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

• Anne5632
In summary, the given polynomials p1(x), p2(x), p3(x), and p4(x) form a basis for the vector space R3[x] since they are linearly independent and there are four of them, which is the same as the dimension of R3[x]. This can be determined by putting the coordinates of the functions into a matrix and solving for the constants, which results in an identity matrix indicating linear independence. This basis can then be used to find the coordinates of other equations, such as q(x) = x^2 - 1, by solving for the constants using the basis functions. The coordinates can be written in vector form with angle brackets.
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.

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 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.

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

Anne5632 said:
Do i need to determine linear independance?
For a vector space or a function space of dimension n, you need n linearly independent vectors/functions.

Anne5632 said:
if a vector space has a dim 4, I'm guessing you need 4 for a basis?
Yes.

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

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
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
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.
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?

Last edited by a moderator:
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?
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:
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##.

Anne5632 said:
(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>.

## 1. What is a basis of R3[x]?

A basis of R3[x] is a set of polynomials that can be used to represent any polynomial in R3[x] as a linear combination of those polynomials. In other words, the basis forms a "building block" for all polynomials in R3[x].

## 2. How do I check if a set of polynomials forms a basis of R3[x]?

To check if a set of polynomials forms a basis of R3[x], you need to make sure that the set is linearly independent and spans R3[x]. This means that none of the polynomials in the set can be written as a linear combination of the others, and that the set contains enough polynomials to represent any polynomial in R3[x].

## 3. Why is it important to check if polynomials form a basis of R3[x]?

Checking if polynomials form a basis of R3[x] is important because it ensures that the set of polynomials can be used to represent any polynomial in R3[x]. This is crucial in many applications, such as solving systems of linear equations and finding solutions to differential equations.

## 4. Can any set of polynomials form a basis of R3[x]?

No, not every set of polynomials can form a basis of R3[x]. The set must be linearly independent and span R3[x] in order to be a basis. This means that the set must have the same number of polynomials as the dimension of R3[x], which is 3.

## 5. How can I determine the dimension of R3[x]?

The dimension of R3[x] is 3, as it is a vector space of polynomials with 3 variables. This means that any basis of R3[x] must contain 3 linearly independent polynomials in order to span the entire space.

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