You're going at this the wrong way around. You want the vector c = <c1, c2, c3>T such that Ac = <-2, 0, -1>T, where A is the matrix you have above.hocuspocus102 said:Homework Statement
The set B = {-4-x^2, -8+4x-2x^2, -14+12x-4x^2} is a basis for P2. Find the coordinates of p(x) = (-2 +0x -x^2) relative to this basis.
Homework Equations
n/a
The Attempt at a Solution
so the set would be in a matrix like this:
|-4 0 -1|
|-8 4 -2|
|-14 12 -4| and multiplying this by
|-2|
| 0|
|-1| as the vector for p(x) gives
| 9|
|18|
|32| but that's not the right answer and I thought that's how you'd do it.
In linear algebra, coordinates relative to a basis refer to a system used to represent a vector or point in a multi-dimensional space. It involves expressing the vector or point in terms of a set of basis vectors that span the space.
Coordinates relative to a basis are essential in linear algebra because they allow us to represent and manipulate vectors and points in a multi-dimensional space in a more efficient and organized manner. It also helps in solving systems of linear equations and performing transformations on vectors and points.
A basis refers to a set of linearly independent vectors that span a vector space, while a coordinate system is a way of assigning numbers or coordinates to vectors in that space. In other words, a basis serves as the building blocks for a coordinate system.
To find coordinates relative to a basis, you need to express the vector or point as a linear combination of the basis vectors. This involves finding the coefficients or weights for each basis vector that, when multiplied with the basis vector, will result in the original vector or point.
Yes, coordinates relative to a basis can be negative. The coordinates represent the distance or magnitude of the vector in each direction, so they can be positive, negative, or zero depending on the direction and length of the vector.