Question

Let P₃ denote the space of all polynomials of degree not greater than 3 . Find a basis and dimension of this vector space. Prove that T=d² /dx² +1 is a linear operator on this space. b) Find the matrix of this linear operator in the basis you found in the part a).

Answer 1

The vector space P₃ has a basis consisting of the set {1, x, x², x³}, making its dimension 4. T=d²/dx² + 1 is a linear operator on P₃, which can be proven by demonstrating linearity. The matrix of T can be found by applying T to the basis vectors and using the resulting coefficients to form the matrix columns.

Step-by-step explanation:

Finding a Basis and Dimension of P₃

The space P₃ of all polynomials of degree not greater than 3 can be represented by a basis that includes the set of polynomials {1, x, x², x³}. This set forms a basis because it is linearly independent and spans the vector space. Consequently, the dimension of P₃ is 4, as there are four polynomials in the basis.

Proving T=d²/dx² + 1 is a Linear Operator

To prove that T is a linear operator on P₃, consider two arbitrary polynomials f(x) and g(x) in P₃ and scalars α and β. We must show that T(αf + βg) = αT(f) + βT(g), which maintains the properties of linearity (additivity and homogeneity).

Finding the Matrix of the Linear Operator T

The matrix of T in the basis {1, x, x², x³} can be found by applying T to each basis vector and expressing the result as a linear combination of the basis vectors. The resulting coefficients form the columns of the matrix.