

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.

Do i need to determine linear independence?
and if a vector space has a dim 4, I'm guessing you need 4 for a basis?

 

[Mark44]

Do i need to determine linear independence?

For a vector space or a function space of dimension n, you need n linearly independent vectors/functions.

if a vector space has a dim 4, I'm guessing you need 4 for a basis?

Yes.

 

[Anne5632]

so does leaving it in echelon form help?
as I got a 4x4 identity matrix so all the corresponding vectors are linearly independent

 

[Mark44]

so does leaving it in echelon form help?
as I got a 4x4 identity matrix so all the corresponding vectors are linearly independent

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]

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.

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?

 

[Mark44]

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?

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.

 

[Mark44]

can i leave my answer like that or do i need to substitute back into the formula which = -1+x^2

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.
You can check your work, by seeing whether $c_1p_1(x) + c_2p_2(x) +c_3p_3(x) +c_4p_4(x)$ turns out to be $-1 + x^2$.

(is there a format to show the coordinates?)

Not sure what you're asking. I usually write vector coordinates in angle brackets like this: <1, 0, 2, 3>.